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Whitepunk [10]
3 years ago
15

The tape in a videotape cassette has a total

Physics
1 answer:
xxMikexx [17]3 years ago
6 0

Answer:

w=19.76 \ rad/s

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

w=w_o+\alpha \ t

Where \alpha is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

\displaystyle v_t=w.r

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call w_1 the angular speed of the loaded reel and w_2 that from the empty reel. We have

w_1=w_{o1}+\alpha_1 \ t

w_2=w_{o2}+\alpha_2 \ t

If r_f=35\ mm=0.035\ m is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

\displaystyle w_{01}=\frac{v_t}{r_f}

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

t=1.8\ h=1.8*3600=6480\ sec

\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s

\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s

If r_e=10\ mm=0.001\ m is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s

The full reel goes from w_{01} to w_{02} in 6480 seconds, so we can compute the angular acceleration:

\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}

\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2

The empty reel goes from w_{02} to w_{01} in 6480 seconds, so we can compute the angular acceleration:

\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2

So the equations for both reels are

w_1=1.098+0.00576 \ t

w_2=38.426-0.00576 \ t

They will be the same when

1.098+0.00576 \ t=38.426-0.00576 \ t

Solving for t

\displaystyle t=\frac{38.426-1.098}{0.0115}

t=3240 \ sec

The common angular speed is

w_1=1.098+0.00576 \ 3240=19.76 \ rad/s

w_2=38.426-0.00576 \ 3240=19.76 \ rad/s

They both result in the same, as expected

\boxed{w=19.76 \ rad/s}

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How far from the surface of Earth is the magnitude of Earth's gravitational field equal to 7.86 N/kg?
AfilCa [17]

Answer:

7.48 x 10⁵ m

Explanation:

g = 7.86 N/kg

M = 5.97 x 10²⁴ kg, R = 6.37 x 10⁶ m.

Find height h

g = GM/(R + h)²

(R + h)² = GM/g = 6.67 x 10⁺¹¹ x 5.97 x 10²⁴ /7.86 = 5.066 x 10¹³

R + h = 7.12 x 10⁶ m

so

h =  7.12 x 10⁶ - 6.37 x 10⁶ = 7.48 x 10⁵ m

4 0
2 years ago
3. A worker pushes a box across the floor to the right at a constant speed with a force of 25N. What
WITCHER [35]

Answer:

Friction between the box and the floor is 25N to the left.

Explanation:

According to Newton's second law of motion, the net force acting on an object is equal to the produce between the object's mass and its acceleration:

F_{net}=ma

where

m is the mass of the object

a is its acceleration

In this problem, we have two forces acting on the object:

- The applied force, F = 25 N, to the right

- The force of friction F_f, opposing the motion of the box, so to the left

So we can write the net force as

F_{net}=F-F_f

Also, we know that the box is moving at constant speed: this means its acceleration is zero, so

a=0

Therefore

F_{net}=0

WHich means:

F-F_f=0

And therefore,

F_f=F=25 N

which means that the force of friction is also 25 N.

6 0
3 years ago
To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth.
Karolina [17]

Answer:

a)  T² = (\frac{4\pi ^2}{GM})  r³

b) veloicity the dependency is the inverse of the root of the distance

kinetic energy  depends on the inverse of the distance

potential energy dependency is the inverse of distance

angular momentum depends directly on the root of the distance

Explanation:

1) for this exercise we will use Newton's second law

            F = ma

in this case the acceleration is centripetal

            a = v² / r

the linear and angular variable are related

           v = w r

we substitute

           a = w² r

force is the universal force of attraction

           F = G \frac{m M}{r^2}

we substitute

         G \frac{m M}{r^2} = m w^2 r

         w² = \frac{GM}{r^3}

angular velocity is related to frequency and period

         w = 2π f = 2π / T

we substitute

            ( \frac{2\pi }{T} ) = \frac{GM}{r^3}

the final equation is

             T² = ()  r³

b) the speed of the orbit can be found

           v = w r

            v = \sqrt{\frac{GM}{r^3} } \ r

            v = \sqrt{\frac{GM}{r} }

in this case the dependency is the inverse of the root of the distance

Kinetic energy

           K = ½ M v²

           K = ½ M GM / r

           K = ½ GM² 1 / r

the kinetic energy depends on the inverse of the distance

Potential energy

          U =

          U = -G mM / r

dependency is the inverse of distance

Angular momentum

          L = r x p

for a circular orbit

           L = r p = r Mv

           L =

         L =

The angular momentum depends directly on the root of the distance

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What is the role of electrical forces in nuclear fission
77julia77 [94]
The electrical forces pulls nucleus apart
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The free-body diagram of a crate is shown. A free body diagram with 4 force vectors. The first vector is pointing downward, labe
satela [25.4K]

The net force acting on the crate is determined as 176 N to the left.

<h3>Net force acting on the crate</h3>

The net force acting on the crate is calculated as follows;

∑F = F1 + F2 + F3 + F4

F(net) = -440y + 176x + 440y - 352x

F(net) = -176 x

The resultant force is pointing in negative x direction.

Thus, the net force acting on the crate is determined as 176 N to the left.

Learn more about net force here: brainly.com/question/14361879

#SPJ1

8 0
2 years ago
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