A projectile is launched with an initial speed of 48.0 m/s at an angle of 34.0° above the horizontal. The projectile lands on a hillside 3.65 s later. Neglect air friction. (Assume that the +x-axis is to the right and the +y-axis is up along the page.) Answer parts a-b.

(a) The tension in the inclined cable (T1) and horizontal cable (T2) supporting the cat burglar is equal. The tension in the vertical cable (T3) is 524 N.

(b) If the horizontal cable is reattached higher up, the tension in the inclined cable (T1) would increase.

(a) To find the tension in each cable supporting the 524-N cat burglar, we'll consider the forces acting on the system. Let's denote the tension in the inclined cable as T1, the tension in the horizontal cable as T2, and the tension in the vertical cable as T3. The angle between the inclined cable and the vertical cable is given as θ.

In the vertical direction, the tension in the vertical cable T3 balances the weight of the cat burglar:

T3 - 524 N = 0

T3 = 524 N

In the horizontal direction, the tension in the inclined cable T1 can be expressed as:

T1 * cos(θ) = T2

Now, we need to determine the value of θ to calculate T1 and T2. Let's assume that θ is the given angle of θ = 0.

Substituting the angle and rearranging the equation, we have:

T1 = T2 / cos(θ)

T1 = T2 / cos(0)

T1 = T2 / 1

T1 = T2

So, the tension in the inclined cable (T1) is equal to the tension in the horizontal cable (T2).

Therefore, the tension in each cable is as follows:

T1 (inclined cable) = T2 (horizontal cable)

T1 = T2

T3 (vertical cable) = 524 N

(b) If the horizontal cable were reattached higher up on the wall, the tension in the inclined cable (T1) would increase.

The correct answer is option A.  

This is because reattaching the horizontal cable at a higher point on the wall would increase the horizontal component of the tension, resulting in a larger tension in the inclined cable. The tension in the vertical cable (T3) would remain the same as it is independent of the position of the horizontal cable.

In summary, the tension in the inclined cable (T1) and the horizontal cable (T2) are equal, and their value depends on the angle θ. The tension in the vertical cable (T3) is 524 N. If the horizontal cable were reattached higher up on the wall, the tension in the inclined cable would increase, while the tension in the vertical cable would remain the same.

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The acceleration of the two blocks is[tex]2.14 m/s^{2[/tex]} and the distance does block M2 move in 2.00 s is 4.27 m.

Now we need to find the acceleration of the two blocks and the distance does block M2 move in 2.00 s.

We know that: mass of M1, m1 = 5.00 kg mass of M2, m2 = 19.0 kgθ = 25.0°Taking upward direction as positive for block M1 and downwards as positive for block M2.

Therefore, we can write the following equation of motion for the two blocks:

For M2: m2g - T = m2a ...(1)

For M1: T - m1g = m1a ...(2)

We can see from the figure that M2 is on an inclined plane making an angle θ with the horizontal.

We can resolve the weight of M2 into two components:

Perpendicular to the plane = m2gcosθParallel to the plane = m2gsinθ

The component parallel to the plane will tend to make the block move downwards.

Therefore, the effective weight will be:

mg = m2gsinθ ...(3)

From equation (1) we can write:

T = m2g - m2a ...(4)

Substituting equation (4) in equation (2), we get:

m2g - m2a - m1g = m1a ...(5)

On solving equation (5), we get the acceleration as:

a = g(m2sinθ - m1) / (m1 + m2)

On substituting the given values, we get:

[tex]a = 2.14 m/s^{2}[/tex]

The distance moved by M2 in 2 seconds can be found out using the formula:[tex]s = ut + \frac{1}{2} at^{2}[/tex]

Here, initial velocity, u = 0m/s Time, t = 2s Acceleration, [tex]a = 2.14 m/s^{2}[/tex]

On substituting these values, we get the distance travelled by M2 as: s = 4.27 m

Therefore, the acceleration of the two blocks is [tex]2.14 m/s^{2}[/tex]. And the distance does block M2 move in 2.00 s is 4.27 m.

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The cardinality of R₁ \ (R₁ intersection A) is 4. The given polynomial (2x² + y² + 22 - 8x + 2y - 2xy + 2xz-16z + 35) cannot be solved for x, y, and z due to insufficient equations.

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The water from a fire hose is aimed at an angle of 34.0° above the horizontal as it is directed towards a building located 54.0 meters away. Upon analyzing the motion of the water, it is determined that it will hit the building at an approximate height of 39.586 meters.

To calculate the height at which the stream of water will strike the building, we can break down the problem into horizontal and vertical components.

