At the beach in San Francisco (0 meters) the pressure of the atmosphere is 101.325 kPa
(kilopascals) and in Denver, 1609.344 meters above sea level, the pressure of the atmosphere
is about 83.437 kPa. Using this data, find a linear equation for pressure P in terms of
altitude h. (Hint: write the pressure and altitude in each location as a point (h, P). Then
use point-slope form to find the equation of the line.)

Answer:

[tex]545.89\ \text{Hz}[/tex]

Explanation:

[tex]v_o[/tex] = Velocity of observer = 0

[tex]v_s[/tex] = Velocity of source = 34 m/s

v = Speed of sound in air = 343 m/s

f = Emitted frequency = 600 Hz

Doppler effect will be observed

[tex]f_o=f(\dfrac{v+v_o}{v+v_s})\\\Rightarrow f_o=600(\dfrac{343+0}{343+34})\\\Rightarrow f_o=545.89\ \text{Hz}[/tex]

The frequency heard will be [tex]545.89\ \text{Hz}[/tex].

Answer:

The correct answer is a

Explanation:

The electric field is given by the relation

         F = q E

where The force is the Coulomb force and q is a positive test charge, therefore the electric field has the same direction as the electric force.

Consequently if the charge is positive the field must go out of the charge, if the charge is negative the electric field must be directed towards the charge.

Consequently, in this exercise we are told that the lines are directed towards the load, therefore the load must have a negative sign.

The correct answer is a

Answer:

No.

Explanation:

Acceleration of an object is defined as the rate of change of velocity per unit time. It can be given by :

a = dv/dt

dv is the change in velocity i.e. final velocity-initial velocity

dt is change in time

At a particular time, its change in velocity can be one only. It would imply that a body cannot have two acceleration at a particular time.

Answer:

500 N

Explanation:

Answer:

1.5* 10¹⁵ Hz

Explanation:

        [tex]c = \lambda * f (1)[/tex]

       [tex]f = \frac{c}{\lambda} = \frac{3e8m/s}{200e-9m} = 1.5e15 1/sec = 1.5e15 Hz (2)[/tex]

Answer:

Median

Explanation:

Extra information:-

Median formula:-

[tex]\boxed{\begin{minipage}{6cm}$\bigstar$\:\:\sf Median = l + $\sf\dfrac{\frac{n}{2}-C.f.}{f}\times h\\\\Here: \\1)\:n = \sum f =\\2)\:l=Lower\:limit\:of\:median\:class=\\3)\:C.f.=Cumulative\:frequency\:of\:class\\preceeding\:the\:median\:class=\\4)\:f= frequency\:of\:median\:class=\\5)\:h= Class\:interval = \end{minipage}}[/tex]

Answer:

The answer is above this

Explanation:

Answer:

All true statements are shown: a) The average speed of the car is [tex]\frac{v}{2}[/tex], c)The magnitude of the acceleration of the car is [tex]\frac{v}{t}[/tex].

Explanation:

Let prove the validity of each statement:

a) The average speed of the car is [tex]\frac{v}{2}[/tex].

The average speed ([tex]\bar v[/tex]) is defined by the following formula:

[tex]\bar v = \frac{v_{o}+v_{f}}{2}[/tex] (1)

Where:

[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the racing car.

If we know that [tex]v_{o} = 0[/tex] and [tex]v_{f} = v[/tex], then the average speed of the racing car:

[tex]\bar v = \frac{0+v}{2}[/tex]

[tex]\bar v = \frac{v}{2}[/tex]

The statement is true.

b) The car travels a distance [tex]v\cdot t[/tex].

Since the racing car is accelerating uniformly, the distance travelled by the car is represented by the following kinematic formula:

[tex]x - x_{o}=v_{o}\cdot t + \frac{1}{2}\cdot a\cdot t^{2}[/tex] (2)

Where [tex]a[/tex] is the acceleration of the racing car, measured in meters per square second.

The statement is false.

c) The magnitude of the acceleration of the car is [tex]\frac{v}{t}[/tex].

