Calculate the work done by an applied force of 76.0 N on a crate for the following. (Include the sign of the value in your answer.)
a. The force is exerted horizontally while pushing the crate 5.20 m.
b. The force is exerted at an angle of 41.0
Answer:
a) 400.4Joules
b) 262.69Joules
Explanation:
Work is said to be done if the force applied to an object cause the object to move through a distance
Workdone = Force × Distance
Given
Force = 76N
Distance= 5.2m
Work done = 77 × 5.2
Work done = 400.4Joules
b) If the force is exerted at an angle of 41°
Work done = Fdsin theta
Work done = 77(5.2)sin41
Work done = 400.4sin41
Work done = 262.69Joules
types of curved mirrors
Answer:
When the reflecting surface is instead curved, we call it a curved mirror. There are two types of curved mirrors; concave and convex mirror. Curved mirrors whose reflecting surfaces curve inwards are called concave mirrors while those whose reflecting surfaces bulge outwards are called convex mirrors.
Explanation:
According to the FITT Principle you should exercise how many days ?
(a) When the terminals of a 100-V battery are connected to two parallel plates 1 cm apart, what is the electrical field in the region between the plates
Answer:
The electrical field in the region between the plates is 10,000 V/m.
Explanation:
Given;
potential difference between the two parallel plates, V = 100 V
distance between the two parallel plates, d = 1 cm = 0.01 m
The electrical field in the region between the plates is given as;
E = V / d
where;
E is the electrical field in the region between the plates
E = (100) / (0.01)
E = 10,000 V/m
Therefore, the electrical field in the region between the plates is 10,000 V/m.
Which
type of energy transformation is taking place when natural gas is used to heat water?
O chemical energy into thermal energy
thermal energy into mechanical energy
mechanical energy into electromagnetic energy
electromagnetic energy into chemical energy
Answer:
O chemical energy into thermal energy
Hope this helped!!
Answer:
chemical energy into thermal energy
Explanation:
an object accepts a pressure of 2:00 p.m. on the ground object is kept above it then the dead pressure will become Dash
Correct question is;
An object exerts a pressure of 2 pa on the ground if another object of the same weight is kept above it then the net pressure will become _____
Answer:
4 pa
Explanation:
We know that formula for pressure is;
P = Force/area
Where formula for force is;
F = mg
Thus;
P = mg/A
We are told P = 2 pa
Thus;
mg/A = 2
Now, when we add the same weight, our new Force is;
F_new = mg + mg = 2mg
Area remains the same A.
Thus, net pressure = F_new/A = 2mg/A
From earlier, we saw that mg/A = 2.
Thus;
net pressure = 2mg/A = 2 × 2 = 4 pa
Find the gravitational potential energy of an 85kg person, standing atop Mt. Everest, at an altitude of 8848m. Use sea level as the location for y=0.
Explanation:
Gravitational potential energy
= mgh
= (85kg)(9.81N/kg)(8848m)
= 7,377,904.8J.
A 3.1 kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 16.6 N. Find the speed after it has moved 3.8m.
We know, acceleration is given by :
a = F/m
a = 16.6/3.1 m/s²
a = 5.35 m/s²
Now, we need to find the speed after it has moved 3.8 m.
By equation of motion :
[tex]v^2 -u^2 = 2as\\\\v=\sqrt{2as}\\\\v = \sqrt{2\times 5.35 \times 3.8 }\ m/s\\\\v = 6.38\ m/s[/tex]
Therefore, the speed after it has moved 3.8 m is 6.38 m/s.
1) Determine the magnitude of energy for each of the blanks on the diagram. Give the correct values for 1A, 1B, and 1C.
2)Explain how the energy transformations follow the Law of Conservation of energy throughout the skier's path. Must answer in complete sentences.
3)Explain how the situation shown above would be different if the skier experiences friction while traveling downhill. Include the terms kinetic energy, heat energy, mechanical energy, total energy and friction. Must answer in complete sentences.
Answer:
Explanation: y’all taking the same test as me hahahahah I got the answers but I can’t attach the picture here so hit me up on snap daniela_0789
The total energy at each point in the path of the skier is constant as the
skier travels down the slope.
1) 1A P.E. = 15,000 J, 1B K.E. = 19,000 J, 1C, P.E. = 0 J2) The mechanical energy of the system is constant3) Energy is given off as heat due to friction such that the total energy of the system is not conserved if friction is experiencedReasons:
1) Given that the total mechanical energy, M.E. is constant, we have;
M.E. = Kinetic Energy, K.E. + Potential Energy, P.E. = Constant
M.E. = K.E. + P.E.
