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Harlamova29_29 [7]
3 years ago
5

2. Willingness to take turns is one way we can express our attitudes through

Physics
1 answer:
leva [86]3 years ago
3 0
B because willingness shows confidence in your answer
You might be interested in
A bicycle wheel rotates at a constant 25 rev/min. What is true about its angular acceleration?
Nostrana [21]

Answer:

The angular acceleration is zero

Explanation:

When an object is in rotational motion, it has a certain angular velocity, which is the rate of displacement of its angular position.

This angular velocity can change or remain constant - this is given by the angular acceleration, which is:

\alpha =\frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time elapsed

Therefore, the angular acceleration is the rate of change of angular velocity.

In this problem, the bicycle rotates at a constant angular velocity of

\omega=25 rev/min

This means that the change in angular velocity is zero:

\Delta \omega=0

And so, that the angular acceleration is zero:

\alpha=0

8 0
3 years ago
Calculate the potential energy of a body of a mass 2kg held 4 meters above the ground if g=10m/s?​
Drupady [299]

Answer:

16000

Explanation:

Mass(m)=2Kg (1kg= 1ooo g then 2 kg=2000 g)

Velocity(v)= 4 meter

Acceleration due to gravity (g)=10m/s

We know that,

P.E= 1/2 mv^2

or, 1/2 × 2ooo × 4^2

or, 1/2×2000 ×16

or, 2000×8

Therefore= 16000

7 0
2 years ago
A pendulum of length L=36.1 cm and mass m=168 g is released from rest when the cord makes an angle of 65.4 degrees with the vert
pychu [463]

(a) -0.211 m

At the beginning the mass is displaced such that the length of the pendulum is L = 36.1 cm and the angle with the vertical is

\theta=65.4^{\circ}

The projection of the length of the pendulum along the vertical direction is

L_y = L cos \theta = (36.1 cm)(cos 65.4^{\circ})=15.0 cm

the full length of the pendulum when the mass is at the lowest position is

L = 36.1 cm

So the y-displacement of the mass is

\Delta y = 15.0 cm - 36.1 cm = -21.1 cm = -0.211 m

(b) 0.347 J

The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to

\Delta U = mg \Delta y

where we have

m = 168 g = 0.168 kg is the mass of the pendulum

g = 9.8 m/s^2 is the acceleration due to gravity

\Delta y = 0.211 m is the vertical displacement of the pendulum

So, the work done by gravity is

W=(0.168 kg)(9.8 m/s^2)(0.211 m)=0.347 J

And the sign is positive, since the force of gravity (downward) is in the same direction as the vertical displacement of the mass.

(c) Zero

The work done by a force is:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

In this situation, the tension in the string always points in a radial direction (towards the pivot of the pendulum), while the displacement of the mass is tangential (it follows a circular trajectory): this means that the tension and the displacement are always perpendicular to each other, so in the formula

\theta=90^{\circ}, cos \theta = 0

and so the work done is zero.

5 0
3 years ago
Compare and contrast fluorescent and incandescent light bulbs.
dezoksy [38]

Incandescent light is a glowing white light produced by heat. An incandescent light bulb works by heating a filament in the bulb. Fluorescent light is a bright light produced by electricity flowing through a tube filled with ionized gas. Fluorescent light bulbs are more energy-efficient than incandescent bulbs

5 0
3 years ago
A polarized light that has an intensity I0 = 60.0 W/m² is incident on three polarizing disks whose planes are parallel and cente
nikitadnepr [17]

Answer:

The transmitted intensity through all polarizers is I_3 =41.31 W/m^2

Explanation:

 According to Malu's law the intensity of a polarized light having an initial intensity I_0 is mathematically represented as

               I = I_0cos^2 \theta

Now  considering the polarizer(The polarizing disk) the equation above becomes

          I = I_0 (cos^2 \theta)^n

Where n is the number of polarizers

       Substituting  60.0W/m^2 for the initial intensity 3 for the n and 20° for the angle of rotation

           I_3 = 60 (cos^220)^3

               =41.31 W/m^2

             

     

                         

6 0
3 years ago
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