Since the racing car is accelerating uniformly, the velocity of the racing car is represented by the following kinematic formula:

[tex]v_{f} = v_{o}+a\cdot t[/tex] (3)

Then, we clear the acceleration of the expression:

[tex]a = \frac{v_{f}-v_{o}}{t}[/tex]

If we know that [tex]v_{o} = 0[/tex] and [tex]v_{f} = v[/tex], then the acceleration of the car is:

[tex]a = \frac{v-0}{t}[/tex]

[tex]a = \frac{v}{t}[/tex]

The statement is true.

d) The velocity of the car remains constant.

Since the car accelerates uniformly, the vehicle does not travel at constant velocity.

The statement is false.

Answer:

Magnitude of the force = 1.25 N

Explanation:

Given:

Spring constant k = 50.0 N/m

Distance s = 2.50 cm = 0.025 m

Find:

Magnitude of the force

Computation:

Magnitude of the force = Ks

Magnitude of the force = 50 x 0.025

Magnitude of the force = 1.25 N

Answer:

C. Gas

Explanation:

Segment CD represents gaseous phase of matter as segment CD shows highest energy which is in case of gas.

Matter in chemistry, is defined as any kind of substance that has mass and occupies space that means it has volume .Matter is composed up of atoms which may or not be of same type.

Atoms are further made up of sub atomic particles which are the protons ,neutrons and the electrons .The matter can exist in various states such as solids, liquids and gases depending on the conditions of temperature and pressure.

The states of matter are inter convertible into each other by changing the parameters of temperature and pressure.The intermolecular forces of attraction are different in different states of matter.Particles of matter have different sizes.

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Answer:

30 mm

Explanation:

[tex]\lambda[/tex] = Wavelength of light = 503 nm

d = Diameter of hole = 0.2 mm

D = Distance of screen from light source = 4.9 m

Diameter of the central bright area is given by

[tex]x=\dfrac{2.44\lambda D}{d}\\\Rightarrow x=\dfrac{2.44\times 503\times 10^{-9}\times 4.9}{0.2\times 10^{-3}}\\\Rightarrow x=0.03\ \text{m}=30\ \text{mm}[/tex]

The diameter of the central bright area is 30 mm.

Answer:

(a) 0.92 cm= 0.092 m.

(b) 141.64 g=0.14164 kg.

(c) 15. 8 cm³=0.0000158 m³

(d) 63.6 g/cm³= 63600 kg/m³

Explanation:

The International System of Units, abbreviated S.I., also called the International System of Measurements is a system of measurements in which its units are based on fundamental physical phenomena. The units of the S.I. They are the international reference for the indications of all measuring instruments.

The International System of Units (SI) arose from the need to unify and give coherence to a great variety of unit subsystems.

The International System of Units consists of seven basic units, also called fundamental units, which define the corresponding fundamental physical quantities and which allow any physical quantity to be expressed in terms or as a combination of them. The fundamental physical quantities are complemented by two more physical quantities, called supplementary ones.

By combining the basic units, the other units are obtained, called units derived from the International System, and which allow defining any physical quantity.

(a)  The SI unit of length is the meter. Being 1 cm = 0.01 m, then 0.92 cm= 0.092 m.

(b)  The SI unit of mass is kg. Being 1 g = 0.001 kg, then 141.64 g=0.14164 kg.

(c) Being 1 cm³ = 0.000001 m³, then 15. 8 cm³=0.0000158 m³

(d) Being 1 g/cm³= 1000 kg/m³, then 63.6 g/cm³= 63600 kg/m³

Given :

Force provided, F = 112 N.

Acceleration of the bag, a = 0.79 m/s².

To Find :

a. What is the mass of the fertilizer bag?

b. How much does the fertilizer bag weigh?

Solution :

We know, force is given by :

F = ma

m = F/a

m = 112/0.79 kg

m = 141.77 kg

Now, weight is given by :

W = mg

W = 141.77 × 9.8 N

W = 1389.35 N

Therefore, the mass of fertilizer bag is 141.77 kg and weight us  1389.35 N.

Answer:

Stationary listener frequency = 77.68 kHz

Explanation:

Given:

Speed of airplane = 344/2 = 172 m/s

Sound of frequency = 5.84 kHz

Computation:

f0 = fs[(v+v0)/(v-vs)]

f0 = 5.84[(344+0)/(344-172)]

f0 = 5.84[(344)/172]

Stationary listener frequency = 77.68 kHz

It is given that up direction be positive.

