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3 years ago

1. A record with a radius of 0.3m spins in a clockwise circle with a centripetal

Physics

1 answer:

Hitman42 [59]3 years ago

6 0

Solve for the linear/tangential speed:

a = v²/r

where a = centripetal acceleration, v = speed, and r = radius.

4.7 m/s² = v²/(0.3 m)

v² = (0.3 m) (4.7 m/s²)

v ≈ 3.96 m/s

For every time the record completes one revolution, a fixed point on the edge of the record travels a distance equal to its circumference, which is 2π (0.3 m) ≈ 1.88 m. So if 1 rev ≈ 1.88 m, then the angular speed of the record is

(3.96 m/s) (1/1.88 rev/m) ≈ 7.46 rev/s

Take the reciprocal of this to get the period:

1 / (7.46 rev/s) ≈ 0.134 s/rev

So it takes the record about 0.134 seconds to complete one revolution.

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The cross section of a copper strip is 1.2 mmthick and 20 mm wide. There is a 25-A current through this cross section, with the

Naily [24]

To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.

The drift velocity is given by the equation:

$V_d = \frac{I}{nAq}$

Where

I = current

n = Number of free electrons

A = Cross-Section Area

q = charge of proton

Our values are given by,

$I = 25 A$

$A= 1.2*20 *10^{-6} m^2$

$q= 1.6*10^{-19}C$

$N = 8.47*10^{19} mm^{-3}$

$V_d =\frac{25}{(1.2*20 *10^{-6})(1.6*10^{-19})(8.47*10^{19} )}$

$V_d = 7.68*10^{-5}m/s$

The hall voltage is given by

$V=\frac{IB}{ned}$

Where

B= Magnetic field

n = number of free electrons

d = distance

e = charge of electron

Then using the formula and replacing,

$V=\frac{(2.5)(25)}{(8.47*10^{28})(1.6*10^{-19})(1.2*10^{-3})}$

$V = 3.84*10^{-6}V$

5 0

3 years ago

A man wishes to lift a stone weighing 1440 N, using a first-class lever that measures 5 meters. What force should it perform if

Crazy boy [7]

Answer:

3360 N

Explanation:

In a first-class lever, the effort force and load force are on opposite sides of the fulcrum.

The lever is 5 m long. The load force is 1.50 m from the fulcrum, so the effort force must be 3.50 m from the fulcrum.

The torques are equal:

Fr = Fr

(1440 N) (3.5 m) = F (1.5 m)

F = 3360 N

4 0

3 years ago

How do professional bicycle riders reduce friction. Name two ways. Thx

wlad13 [49]

A well maintained bicycle is key. Oil the chain so that it's less likely to get hung up. It will have a smoother flow. Another could be in their clothing. Notice they usually wear good fitting clothing? I believe the more from fitting pants help clothing friction. Another way could be keeping the tires well inflated so that the tires aren't dragging. Rolling along smoothly on properly inflated tires seems like a must. Get a few more answers along with mine, so that you have a variety to choose from.

4 0

3 years ago

What was the direction of the ball’s velocity

Tju [1.3M]

Part of the question is missing. Here it is:

A 72 g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of 0.7m away from the table. The acceleration of gravity is 9.81 m/s2. What was the direction of the ball’s velocity just before it hit the floor?

Answer:

$\theta=-75.7^{\circ}$

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A horizontal motion at constant velocity

- A vertical motion at constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time of flight of the ball. This can be done by using the suvat equation

$s=ut+\frac{1}{2}at^2$

where, choosing downward as positive direction:

s =1.3 m is the vertical displacement of the ball

u = 0 is the initial vertical velocity

$a=g=9.8 m/s^2$ is the acceleration of gravity

t is the time

Solving for t,

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.3)}{9.8}}=0.52 s$

Now we can find the final vertical velocity of the ball, using:

$v_y=u+at$

And susbtituting t = 0.52 s, we find

$v_y = 0 +(9.8)(0.52)=5.1 m/s$

It is important to keep in mind that the direction of this velocity is downward, since we chose downward as positive direction.

The horizontal velocity of the ball instead is constant; we know that the ball covers a horizontal distance of

d = 0.7 m

In a time of

t = 0.52 s

So, the horizontal velocity is

$v_x = \frac{0.7}{0.52}=1.3 m/s$

So now we can find the direction of the ball's velocity using:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.1}{1.3})=75.7^{\circ}$

And since the vertical direction is downward, this means that this velocity is below the horizontal, so the answer is

$\theta=-75.7^{\circ}$

8 0

3 years ago

The following are connected in parallel circuit at home EXCEPT:

nikdorinn [45]

Light bulbs, think about parallel circuits as something you plug into, so for example you plug in a tv for it to work, same with a refrigerator and Christmas lights

8 0

2 years ago