At the start, we have;
P.E. = 25,000 J
K.E. = 0 J
Therefore;
M.E. = 25,000 J + 0 J = 25,000 J
At point 1A, we have, K.E = 10,000 J
P.E. = M.E. - K.E.
Therefore;
P.E. = 25,000 - 10,000 = 15,000
The potential energy at point 1A, P.E. = 15,000 J
At point 1B, we have; PE = 6000 J
K.E. = M.E. - P.E.
Therefore;
K.E. = 25,000 J - 6,000 J = 19,000 J
At 1B, K.E. = 19,000 J
At point 1C, we have; K.E. = 25,000 J
Therefore;
P.E. = 25,000 J - 25,000 J = 0 J
At 1C P.E. = 0 J
2) The Law of Conservation of Energy states that the energy in a closed
system is constant.
The energy transformation follows the Law of Conservation of energy
given that the total mechanical energy is constant at all points along the
path.
3) The energy of a system is not conserved when a external force is
applied.
If the skier experiences friction, the force of friction does work to reduce
the speed of the skier, thereby reducing the kinetic energy at a point
downslope, where the potential energy has already been reduced,
resulting in an reduction in the mechanical energy of the system.
Therefore, if the skier experiences friction, the total energy of the system is
not conserved, as energy will be consumed in the work done by friction
which is converted to heat and sound energies.
Learn more about the Law of Conservation of Energy here:
https://brainly.com/question/2950237
*PLEASE HELP*
WILL MARK BRAINLIEST
Light passing through a double slit with
separation d = 6.42x10-5 m creates its
first maximum (m=1) at an angle of
0.396 deg. What is the wavelength of
the light, IN NANOMETERS?
(Hint: The answer will be between 400 and
700.) (Unit = nm)
Answer:
443.8 nm
Explanation:
mλ = yd/D = d(y/D)
m = 1
d = 6.42 x 10-5 m
θ = 0.396°
y/D = tanθ = tan(0.396°) ≅ 6.912 x 10-3
λ = (6.42 x 10-5 m) tan(0.396°) = (6.42 x 10-5 m)(6.912 x 10-3) ≅ 4.438 x 10-7 m = 443.8 nm
I JUST WANT TO MAKE CLEAR THAT d=6.42*10^-5 NOT d = 6.42x10-5 right?
PLEASE HELP!!
To conclude, how would you explain
electromagnetic induction to a 5 year old?
words to include: magnets, electricity, and
change.
Answer:
fgdyauqushxhxxgxggxgxgxgsgsysysy
sysydyydydysysysysyxhxyt56fys7w727ueudu
BERE
Which describes the positions on a horizontal number line?
0
O All points to the left of one are positive.
O All points to the right of one are positive.
O All points to the left of zero are negative.
O All points to the right of zero are negative.
Mark this and return
Save and Exit
Next
Submit
Answer:
All points to the left of zero are negative
Explanation:
Answer:
C
Explanation:
on edge
Which of the following is not a metric base unit? *
Meter
Liter
Inch
Gram
Answer:
Liter
Explanation:
A block of mass 3.20 kg is placed against a horizontal spring of constant k = 865 N/m and pushed so the spring compresses by 0.0650 m. HINT (a) What is the elastic potential energy of the block-spring system (in J)? (b) If the block is now released and the surface is frictionless, calculate the block's speed (in m/s) after leaving the spring. n/s
Answer:
a) The initial elastic potential energy of the block-spring system is 28.113 joules.
b) The final speed of the block is approximately 4.192 meters per second.
Explanation:
a) By applying Hooke's law and definition of work, we define the elastic potential energy ([tex]U_{g}[/tex]), measured in joules, by the following formula:
[tex]U_{g} = \frac{1}{2}\cdot k\cdot x^{2}[/tex] (1)
Where:
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]x[/tex] - Deformation of the spring, measured in meters.
If we know that [tex]k = 865\,\frac{N}{m}[/tex] and [tex]x = 0.065\,m[/tex], then the elastic potential energy is:
[tex]U_{g} = \frac{1}{2}\cdot \left(865\,\frac{N}{m} \right) \cdot (0.065\,m)[/tex]
[tex]U_{g} = 28.113\,J[/tex]
The initial elastic potential energy of the block-spring system is 28.113 joules.
b) According to the Principle of Energy Conservation, the initial elastic potential energy of the block-spring system becomes into translational kinetic energy, that is:
[tex]U_{g} = \frac{1}{2}\cdot m\cdot v^{2}[/tex] (2)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v[/tex] - Final speed, measured in meters per second.