Ball velocity at the time is was dropped is 3 m/s.

g = -9.8 m/s²

s = -50 m

Now, using equation of motion for time :

s = ut + at²/2

3t - 4.9t² + 50 = 0

4.9t² - 3t - 50 = 0

Solving above equation, we get :

t = 3.52 s

Therefore, the rock will splash into the water after 3.52 s.

Answer:

981N

Explanation:

The gravitational force between the football player and the Earth at the given distance is 979.91 N.

The given parameters:

The gravitational force between the football player and the Earth at the given distance is calculated by applying Newton's law of universal gravitation as follows.

[tex]F = \frac{GmM}{r^2} \\\\F = \frac{6.67 \times 10^{-11} \times 100 \times 5.98 \times 10^{24}}{(6.38 \times ^6)^2} \\\\F = 979.91 \ N[/tex]

Thus, the gravitational force between the football player and the Earth at the given distance is 979.91 N.

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Answer:

A) The disk arrives first

Explanation:

The disk arrives first, and the reason behind this is that the disk posses smaller moment of inertia, picking up quickly of angular speed of an object for a particular torgue can be determined by ts moment of inertia . So small quantity of energy will be used up as rotational energy, and with the rotational energy the disk will move faster that the hoop.

The option that illustrates the object that reaches the bottom first is A. The disk arrives first.

From the information given, we're told that a solid disk and a hoop are simultaneously released from rest at the top of an incline and rolled down without slipping.

It should be noted that the disk will arrive first due to the fact that it has smaller inertia. In such a case, the disk will arrive first while the hoop arrives later.

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Answer: Option C.

Explanation:

When the box is moving, he pushes with a force F.

in this case, the coefficient is μk, and the box is accelerating, this means that the net force is not zero, so F is larger than the friction force.

When the box is still, the man pushes again with a force F, but now the box does not move, so there is no acceleration, which means that the net force is zero, then F is not greater than the maximum static friction force.

Now, the friction force between an object of mass M, and a surface with a coefficient of friction  μ is:  μ*m*g

where g is the gravitational acceleration.

Then, from the first part, we can conclude that:

μk*m*g < F

(the force F is larger than the kinetic friction force)

and from the second part, we know that:

F ≤  μs*m*g

(The force F is not greater than the static friction force)

If we write those two together, we have:

μk*m*g < F ≤  μs*m*g

Then the correct option is c.

Answer:

D

Explanation:

a neutron does not have a positive nor negative charge it remains neutral

Answer:

a) The translational kinetic energy of the particle at point A is 17.28 joules.

b) The speed of the particle at point B is approximately 5.270 meters per second.

c) The total work done on the particle as it moves from A to B is - 9.78 joules.

Explanation:

Let be this particle a conservative system, that is, that non-conservative forces (i.e. friction, viscosity) are negligible.

a) The translational kinetic energy of the particle ([tex]K[/tex]), measured in joules, is determined by the following formula:

[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1)

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]v[/tex] - Speed, measured in meters per second.

If we know that [tex]m = 0.54\,kg[/tex] and [tex]v = 8\,\frac{m}{s}[/tex], the translational kinetic energy at point A is:

[tex]K = \frac{1}{2}\cdot (0.54\,kg)\cdot \left(8\,\frac{m}{s} \right)^{2}[/tex]

[tex]K = 17.28\,J[/tex]

The translational kinetic energy of the particle at point A is 17.28 joules.

b) The speed of the particle is clear in (1):

[tex]v = \sqrt{\frac{2\cdot K}{m} }[/tex]

If we know that [tex]K = 7.5\,J[/tex] and [tex]m = 0.54\,kg[/tex], then the speed of the particle at point B:

[tex]v = \sqrt{\frac{2\cdot (7.5\,J)}{0.54\,kg} }[/tex]

[tex]v\approx 5.270\,\frac{m}{s}[/tex]

The speed of the particle at point B is approximately 5.270 meters per second.

c) According to the Work-Energy Theorem, the total work done on the particle as it moves from A to B ([tex]W_{A\rightarrow B}[/tex]), measured in joules, is equal to the change in the translational kinetic energy of the particle. That is:

[tex]W_{A\rightarrow B} = K_{B}-K_{A}[/tex] (2)

If we know that [tex]K_{A} = 17.28\,J[/tex] and [tex]K_{B} = 7.50\,J[/tex], then the change in the translational kinetic energy of the particle is:

[tex]W_{A\rightarrow B} = 7.50\,J-17.28\,J[/tex]

[tex]W_{A\rightarrow B} = -9.78\,J[/tex]

The total work done on the particle as it moves from A to B is - 9.78 joules.