Given:

- Distance from the fireman to the building (horizontal distance): d = 54.0 m

- Angle of elevation above the horizontal: θ = 34.0°

- Initial speed of the water stream: [tex]v_i[/tex] = 40.0 m/s

- Acceleration due to gravity: g = 9.8 m/s²

1. Horizontal Component:

Using the horizontal distance and the angle of elevation, we can calculate the time it takes for the water stream to reach the building.

t = d / ([tex]v_i[/tex] * cosθ)

Substituting the values:

t = 54.0 / (40.0 * cos34.0°)

t ≈ 1.331 seconds

2. Vertical Component:

Next, we can determine the vertical component of the initial velocity.

[tex]v_y[/tex] = [tex]v_i[/tex] * sinθ

[tex]v_y[/tex] = 40.0 * sin34.0°

[tex]v_y[/tex]≈ 22.148 m/s

3. Height Calculation:

To find the height at which the water stream strikes the building, we can use the time and vertical velocity components.

h = [tex]v_y[/tex] * t + (1/2) * g * t²

Substituting the values:

h = 22.148 * 1.331 + (1/2) * 9.8 * (1.331)²

h ≈ 30.882 + 8.704

h ≈ 39.586 meters

Therefore, the stream of water will strike the building at a height of approximately 39.586 meters.

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(a) The average velocity of the tennis player at 0 to 1 s is  4 m/s.

(b) The average velocity of the tennis player at 0 to 4 s is -0.5 m/s.

(c) The average velocity of the tennis player at 1 to 5 s is  1 m/s.

(d) The average velocity of the tennis player at 0 to 5 s is 0.8 m/s.

The average velocity of the tennis player at the given time, is calculated by applying the formula for average velocity as follows;

average velocity = total displacement / total time

(a) The average velocity at 0 to 1 s;

average vel. = (4 m - 0 m ) / (1 s ) = 4 m/s

(b) The average velocity at 0 to 4 s;

average vel. = (-2 - 0 )m / 4 s = -0.5 m/s

(c) The average velocity at 1 to 5 s;

average vel. = (4 - 0 )m / (5 - 1) s =  1 m/s

(d) The average velocity at 0 to 5 s;

average vel. = (4 - 0 )m / (5 - 0) s =  0.8 m/s

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The velocity of the third piece is (81.25 m/s, -43.3 m/s).

To determine the velocity of the third piece, we can use the principle of conservation of momentum.

Given:

Mass of the first piece (m1) = 150 kg

Velocity of the first piece (v1) = 150 m/s (to the East)

Mass of the second piece (m2) = 100 kg

Velocity of the second piece (v2) = 200 m/s at a direction of south 60° West

Let's break down the velocities into their respective horizontal (x) and vertical (y) components.

For the first piece:

v1x = 150 m/s (since it's moving to the East)

v1y = 0 m/s (no vertical component)

For the second piece:

v2x = 200 m/s * cos(60°) = 200 m/s * 0.5 = 100 m/s (horizontal component)

v2y = -200 m/s * sin(60°) = -200 m/s * 0.866 = -173.2 m/s (vertical component, negative since it's moving downward)

Now, let's calculate the momentum of the first and second pieces:

The momentum of the first piece (p1) = m1 * v1

= 150 kg * 150 m/s

= 22,500 kg·m/s

The momentum of the second piece (p2) = m2 * v2

= 100 kg * (100 m/s, -173.2 m/s)

= (10,000 kg·m/s, -17,320 kg·m/s)

To find the total momentum after the explosion, we can add the momenta of the individual pieces:

Total momentum after the explosion = p1 + p2

= (22,500 kg·m/s, 0 kg·m/s) + (10,000 kg·m/s, -17,320 kg·m/s)

= (32,500 kg·m/s, -17,320 kg·m/s)

The total momentum after the explosion should also be equal to the momentum of the third piece:

The momentum of the third piece (p3) = m3 * v3

Given:

Mass of the third piece (m3) = 400 kg (calculated from the given mass of the bomb)

Let's assume the velocity of the third piece is (v3x, v3y).

Therefore, we have the equation:

(32,500 kg·m/s, -17,320 kg·m/s) = 400 kg * (v3x, v3y)

By equating the x and y components separately, we can solve for the velocity components of the third piece:

32,500 kg·m/s = 400 kg * v3x

-17,320 kg·m/s = 400 kg * v3y

Solving these equations, we find:

v3x = 81.25 m/s

v3y = -43.3 m/s

Therefore, the velocity of the third piece is approximately (81.25 m/s, -43.3 m/s).

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a. The water will land 0.30 meters from the wall.

b.  The water should flow at 0.042 m/s in the model.

(a)

Horizontal distance = velocity × time

h = (1/2) × g × t²

h = vertical displacement (2.30 m)

g = acceleration due to gravity (9.8 m/s²

t = time

t = √(2h/g)

t = √(2 × 2.30 / 9.8) = 0.478 s

Now, we can calculate the horizontal distance:

Horizontal distance = velocity × time

Horizontal distance = 0.628 m/s × 0.478 s = 0.30 m

The water will land less than 2 m from the wall, the space behind the waterfall will not be wide enough for a pedestrian walkway.