Then, the final speed is cleared:
[tex]v = \sqrt{\frac{2\cdot U_{g}}{m} }[/tex]
If we know that [tex]U_{g} = 28.113\,J[/tex] and [tex]m = 3.20\,kg[/tex], then the final speed of the block is:
[tex]v = \sqrt{\frac{2\cdot (28.113\,J)}{3.20\,kg} }[/tex]
[tex]v \approx 4.192\,\frac{m}{s}[/tex]
The final speed of the block is approximately 4.192 meters per second.
What is the mass of an object accelerated at 2 m/s2 by a net force of 12 N?
Please help I’m struggling !!
Answer: the answer is 6kg.
Explanation:
Mass= force divided by acceleration, which would be 12 divided by 2.
. Momentum is ____________ proportional to an objects mass and its velocity, meaning that if either the mass or the velocity increases, the momentum will _______________.
Answer:
1) directly
2) increase or double
Explanation:
Momentum is directly proportional to an objects mass and its velocity, meaning that if either the mass or the velocity increases, the momentum will double/increase.
In schematic diagrams, currents are indicated using arrows. What do the arrows indicate? a) the direction of motion of the electrons b) the direction of the current vector c) the direction of motion of the charge carriers d) the direction that positive charge carriers would move e) nothing; they are just a convenient drawing tool
Answer:
D
Explanation:
The direction that positive charge would move
A uniform rod of length L, mass M, is suspended by two thin strings. Which of the following statements is true
regarding the tensions in the strings? Please explain
(a) T2 = 0.8 T1
(b) None of the above
(c) T2 = T1
(d) Not enough information to determine
(e) T2 = 0.6 T1
(f) T2 = 2.5 T1
Answer:
(d) not enough info
Explanation:
because it doesn't specify where the strings are attached
if it was the two ends of the rod then T1 would be equal to T2
Under which of the following conditions is the magnitude of the average velocity of a particle moving in one dimension smaller than the average speed over some time interval?
Answer:
A particle moves in the +x direction and then reverses the direction of its velocity
Explanation:
This is illustrated in that when a particle moves in a straight route with no alterations in direction, this will lead to displacement and distance being equal at any point in time during the movement. Thereby the quantity of average speed and average velocity will equal.
On the other hand, should the particle reverses direction, the distance traveled will be greater than it's displacement, thereby, the average speed will be greater than the average velocity.
______ is the limiting factor that can keep some people from exercising if they can’t afford to purchase certain types of exercise equipment or pay for activities. Money Understanding Motivation Location Family behaviors
Answer:
I think it is Location.
Explanation:
If im wrong im sorry
Answer:
Money
Explanation:
______ is the limiting factor that can keep some people from exercising if they (can’t afford) to purchase certain types of exercise equipment or pay for activities.
Says they cant afford it, therefore it would obviously be money
In a test run, a certain car accelerates uniformly from zero to 20.4 m/s in 2.60 s.
Required:
a. What is the magnitude of the cars acceleration?
b. How long does it take the car to change speed from 10.0 m/s to 20 m/s.
c. Will doubling the time always double the change in speed? why?
Answer:
(a) The acceleration is 7.85 m/s²
(b) It takes the car to change speed from 10.0 m / s to 20 m / s in a time of 1.27 seconds.
(c) Doubling the time will double the change in velocity if the acceleration is kept constant.
Explanation:
(a) Acceleration is the physical quantity that measures the rate of change of velocity with time. That is, acceleration relates changes in speed with the time in which they occur, that is, it measures how fast the changes in speed are.
The average acceleration is calculated using the following expression:
[tex]a=\frac{vf-vi}{t}[/tex]
where a is the acceleration, vf is the final velocity, vi is the initial velocity and t is the time.
In this case:
vf= 20.4 m/svi=0 m/st= 2.60 sReplacing:
[tex]a=\frac{20.4 \frac{m}{s} - 0\frac{m}{s} }{2.60 s}[/tex]
a= 7.85 m/s²
The acceleration is 7.85 m/s²
(b) In this case you know:
a= 7.85 m/s²vf= 20 m/svi= 10 m/sReplacing:
[tex]7.85 \frac{m}{s^{2} } =\frac{20 \frac{m}{s} - 10\frac{m}{s} }{t}[/tex]
and solving you get:
[tex]t=\frac{20 \frac{m}{s} - 10\frac{m}{s} }{7.85 \frac{m}{s^{2} } }[/tex]
t=1.27 s
It takes the car to change speed from 10.0 m / s to 20 m / s in a time of 1.27 seconds.
(c) Being:
[tex]a=\frac{vf-vi}{t}[/tex]
Then:
a*t= vf - vi
vf - vi represents the change in velocity. You can see that, if a (acceleration) is constant, then (vf - vi) is directly proportional to the time t: therefore, if t doubles, the change in velocity doubles as well.