Answer:

D

Explanation:

Valley Formations

The dive profile used when planning our dive would be 22 metres/70 feet for 30 minutes.

Diving is the sport or activity of swimming for exploring under water.

My buddy and I are planning to dive at a site where the water and air temperature are near freezing. We plan to dive to 18 meters for 30 minutes.

A General rule as printed on the RDP Table and eRDPML is that if someone planning a dive in cold water that might be harmful. Plan the dive assuming the depth is 4 meters/10 feet deeper than actual. For any other information, see the RDP/eRDPML Instructions for Use booklet.

Thus, dive profile used would be 22 metres/70 feet for 30 minutes.

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Answer:

68 kg/m

Explanation:

bro

Total distance = 36500 m

The average velocity = 19.73 m/s

Given

vo=initial velocity=0(from rest)

a=acceleration= 1 m/s²

t₁ = 20 s

t₂ = 0.5 hr = 1800 s

t₃= 30 s

Required

Total distance

Solution

State 1 : acceleration

[tex]\tt d=vo.t+\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 1\times 20^2\rightarrow vo=0\\\\d=200~m[/tex]

[tex]\tt vt=vo+at\\\\vt=at\rightarrow vo=0\\\\vt=1\times 20\\\\vt=20~m/s[/tex]

State 2 : constant speed

[tex]\tt d=v\times t\\\\d=20\times 1800\\\\d=36000~m[/tex]

State 3 : deceleration

[tex]\tt vt=vo+at\rightarrow vt=0(stop)\\\\vo=-at\\\\20=-a.30~s\\\\a=-\dfrac{2}{3}m/s^2(negative=deceleration)[/tex]

[tex]\tt d=vot+\dfrac{1}{2}at^2\\\\d=20.30-\dfrac{1}{2}.\dfrac{2}{3}.30^2\\\\d=300~m[/tex]

Total distance : state 1+ state 2+state 3

[tex]\tt 200 + 36000 + 300=36500~m[/tex]

the average velocity = total distance : total time

[tex]\tt avg~velocity=\dfrac{36500}{20~s+1800~s+30~s}=19.73~m/s[/tex]

The total distance covered by the car will be 36500m and the average velocity of the car for the entire trip will be 19.73 m/sec.

What is average velocity?

The average velocity is the ratio of the total distance traveled to the total time taken by the body. Its unit is m/sec.

The given data in the problem is;

v₀ is the initial velocity=0(from rest)

a is the acceleration= 1 m/s²

t₁ is the time period 1= 20 s

t₂ is the time period 2= 0.5 hr = 1800 s

t₃ is the time period 3= 30 s

For part 1

The distance travelled is found as;

[tex]\rm d=v_0t+\frac{1}{2} at^2 \\\\ \rm d=\frac{1}{2}\times 1 \times (20)^2 \\\\ \rm d=200\ m[/tex]

The initial velocity is found as ;

[tex]\rm v_t=v_0+at\\\\ \rm v_t=0+1 \times 20\\\\ \rm v_t=20 \ m/sec[/tex]

For part 2

The distance traveled is found as;

[tex]\rm d= v \times t \\\\ \rm d= 20 \times 1800 \\\\ \rm d= 36000 \ m[/tex]

For part 3

The acceleration is found as;

[tex]\rm v_t=v_0+at\\\\ \rm 0=v_0+a \times 30\\\\ \rm 20=-30 a \\\\ a= - \frac{2}{3}\ m/sec^2[/tex]

The distance traveled is found as;

[tex]\rm d=v_0t+\frac{1}{2} at^2 \\\\ \rm d=20.30 +\frac{1}{2} \times \frac{2}{3} (30)^2 \\\\ \rm d=300 \ m[/tex]

Total distance travelled= distance travelled( part 1+ part 2+ part3)

Total distance travelled=200+3600+300 = 36500 m

The average velocity is found as;

[tex]v_{avg}= \frac{36500}{20+1800+30} \\\\ v_{avg}= 19.73 \ sec[/tex]

Hence the total distance covered by the car will be 36500m and the average velocity of the car for the entire trip will be 19.73 m/sec.