The answer is "No."

(b)

Actual speed of water = 0.628 m/s

Speed of water in the model = Actual speed / Scale factor

Speed of water in the model = 0.628 m/s / 15

= 0.042 m/s

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The electric field between the plates is 1.65 × 10⁵ N/C.

Given:Area of two parallel plates, A = 5.68 × 10⁻⁴ m² Charge on each plate, q = 8.38 × 10⁻¹¹ C

We know that the electric field due to the charged plates is given by:E = σ / ε₀where σ = charge per unit area and ε₀ = permittivity of free space.

σ = q / AA = 5.68 × 10⁻⁴ m²q = 8.38 × 10⁻¹¹ C

σ = q / A = 8.38 × 10⁻¹¹ / 5.68 × 10⁻⁴

σ = 1.47 × 10⁻⁷ C/m²ε₀ = 8.85 × 10⁻¹² F/m²

Now, substituting the values in the equation,

E = σ / ε₀E = (1.47 × 10⁻⁷) / (8.85 × 10⁻¹²)

E = 16.5 × 10⁴ N/C≈ 1.65 × 10⁵ N/C

Therefore, the electric field between the plates is 1.65 × 10⁵ N/C.

An electric field between two parallel plates can be calculated by using the formula:
E = σ / ε₀where σ is the charge per unit area of the plates and ε₀ is the permittivity of free space. In this particular question, the area of two parallel plates, A = 5.68 × 10⁻⁴ m², and charge on each plate, q = 8.38 × 10⁻¹¹ C was given. Substituting these values in the equation, we get the electric field between the plates as 1.65 × 10⁵ N/C.

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The height of the building is approximately 32.34 meters.

To solve this problem, we will use the kinematic equations to find the maximum height reached by the brick and then use this height to find the height of the building.

We can start by breaking the initial velocity of the brick into its horizontal and vertical components as follows:

v₀x = v₀cos(θ) = 17cos(25°) ≈ 15.84 m/s

v₀y = v₀sin(θ) = 17sin(25°) ≈ 7.23 m/s

where θ is the angle of the initial velocity to the horizontal.

Next, we can use the following kinematic equation to find the maximum height reached by the brick:

y = y₀ + v₀yt - 1/2gt²

where y₀ is the initial height (height of the building), t is the time of flight, and g is the acceleration due to gravity (9.81 m/s²).

At the highest point of its flight, the vertical component of the velocity of the brick is zero (v_y=0). We can use this fact to find the time taken to reach maximum height:

v_y = v₀y - gt

0 = v₀y - gt_max

t_max = v₀y / g ≈ 0.738 s

We can then substitute this value of t_max into the expression for y to obtain the maximum height:

y_max = y₀ + v₀y t_max - 1/2 g t_max²

where we set y = y_max and t = t_max.

Next, we can use the total flight time of the brick (3.1 s) to find the initial height of the building:

3.1 = t_max + t_down

where t_down is the time taken by the brick to fall from the maximum height to the ground. Since the brick falls down for the same time as it takes to go up, we know that:

t_down ≈ t_max ≈ 0.738 s

Substituting this value into the equation above, we find:

3.1 ≈ 2 × 0.738 s

Finally, we can use the value of y_max obtained earlier to calculate the height of the building:

y₀ = y_max - v₀y t_down + 1/2 g t_down²

y₀ = y_max - v₀y t_max + 1/2 g t_max²

y₀ ≈ 32.34 m

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Answer:

[tex]\displaystyle \frac{M + m}{m}\, \sqrt{2\, x\, u\,g}[/tex].

Explanation:

This question can be solved in the following steps:

Assume that the table is level. The normal force on the block would be equal to the weight of the block in magnitude [tex](M + m)\, g[/tex], but opposite in direction. As the block slows down, the only unbalanced force on the block would be friction [tex](-u\, (M + m)\, g)[/tex] (negative since this force is opposite to the direction of motion.)

The acceleration of the block would be:

[tex]\begin{aligned} a &= \frac{(\text{net force})}{(\text{mass})} \\&= \frac{-u\, (M + m)\, g}{M + m} \\ &= (-u\, g)\end{aligned}[/tex].

Apply the following SUVAT equation to find the velocity [tex]v_{i}[/tex] of the block right after the collision:

[tex]\displaystyle {v_{2}}^{2} - {v_{1}}^{2} = 2\, a\, x[/tex],

Where:

Rearrange and solve for [tex]v_{1}[/tex], the velocity right after collision:

[tex]\begin{aligned}v_{1} &= \sqrt{{v_{2}}^{2} - 2\, a\, x} \\ &= \sqrt{0^{2} - 2\, (-u\, g)\, x} \\ &= \sqrt{2\, x\, u\, g}\end{aligned}[/tex].