In other words, doubling the time will double the change in velocity if the acceleration is kept constant.
A standard inverting op-amp circuit has an R1 of 10 kΩ and an Rf of 220 kΩ. If the offset current is 100 nA the output offset voltage due to this current is ________.
Answer:
The value is [tex]V_{os} = 0.001 \ V[/tex]
Explanation:
From the question we are told that
The circuit resistance is [tex]R_1 = 10 \ k \Omega[/tex]
The feedback resistance is [tex]R_f = 220 \ k \Omega[/tex]
The offset current is [tex]I_{os } = 100 \ nA = 100 * 1)^{-9} \ A[/tex]
Generally the offset voltage is mathematically reparented as
[tex]V_{os} = R_f * I_{os}[/tex]
=> [tex]V_{os} = 10 *10^{3}* 100 *10^{-9}[/tex]
=> [tex]V_{os} = 0.001 \ V[/tex]
When a punter kicks a football, is he doing any work on the ball while the toe of his foot is in contact with it?
a. Is he doing any work on the ball after it loses contact with his toe?
b. Are any forces doing work on the ball while it is in flight?
Answer:
a) W=0 b) force of gravity and friction force
Explanation:
a) When the kicker touches the soccer ball, his foot exerts a force on the ball in the same direction of its movement, but the ball is stationary, therefore the displacement of the ball during the very short contact is zero and as the work is the product of force and distance, we conclude that there is no work
W = F. d
d = 0
W = 0
b) When the ball is in flight there is mainly the force of gravity that has a vertical direction and the friction force that opposes the movement at all times
a.He does not do any work on the ball after it loses contact with his toe
b.The force of gravity and friction force doing work on the ball while it is in flight.
What is work?Work is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.
Work may be zero, positive and negative.it depend on the diection of body displaced . if the body is displaced in the same direction of force it will be positive.
While if the displacement is in the opposite direction of force applied the work will be negative work . if their is no displacement of the body the work done will be zero.
a.He does not do any work on the ball after it loses contact with his toe.
When the kicker makes contact with the soccer ball, his foot exerts a force on the ball in the same direction as its movement, but the ball is stationary,
So the displacement of the ball during the very short contact is zero, and since work is the product of force and distance, we conclude that there is no work.
[tex]\rm W= F \times d \\\\ \rm W= F \times 0 \\\\ \rm W=0[/tex]
Hence he does not do any work on the ball after it loses contact with his toe.
b.The force of gravity and friction force doing work on the ball while it is in flight.
When the ball is in flight, there is primarily a vertical force of gravity and a friction force that resists movement at all times.]
Hence the force of gravity and friction force doing work on the ball while it is in flight.
To learn more about the work refer to the link ;
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A sports car is accelerated from 0
car?
100 km per hour in 3 s. What is the acceleration of the?
A. 0.1g
100
B. 0.3 g
3s 1 1m 11h
160, 160m
C. 0.9 g
D. 3 g
23. [1 mark]
What is the net force on a truck if the force of friction is 31 N and the force of the engine is 79 N?
48 N backword
79 N forward
48 N forward
110 N forward
Answer:
Fr = 48 [N] forward.
Explanation:
Suppose the movement is on the X axis, in this way we have the force of the engine that produces the movement to the right, while the force produced by the brake causes the vehicle to decrease its speed in this way the sign must be negative.
∑F = Fr
[tex]F_{engine}-F_{brake} =F_{r}\\F_{r}=79-31\\F_{r}=48[N][/tex]
The movement remains forward, since the force produced by the movement is greater than the braking force.
I have a cylinder shaped glass container that is 15 cm tall. It holds 100 cm^3 of nitrogen gas at 1 atm pressure (101.3 kPa) and 22 degrees C. I place a rubber stopper in the top so that no gas can escape. If a 40 N force is required to force the rubber stopper off the top of the flask, what temperature can I heat the nitrogen to with a Bunsen burner before the rubber stopper pops off
Answer:
T₂ = 469.73 K = 196.73 °C
Explanation:
First we will find the surface area of rubber stop:
[tex]Area = A= \frac{Volume}{Length} \\\\A = \frac{100\ cm^3}{15\ cm}\\\\A = 6.67\ cm^2 = 6.67 \ x\ 10^{-4}\ m^2[/tex]
Now, we will find the final pressure required to remove the rubber stop:
[tex]Final\ Pressure\ = P_{2} = \frac{Force}{Area}+Atmospheric Pressure \\\\P_{2} = \frac{40\ N}{6.67\ x\ 10^{-4}\ m^2} + 101.3 KPa\\\\ P_{2} = 60000\ Pa + 101.3 KPa = 60\ KPa + 101.3 KPa\\\\P_{2} = 161.3\ KPa[/tex]
Now, we use equation of state:
[tex]\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}[/tex]
for constant volume due to rigid cylinder:
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\\T_{2} = \frac{P_{2} T_{1}}{P_{1}}[/tex]
where,
P₁ = initial pressure = 101.3 KPa
P₂ = final pressure = 161.3 KPa
T₁ = Initial Temperature = 22°C = 295 K
T₂ = Final Temperature = ?