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Answer:

[tex]149.112\ \text{N}[/tex]

[tex]120.94\ \text{N}[/tex]

[tex]302.77\ \text{N}[/tex]

[tex]58.44\ \text{N}[/tex]

Explanation:

m = Mass of block = 15.2 kg

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

[tex]\theta[/tex] = Angle

a = Acceleration

F = Applied force

Normal force is given by

[tex]N=mg\\\Rightarrow N=15.2\times 9.81\\\Rightarrow N=\boldsymbol{149.112\ \mathbf{N}}[/tex]

[tex]\theta=35^{\circ}[/tex]

[tex]N=mg\cos\theta\\\Rightarrow N=15.2\times 9.81\times \cos35.8^{\circ}\\\Rightarrow N=\mathbf{120.94\ N}[/tex]

[tex]a=3.53\ \text{m/s}^2[/tex]

[tex]N=m(g+a)\\\Rightarrow N=15.2(9.81+3.53)\\\Rightarrow N=\boldsymbol{302.77\ \mathbf{N}}[/tex]

[tex]F=155\ \text{N}[/tex]

[tex]N=mg-F\sin35.8^{\circ}\\\Rightarrow N=15.2\times 9.81-155\sin35.8^{\circ}\\\Rightarrow N=\boldsymbol{58.44\ \mathbf{N}}[/tex]

Answer:

1962 joules

Explanation:

m = 10 kg

h = 20 m

g = 9.81 ms^-2

PE = ?

PE = MGH

PE = 10 x 9.81 x 20

PE = 1962 joules

Answer:

90J

Explanation:

The only time work is being done is when he lifts the box off the ground. Therefore, using the work formula, 2 x 45, you get 90J. Hope this helps someone.

Explanation:

A force involves an interaction between two or more objects, and it causes a push or pull between the objects. An example of opposing force include drag due to interaction with an air mass and the force due to friction between two objects.

hope it helps!

Answer:

-4.35 × 10^-6 N

Explanation:

i just answered it on ap3x :)

Answer:

a)  v = 0.515 km / min ,   b)  x_total = 52 km

Explanation:

The measured speed is defined by the distance traveled between the time

         v = [tex]\frac{\Delta x}{\Delta t}[/tex]

In this case they give us the speed in several time intervals

let's find the distance traveled in each interval

a) Goes at 75 km/h for 10 min

        v = [tex]\frac{x}{t}[/tex]x / t

         x₁ = v t

let's reduce speed to km / min

         v₁ = 75 km / h (1h / 60 min) = 1.25 km / min

the distance traveled in this time is

        x₁ = 1.25 10

        x₁ = 12.5 km

b) goes to v = 95 km / h for 6 min

        v = 95 km / h (1h 60 min) = 1.5833 km / min

the distance traveled is

        x₂ = v₂2 t

        x₂ = 1.58333 6

        x₂ = 9.5 km

c) goes to v = 40 km / h for 45.0 min

         v₃ = 40 km / h (1 h / 60min) = 0.6667 km / min

         x₃ = 0.6667 45

         x₃ = 30 km

d) t = 40 min, stopped

         x₄ = 0

A) let's calculate the average speed of the trip

          v =[tex]\frac{x_{1}+x_{2}+x_{3}+x_{4} }{t_{1}+t_{2}+t_{3}+t_{4} }[/tex]

          v = (12.5 +9.5 +30 +0) / (10 +6 +45 +40)

          v = 52/101

         v = 0.515 km / min

B) the distance between the two cities is

        x_total = x₁ + x₂ + x₃

        x_total = 12.5 +9.5 + 30

        x_total = 52 km