Apply the conservation of momentum to find the velocity of the bullet before the collision. Right after the collision, sum of momentum would be:

[tex](M + m)\, \sqrt{2\, x\, u\, g}[/tex].

Right before the collision, sum of momentum would be:

[tex]m\, v_{0}[/tex].

By the conservation of momentum:

[tex]m\, v_{0} = (M + m)\, \sqrt{2\, x\, u\, g}[/tex].

Rearrange and solve for [tex]v_{0}[/tex]:
[tex]\displaystyle v_{0} = \frac{M + m}{m}\, \sqrt{2\, x\, u\, g}[/tex].

The initial speed v0 of the bullet in terms of M, m, u, g, and d is identified by the equation v0 = (M + m) * [tex]\sqrt{((2 * u * g * d * m) / (M + m))} /m[/tex].

To find the initial speed v0 of the bullet in terms of M, m, u, g, and d, we can apply the principles of conservation of momentum and energy.

First, let's consider the conservation of momentum. Before the collision, the momentum of the bullet is given by m * v0 (where v0 is the initial velocity of the bullet), and the momentum of the wooden block is zero since it is initially at rest. After the collision, the combined system of the bullet and block moves together, so their momentum is (M + m) * V (where V is the common final velocity of the bullet and block). Since momentum is conserved, we have:

m * v0 = (M + m) * V

Next, let's consider energy conservation. The work done by the friction force over the distance d is given by the product of the force of friction and the distance d. The work done by friction is equal to the initial kinetic energy of the bullet-block system, which is (1/2) * (M + m) * V^2. Thus, we have:

(1/2) * (M + m) * V² = u * (M + m) * g * d

Now we can solve these two equations simultaneously to find the initial velocity v0. Rearranging the first equation, we have:

v0 = (M + m) * V / m

Substituting this expression for v0 into the second equation, we get:

(1/2) * (M + m) * [(M + m) * V / m]² = u * (M + m) * g * d

Simplifying and solving for V, we obtain:

V = [tex]\sqrt{((2 * u * g * d * m) / (M + m))}[/tex]

Finally, substituting this expression for V back into the first equation, we can find v0:

v0 = (M + m) * [tex]\sqrt{(2 * u * g * d * m) / (M + m)}[/tex] / m

Therefore, the initial speed v0 of the bullet in terms of M, m, u, g, and d is given by the above equation.

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Answer: the material involved in the change is structurally the same before and after the change. Types of some physical changes are texture, shape, temperature, and a change in the state of matter. A change in the texture of a substance is a change in the way it feels

Explanation:

The six digit grid coordinates for the windtee  should be 100049.

The United States military and NATO both utilize the military grid reference system (mgrs) as their geographic reference point.

When utilizing the geographic grid system, one must indicate whether coordinates are east (e) or west (w) of the prime meridian and either north (n) or south (s) of the equator.

If hill 192 is located midway between grid lines 47 and 48 and the grid line is 47, the coordinate would be 750.

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Answer:

D.  1.4 m/s

Explanation:

forward direction is +

back direction is -

Momentum = P = mass x velocity = mv

let v =  velocity of ball after collision

Law of Conservation of Momentum:  total momentum before the collision must equal the total momentum after the collision

(0.20 kg)(5 m/s) + (1 kg)(0 m/s) = (0.2 kg)(-2 m/s) + (1 kg)v

1 kg·m/s + 0 = -0.4 kg·m/s + (1 kg)v    

1 kg·m/s + 0.4 kg·m/s =  (1 kg)v    rearrange the equation and solve for v

(1 kg)v  = 1.4 kg·m/s

v = (1.4 kg·m/s) / (1 kg) = 1.4 m/s

If two waves with equal amplitudes and wavelengths travel through a medium in such a way that a particular particle of the medium is at the crest of one wave and at the trough of the other wave at the same time then D) The particle will remain stationary due to interference.

When two waves with equal amplitudes and wavelengths pass through a medium, they undergo interference. Interference occurs when the crests and troughs of the waves overlap. In this case, if a particular particle of the medium is at the crest of one wave and at the trough of the other wave at the same time, it experiences what is called destructive interference.

Destructive interference happens when the peaks (crests) of one wave align with the troughs of the other wave. In this situation, the positive displacements caused by the crest are canceled out by the negative displacements caused by the trough. As a result, the net displacement of the particle is zero, and it remains stationary.

This phenomenon occurs due to the principle of superposition, which states that the total displacement of a particle at any point in a medium is the vector sum of the individual displacements caused by each wave. Therefore, in this scenario, the particle will remain stationary due to the destructive interference between the two waves. Therefore, Option D is correct.

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Answer: D

Explanation:

Answer:

3.79 m/s^2

Explanation:

We know the small part loses 0.932 MJ of gravitational potential energy during its fall.

