Planet-solar system-galaxy-local group. what comes next

Which best describes the motion of air particles when a transverse wave passes through them?

Minchanka [31]

C.
The particles move perpendicular to the direction of the wave.

3 0

3 years ago

Rahul wants to change the motion map shown so that it shows uniform circular motion. What change should Rahul make?

saw5 [17]

Answer:

B. He should change the lengths of the vectors that point tangent to the circle so that each is the same length.

Explanation:

A uniform circular motion is a motion in a circle where the tangential speed of the object is constant.

In the motion map:

- The arrows pointing towards the centre of the circle represent the centripetal acceleration, and their length represent the magnitude of the acceleration

- The arrows pointing tangential to the circle represent the tangential speed, and their length represent the magnitude of the speed

In this motion map, we see that the length of the vectors pointing tangent to the circle is not constant: this means that the speed is not constant. In order to have a uniform circular motion, the speed must be constant, therefore the lengths of the vectors that point tangent to the circle must be the same.

6 0

3 years ago

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In a nuclear power plant, _____. energy is released from the nuclei of atoms energy is released from the bonds of molecules ener

oksian1 [2.3K]

In a nuclear power plant, energy is released from the nuclei of atoms. The correct option among all the options given in the question is the first option. Huge amount of thermal energy is released by the breaking of the uranium atoms. This energy is used for turning a turbine that produces electricity. It is a very clean method of producing electricity.

8 0

3 years ago

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3.00 textbook rests on a frictionless, horizontal tabletop surface. A cord attached to the book passes over a pulley whose diame

sammy [17]

Answer:

a1 = 3.56 m/s²

Explanation:

We are given;

Mass of book on horizontal surface; m1 = 3 kg

Mass of hanging book; m2 = 4 kg

Diameter of pulley; D = 0.15 m

Radius of pulley; r = D/2 = 0.15/2 = 0.075 m

Change in displacement; Δx = Δy = 1 m

Time; t = 0.75

I've drawn a free body diagram to depict this question.

Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;

ΣF_x = T1 = m1 × a1

a1 is acceleration and can be calculated from Newton's 2nd equation of motion.

s = ut + ½at²

our s is now Δx and a1 is a.

Thus;

Δx = ut + ½a1(t²)

u is initial velocity and equal to zero because the 3 kg book was at rest initially.

Thus, plugging in the relevant values;

1 = 0 + ½a1(0.75²)

Multiply through by 2;

2 = 0.75²a1

a1 = 2/0.75²

a1 = 3.56 m/s²

6 0

3 years ago

A uniform cylindrical grindstone has a mass of 10 kg and a radius of 12 cm. (a) What is the rotational kinetic energy of the gri

Drupady [299]

Answer:

a) KE = 888.26J

b) N = 294.5 turns

Explanation:

For the kinetic energy:

$KE = I/2*\omega_o^2$

The inertia is:

$I=m/2*R^2=0.072kg.m^2$

So, the kinetic energy will be:

$KE = 888.26J$

Now, friction force is:

Ff = μ*N = 0.80*5N = 4N

The energy balance would be:

Kf - Ko = Wf where Kf=0; Ko = 888.26J; and Wf is the work done by friction force.

Wf = -Ff*d = -Ff*N*2*π*R where N is the amount of turns it gives.

Replacing these values into the energy balance:

0-888.26=-4*N*2*π*0.12

-888.26=-0.96*π*N

N=294.5 turns

6 0

3 years ago