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3 years ago

How would you best define moving charges

Physics

1 answer:

svet-max [94.6K]3 years ago

8 0

Current.A moving charge is an object that changes position to one particular obsever.

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What is the magnitude of the torque that the axle must apply to prevent the disk from rotating?

mihalych1998 [28]

The required torque at the axle, is given by the difference between the

moments of the applied forces.

The torque required is 19.62 N·m counterclockwise

Reasons:

The given parameters are;

Mass of the disk, m = 5.0 kg

Location of the axle = Half the radius of the disk

Diameter of the disk, D = 40 cm = 0.4 m

Applied mass, 0.1 m from the axle = 15 kg

Applied mass, 0.3 m from the axle = 10 kg

Required:

Magnitude of torque at the axle that prevent the disk from rotating

Solution:

Torque needed = Clockwise moment - Counterclockwise moment

Clockwise moment = (10 kg × 0.3 m + 5 kg × 0.1 m) × 9.81 m/s² = 34.335 N·m

Counterclockwise moment = 15 kg × 0.1 m × 9.81 m/s² = 14.715 N·m

τ + Counterclockwise moment = Clockwise moment

τ + 14.715 N·m = 34.335 N·m

Torque required, τ = 34.335 N·m - 14.715 N·m = 19.62 N·m

Torque required, τ = 19.62 N·m counterclockwise

Learn more here:

brainly.com/question/19044661

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The probable question drawing obtained from a similar question online is attached

7 0

2 years ago

Which of the following will most likely be a sustainable environment?

Andrei [34K]

Answer:

I think A

Explanation:

7 0

3 years ago

Read 2 more answers

You read in a science magazine that on the Moon, the speed of a shell leaving the barrel of a modern tank is enough to put the s

olga_2 [115]

To solve this problem we will use the definition of the kinematic equations of centrifugal motion, using the constants of the gravitational acceleration of the moon and the radius of this star.

Centrifugal acceleration is determined by

$a_c = \frac{v^2}{r}$

Where,

v = Velocity

r = Radius

From the given data of the moon we know that gravity there is equivalent to

$a = 1.62m/s$

While the radius of the moon is given by

$r = 1.74*10^6m$

If we rearrange the function to find the speed we will have to

$v = \sqrt{ar}$

$v = \sqrt{1.6(1.74*10^6)}$

$v = 1.7km/s$

The speed for this to happen is 1.7km/s

3 0

3 years ago

a train moving west with an initial velocity of 20m/s accelerates 4m/s2 for 10 seconds . during this time , the train moves a di

Blababa [14]

X=1/2at^2+vt+x0
x=1/2(4)(100)+20(10) = 400 m

8 0

3 years ago

Suppose that you have a spring gun that you use to launch a small metal ball. You try the first two settings of the gun. The fir

Ivan

Answer:

The distance s of how far the ball will go at the highest setting = 2.25m

Explanation:

Let consider x to be the representative of the compression and the distance to be s

Recall that:

$\dfrac{1}{2}\times K \times x^2 = mgs +c$

By cross multiplying

$K \times x^2 = 2(mgs +c)$

$K \times x^2 = 2\times 9.81(ms) +2c$

$K \times x^2 = 19.62(ms) +2c$

$x^2 = A \times s+B$

Thus, for the first setting

x = 1 , s = 0.25

for the second setting

x = 2, s = 1

1 = 0.25A + B --- (1)

4 = A + B ----- (2)

From (1); let B = 1 - 0.25A and substitute it into (2)

4 = A + 1 - 0.25 A

4 - 1 = A - 0.25 A

3 = 0.75 A

A = 3/0.75

A = 4

From (2)

4 = A + B

4 = 4 + B

B = 4 - 4

B = 0

Therefore, for the highest setting, where x = 3

Then :

$x^2 = A \times s+B$ will be:

3² = 4s + 0

9 = 4s

s = 9/4

s = 2.25 m

∴

The distance s of how far the ball will go at the highest setting = 2.25m

6 0

3 years ago