Potential energy = mass x gravitational field strength x height

Re-arranging to solve for gravitational field strength:

g = Potential energy/(mass x height)

Plugging in the given values:

g = 0.932 MJ / (1.8kg x 140km)

= 0.932 x 10^6 J / (1.8 x 1000kg x 140 x 1000m)

= 3.79 m/s^2

Therefore, the gravitational field strength of Mars is calculated to be 3.79 m/s^2.

Guiseppe's buys supplies to make pizzas at a cost of $4.02. Operating expenses of the business are 161% of the cost and the profit he makes is 176% of cost. The regular selling price of each pizza is $7.33.

Let's denote the cost of supplies as C.

Operating expenses:

The operating expenses of the business are 161% of the cost. Therefore, the operating expenses can be calculated as:

Operating Expenses = 1.61 * C

Profit:

The profit made by Guiseppe is 176% of the cost. Therefore, the profit can be calculated as:

Profit = 1.76 * C

Total cost:

The total cost includes the cost of supplies and the operating expenses:

Total Cost = C + Operating Expenses = C + 1.61 * C = 2.61 * C

Regular selling price:

The regular selling price is the sum of the total cost and the profit:

Regular Selling Price = Total Cost + Profit = 2.61 * C + 1.76 * C = 4.37 * C

Given that the cost of supplies is $4.02, we can substitute this value into the equation:

Regular Selling Price = 4.37 * 4.02 = $17.5674

Rounding the final answer to the nearest cent, the regular selling price of each pizza is approximately $7.33.

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Given that the initial velocity at which the ball is thrown vertically upward is 15m/s. Let us also assume that the value of acceleration due to gravity (g) = 9.8m/s² and in this case, the value will be -9.8m/s² as the ball is moving against gravity.

a) To calculate how high the ball rises, we can use the kinematic equation:

v² = u² + 2gs......(i)

where v ⇒ final velocity

u ⇒ initial velocity

g ⇒ acceleration and,

s ⇒ displacement (the height)

The final velocity will be 0 when the ball reaches its maximum height.

Substituting the values in equation (i), we get

0² = 15² + (2*-9.8*s)

0 = 225 - 19.6s

Thus, s = 225/19.6 = 11.48 m.

Therefore, the ball rises approximately 11.48 meters.

b) To find the time taken to reach the highest point, we can use the kinematic equation,

v = u + gt......(ii)

where t = time

Substituting the values in equation (ii)

0 = 15 - 9.8*t

t = -15/ -9.8 = 1.53 seconds

Thus, the time taken to reach the highest point = 1.53 seconds.

c) To find the time taken for the ball to hit the ground after it reaches its highest point, we can use the equation,

s = ut +1/2gt².....(iii)

As the ball is moving downwards, the initial velocity, u will be 0m/s.

Thus, substituting the values in equation (iii), we get

11.48 = 0*t + 1/2*9.8*t²

11.48 = 4.9t²

t² = 2.34

Therefore t = 1.53 seconds

Thus, the time taken for the ball to hit the ground is 1.53 seconds.

d)  To find the velocity at which the ball returns to the level from which it started, we can use the equation

v = u+ gt.....(iv)

v = 0 + 9.8*1.53

Thus, v = 14.99 ≅ 15 m/s

Therefore, the velocity when it returns to the level from which it started is 15m/s.

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The magnitude of the displacement is 1,097.7 mi, and the angle is 89.9°

To find the displacement from Dallas to Chicago, we can break down the vectors representing the distances and directions into their x and y components. Since the Earth is modeled as flat, we can use basic trigonometry to calculate the components.

Let's start by considering the vector from Dallas to Atlanta. The magnitude of this vector is given as 730 miles, and the direction is 5.00° north of east. To calculate the x and y components, we can use the following equations:

Substituting the values:

x = 730 * cos(5.00°)

y = 730 * sin(5.00°)

Similarly, for the vector from Atlanta to Chicago, with a magnitude of 560 miles and a direction 21.0° west of north:

x = magnitude_AC * sin(angle_AC)

y = magnitude_AC * cos(angle_AC)

Substituting the values:

x = 560 * sin(21.0°)

y = 560 * cos(21.0°)

To find the displacement from Dallas to Chicago, we can sum the x and y components:

Now, we can calculate the magnitude and direction of the displacement using these x and y components:

magnitude_displacement = √(x_displacement² + y_displacement²)

angle_displacement = atan(y_displacement / x_displacement)

Finally, we can substitute the calculated values and solve for the magnitude and direction:

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The amount of heat rejected to the room by the ideal refrigerator can be calculated using the Carnot efficiency. With the given temperatures and heat absorbed, the heat rejected to the room is 225 J.