Therefore,
[tex]T_{2} = \frac{(161.3\ KPa)(295\ K)}{101.3\ KPa}[/tex]
T₂ = 469.73 K = 196.73 °C
A soccer ball with a mass of 0.427 kg approaches a player horizontally with a speed of 16.0 m/s. The player kicks the ball with her foot, which causes the ball to move in the opposite direction with a speed of 21.7 m/s.
Required:
a. What magnitude of impulse (in kg · m/s) is delivered to the ball by the player?
b. What is the direction of the impulse delivered to the ball by the player?
c. If the player's foot is in contact with the ball for 0.0600 s, what is the magnitude of the average force (in N) exerted on the player's foot by the ball?
Explanation:
Given that,
Mass of a soccer ball, m = 0.427 kg
Initial speed, u = 16 m/s
Final speed, v = -21.7 m/s (negative as it is in opposite direction)
(a) Impulse = change in momentum
J = 0.427 × (-21.7 -16)
J = -16.097 kg-m/s
(b) Impulse delivered to the ball by the player is in opposite direction.
(c) Time of contant, t = 0.0600 s
Average force,
J = force×time
[tex]F=\dfrac{J}{t}\\\\F=\dfrac{16.097}{0.0600 }\\\\=268.28\ N[/tex]
Hence, this is the required solution.
Older televisions display a picture using a device called a cathode ray tube, where electrons are emitted at high speed and collide with a phosphorescent surface, causing light to be emitted. The paths of the electrons are altered by magnetic fields. Consider one such electron that is emitted with an initial velocity of 1.85 107 m/s in the horizontal direction when magnetic forces deflect the electron with a vertically upward acceleration of 5.45 1015 m/s2. The phosphorescent screen is a horizontal distance of 5.6 cm away from the point where the electron is emitted. (a) How much time does the electron take to travel from the emission point to the screen? (b) How far does the electron travel vertically before it hits the screen?
Answer:
a) t = 3.027 10⁻⁹ s , b) y = 2.25 10⁻² m
Explanation:
We can solve this problem using the kinematic relations
a) as on the x-axis there is no relationship
vₓ = x / t
t = x / vₓ
We reduce the magnitudes to the SI system
x = 5.6 cm (1m / 100 vm) = 0.056 m
we calculate
t = 0.056 / 1.85 10⁷
t = 3.027 10⁻⁹ s
b) the time is the same for the two movements, on the y axis
y = v₀t + ½ a t²
as the beam leaves horizontal there is no initial vertical velocity
y = ½ a t²
let's calculate
y = ½ 5.45 10¹⁵ (3.027 10⁻⁹)²
y = 2.25 10⁻² m
A honeybee is in flight between a flower and its hive. Which of the following statements is true?
A) The bee exerts a force on the Earth equal to the force the Earth exerts on the bee.
B)The Earth exerts a greater force on the bee than the bee exerts on Earth.
C) The bee is in flight, but not in orbit, so there is no force of gravity on it.
D) The bee exerts a greater force on the Earth than the Earth exerts on the bee.
Answer:
D
Explanation:
The statement i.e. true is option d. The bee exerts a greater force on the Earth than the Earth exerts on the bee.
The Flight that lies between the flower and its hive:When the honeybee is in flight that lies between a flower and its hive.
So here the bee should be exerted a high force on the earth as compared to the earth exerted on the bee.
Therefore, the option d is correct.
And, the rest of the options are wrong.
Learn more about force here: https://brainly.com/question/2566388
12. Why do you think afternoons are much warmer than the mornings or evenings?
Answer:
Afternoons are much warmer than the mornings or evenings is explained below in complete detail.
Explanation:
Because in the afternoon the sun is on the head and the beams of sunshine arrive straight to the earth so we exhibit warmer and in daylight and evening the sun is sparse to the earth so we observe colder than an afternoon. However, in the evening, when the sun is weak, I don't sense or experienced loads of heat and sunburn on body parts that are viewing the sun directly.