To calculate the amount of heat rejected to the room by the ideal refrigerator, we can use the Carnot efficiency, which is given by the formula:

Efficiency = 1 - ([tex]T_c_o_l_d[/tex] / [tex]T_h_o_t[/tex])

where[tex]T_c_o_l_d[/tex]is the temperature of the cold reservoir (freezer compartment) and [tex]T_h_o_t[/tex] is the temperature of the hot reservoir (room temperature).

Given:

[tex]T_c_o_l_d[/tex] = -9 °C (converted to Kelvin: 264 K)

[tex]T_h_o_t[/tex]= 25 °C (converted to Kelvin: 298 K)

Heat absorbed from the freezer compartment ([tex]Q_c_o_l_d[/tex] = 120 J

First, we calculate the Carnot efficiency:

Efficiency = 1 - (264 K / 298 K)

Efficiency ≈ 0.1134

The Carnot efficiency represents the ratio of heat transferred from the cold reservoir to the work done by the refrigerator. Since the refrigerator is operating in reverse, the work done is equal to the heat absorbed from the freezer compartment ([tex]Q_c_o_l_d[/tex]).

[tex]Q_c_o_l_d[/tex] = 120 J

Now, we can calculate the heat rejected to the room ([tex]Q_h_o_t[/tex]) using the equation:

[tex]Q_h_o_t[/tex] = Efficiency * [tex]Q_c_o_l_d[/tex]

[tex]Q_h_o_t[/tex] ≈ 0.1134 * 120 J

[tex]Q_h_o_t[/tex] ≈ 13.61 J

Therefore, the amount of heat rejected to the room by the ideal refrigerator is approximately 13.61 J.

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a) The speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.

b) We cannot calculate the work done by the friction force.

c) The net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.

(a) The work done on the 6.00 kg block by gravity can be calculated using the formula:

Work_gravity = force_gravity * displacement * cos(theta),

where force_gravity is the weight of the block, displacement is the downward displacement of the block, and theta is the angle between the force and displacement vectors (which is 0 degrees in this case).

The weight of the block is given by:

force_gravity = mass * acceleration_due_to_gravity = 6.00 kg * 9.8 m/s^2 = 58.8 N.

Plugging in the values, we get:

Work_gravity = 58.8 N * 0.800 m * cos(0) = 47.04 J.

The work done on the 6.00 kg block by the tension in the rope is given by:

Work_tension = tension * displacement * cos(theta).

Plugging in the values, we get:

Work_tension = 37.0 N * 0.800 m * cos(180) = -29.6 J.

The negative sign indicates that the tension is in the opposite direction of the displacement.

Using the work-energy theorem, we can find the speed of the 6.00 kg block after descending 0.800 m:

Work_net = change_in_kinetic_energy.

Since the block starts from rest, its initial kinetic energy is zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * mass * velocity^2.

Solving for velocity, we get:

velocity = sqrt(2 * Work_net / mass).

The net work done on the block is the sum of the work done by gravity and the tension:

Work_net = Work_gravity + Work_tension = 47.04 J - 29.6 J = 17.44 J.

Plugging in the values, we get:

velocity = sqrt(2 * 17.44 J / 6.00 kg) = 2.07 m/s.

Therefore, the speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.

(b) The total work done on the 8.00 kg block during the 0.800 m displacement can be calculated using the work-energy theorem:

Work_net = change_in_kinetic_energy.

Since the 8.00 kg block is not moving vertically, its initial and final kinetic energies are zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 0.

The work done on the 8.00 kg block by the tension in the rope is given by:

Work_tension = tension * displacement * cos(theta).

Plugging in the values, we get:

Work_tension = 37.0 N * 0.800 m * cos(0) = 29.6 J.

The work done on the 8.00 kg block by the friction force can be calculated using the formula:

Work_friction = force_friction * displacement * cos(theta),

where force_friction is the frictional force on the block. However, the problem statement does not provide the value of the friction force. Therefore, we cannot calculate the work done by the friction force.

(c) The net work done on the system of two blocks during the 0.800 m displacement of the 6.00 kg block can be found using the work-energy theorem:

Work_net = change_in_kinetic_energy.

Since the system starts from rest, the initial kinetic energy of the system is zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * (6.00 kg + 8.00 kg) * velocity^2.

Simplifying, we get:

Work_net = 1/2 * 14.00 kg * velocity^2.

Using the value of velocity calculated in part (a), we get:

Work_net = 1/2 * 14.00 kg * (2.07 m/s)^2 = 29.13 J.

The work done on the system of two blocks by gravity is the sum of the work done on the individual blocks by gravity:

Work_gravity_system = Work_gravity_6kg + Work_gravity_8kg = 47.04 J + 0 J = 47.04 J.

The work done on the system of two blocks by the tension in the rope is the sum of the work done on the individual blocks by the tension:

Work_tension_system = Work_tension_6kg + Work_tension_8kg = -29.6 J + 29.6 J = 0 J.

Therefore, the net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.

Note: The calculations for part (b) and (c) were based on the given information, but the value of the friction force was not provided in the problem statement.

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Formulae to be used here are :

[tex]\quad\displaystyle \circ \: \rm kinetic \: \: energy = \frac{ 1 }{2} m {v}^{2} [/tex]

[tex]\quad\displaystyle \circ \rm \: work \: done = change \: \: in \: \: KE[/tex]

The kinetic energy of cart while moving with velocity 6.0m/s can be calculated as :

[tex]\qquad\displaystyle \tt \dashrightarrow \: \frac{1}{2} (5)(6) {}^{2} [/tex]

[tex]\qquad\displaystyle \tt \dashrightarrow \: \frac{1}{2} (5)(36)[/tex]

Similarly, kinetic energy at velocity 10.0m/s would be :

[tex]\qquad\displaystyle \tt \dashrightarrow \: \frac{1}{2} (5)(10 {}^{} ) {}^{2} [/tex]

[tex]\qquad\displaystyle \tt \dashrightarrow \: \frac{1}{2} (5)(100)[/tex]

Next up ;

[tex]\qquad\displaystyle \tt \dashrightarrow \: work \: done = KE_{final} - KE_{initial} [/tex]

[tex]\qquad\displaystyle \tt \dashrightarrow \: WD = \frac{1}{2} (5)(100) - \frac{1}{2} (5)(36)[/tex]

[tex]\qquad\displaystyle \tt \dashrightarrow \: WD = \frac{1}{2} (5)(100 - 36)[/tex]

[tex]\qquad\displaystyle \tt \dashrightarrow \: WD = \frac{1}{2} (5)(64)[/tex]

[tex]\qquad\displaystyle \tt \dashrightarrow \: WD = 5 \times 32[/tex]

[tex]\qquad\displaystyle \tt \dashrightarrow \: WD = 160 \: \: joules[/tex]

That's our required answer, n matches with choice A.) 160 J

(a) The tension in the string is determined as 19.6 N.

(b) The acceleration of each object is 5.3 m/s².

(c) The distance each object will move in the first second if it started from rest is 2.65 m.

(a) The tension in the string is the resultant weight of the masses and magnitude is calculated as follows;

T = ( 5.7 kg - 3.7 kg ) x 9.8 m/s²

T = 19.6 N

(b) The acceleration of each object is calculated as follows;

a = T / m

where;

a = 19.6 N / 3.7 kg

a = 5.3 m/s²

(c) The distance each object will move in the first second if it started from rest is calculated as;

s = ut + ¹/₂at²

where;

s = 0 + ¹/₂(5.3)(1²)

s = 2.65 m

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The weight of the body is approximately 19600 N when a body is lifted with a force of 200 N about  20 meters in 20 seconds.

We know that distance = initial velocity*time  + (1/2) * acceleration * [tex]time^2[/tex]

Given that, the initial velocity equal to zero

distance= (1/2) * acceleration * [tex]time^2[/tex]

Also given-

distance = 20 meters

time = 20 seconds

Rearranging the formula-

20 = (1/2) * acceleration * [tex](20^2)[/tex]

20 = (1/2)   * acceleration * 400

40 = acceleration * 400

acceleration = 40/400 = 0.1 metre/[tex]sec^{2}[/tex]

Force= mass*acceleration

mass= force/acceleration

=200/0.1= 2000 Kg

But this acceleration is due to applied force but weight only involves gravitational force only.  

Since weight is defined as the force acting on an object due to gravity,

weight = mass * acceleration due to gravity

= 2000 * 9.8 (acceleration due to gravity)

= 19600 N

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Answer:

The weight of the body is 3,924 N.

Explanation:

To solve this problem, we can use the formula for distance as a function of acceleration with initial velocity equal to zero:

distance = (1/2) x acceleration x time^2

We know that the distance the body is lifted is 20 meters, the time taken is 20 seconds, and the force applied is 200 N. We can use this information to calculate the weight of the body.

First, we need to calculate the acceleration:

distance = (1/2) x acceleration x time^2

20 = (1/2) x acceleration x (20)^2

acceleration = 0.5 m/s^2

Now that we know the acceleration, we can use the formula for weight:

force = mass x acceleration

We can rearrange this formula to solve for mass:

mass = force / acceleration

mass = 200 N / 0.5 m/s^2

mass = 400 kg

Finally, we can calculate the weight of the body using the formula:

weight = mass x gravity

Assuming a standard acceleration due to gravity of 9.81 m/s^2, we can calculate the weight:

weight = 400 kg x 9.81 m/s^2

weight = 3,924 N

Therefore, the weight of the body is 3,924 N.

a.  the tension in the rope is  91.5 N.

b.   the moment of inertia of the wheel is  0.1008 kg⋅m².

c.  the angular speed of the wheel 2.30 s after it begins rotating is  38.34 rad/s.

(a)

The tension in the rope can be found by considering the forces acting on the object.

ma = mg*sin(θ) - T

(14.0 kg)(2.00 m/s²)

= (14.0 kg)(9.8 m/s²)*sin(37°) - T

T = (14.0 kg)(9.8 m/s²)*sin(37°) - (14.0 kg)(2.00 m/s²)

T =  91.5 N

(b)

The moment of inertia of a wheel:

I = (1/2)MR²

I = (1/2)(14.0 kg)(0.12 m)²

I = 0.1008 kg⋅m²

(c)

The angular acceleration of the wheel:

α = a/R

α = angular acceleration,

a = linear acceleration of the object,

R =  radius of the wheel.

α = (2.00 m/s²)/(0.12 m)

α = 16.67 rad/s²

The angular speed (ω) of the wheel after time t is :

ω = ω₀ + αt

ω = 0 + (16.67 rad/s²)(2.30 s)

ω = 38.34 rad/s

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For a bomber flying horizontally over level terrain:

(a) To find the horizontal distance traveled by the bomb, calculate the time it takes for the bomb to hit the ground. Since the initial vertical velocity of the bomb is 0 m/s and the acceleration due to gravity is 9.8 m/s², use the equation:

Δy = v₀yt + (1/2)at²

where Δy = change in vertical position (altitude), v₀y = initial vertical velocity, t = time, and a = acceleration due to gravity.

Plugging in the values:

Δy = -2.50 km = -2500 m (negative because it's downward)

v₀y = 0 m/s

a = -9.8 m/s²

Rearrange the equation to solve for t:

t = √(2Δy/a)

t = √(2(-2500 m)/(-9.8 m/s²))

t ≈ 16.07 s

Since the horizontal velocity of the bomb is 290 m/s, calculate the horizontal distance traveled:

Δx = v₀xt = (290 m/s)(16.07 s)

Δx ≈ 4657.3 m = 4.6573 km

Therefore, the bomb travels approximately 4.6573 km horizontally between its release and its impact on the ground.

(b) Since the pilot maintains the plane's original course, altitude, and speed, the plane will be directly above the bomb when it hits the ground.

Answer: directly above the bomb

(c) Since the bomb hits the target seen in the sight at the moment of release, the bombsight must be set at an angle equal to the angle of depression from the horizontal.

Using trigonometry, find this angle:

tan θ = Δy / Δx

θ = tan⁻¹(Δy / Δx)

θ = tan⁻¹(-2500 m / 4657.3 m)

θ ≈ -29.11°

Since the angle is measured forward from the vertical, the bombsight is set at approximately 29.11° forward from the vertical.

Therefore, the bombsight is set at an angle of approximately 29.11° forward from the vertical.

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When white light reflects off of a green surface, only the green wavelengths of light are reflected (option d).

1. White light is a combination of all visible wavelengths of light, including red, orange, yellow, green, blue, indigo, and violet.

2. When white light hits a green surface, the surface absorbs some wavelengths of light and reflects others.

3. The color we perceive as "green" is the result of the green wavelengths of light being reflected by the surface.

4. In this case, the green surface absorbs all the wavelengths of light except for the green wavelengths, which are reflected back.

5. As a result, our eyes detect the reflected green light and interpret it as the color green.

6. This phenomenon occurs because the green surface selectively absorbs and reflects different wavelengths of light based on its molecular structure and the interactions between light and matter.

7. The absorption and reflection of specific wavelengths of light give objects their perceived color.

8. Therefore, when white light reflects off of a green surface, only the green wavelengths of light are reflected, while the other wavelengths are absorbed by the surface.

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The function of the power supply is to provide electrical energy to the device or system that needs it. The power supply converts the incoming voltage from the power source into a form that is usable by the device, such as DC voltage.

The readout is a device or component that displays data or information to the user. The readout could be a simple LED display or a complex graphical display.

A peripheral is a device or component that connects to a computer or other electronic device to provide additional functionality. Examples of peripherals include printers, scanners, and external hard drives.

A microcomputer is a type of computer that is designed to fit on a single microchip. Microcomputers are found in a wide range of devices, including smart phones, tablets, and embedded systems.

A transducer is a device that converts one form of energy to another. In electronics, transducers are commonly used to convert electrical energy into mechanical energy, or vice versa.

The processor is the central component of a computer or electronic device. The processor is responsible for executing instructions and controlling the other components of the system. The performance and capabilities of a device are largely determined by the speed and power of the processor.

Answer:

D - Reduced impact time will increase the impact force

Explanation:

A. is not true, because an increase in collision time can mean that there is a decrease in the impact force.

B. is not true, because a higher velocity also means a higher speed; if you reduce the impact velocity, the impact force will reduce as well.

C. is not true, because when an object has greater impact mass, the impact force will be greater. The impact force will not increase if the object has reduced mass.