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trapecia [35]
3 years ago
5

How much work is done moving an object 5.0 m by a 15 N force? a. 75 J b. 15 J c. 5.0 J d. 3.0 J

Physics
1 answer:
White raven [17]3 years ago
7 0

Answer:

75 Joules

Explanation:

Work=F*D, plugging in the constant force of 15 newtons multiplying it by the displacement of 5 meters, it results in 75 Joules of work.

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The tape in a videotape cassette has a total
xxMikexx [17]

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}

6 0
3 years ago
At a certain place, Earth's magnetic field has magnitude B =0.703 gauss and is inclined downward at an angle of 75.4° to the hor
Irina18 [472]

Answer:

The charge flows in coulombs is

dq=1.843x10^{-5}C

Explanation:

The current magnitude of current is given by the resistance and the induced Emf as:

I=N*\frac{dF}{Rdt}

\frac{dq}{dt}=\frac{dF}{Rdt}=dq=N*\frac{dF}{R}

dq=\frac{N*\beta*A*(Cos(\alpha_f)-Cos(\alpha_i}{R}

N=1300, \beta=0.703, A=\pi*r^2=\pi*0.10^2=0.01\pi m^2, R=99.4+202=301.4Ω

\alpha_f=14.6,\alpha_i=165.4

Replacing :

dq=\frac{1300*0.703x10^{-4}*0.01\pi*(0.9667-(-0.9667))}{202+99.4}

dq=1.843x10^{-5}C

5 0
3 years ago
A ball is dropped from a height of 20 meters. At what helght does the ball have a velocity of 10 meters/second?
Sunny_sXe [5.5K]

Answer:

5.10 meters.

Explanation:

v²=u²+2gh

or, (10)²=(0)²+2×9.8×h

or, 19.6h=100

or, h=5.10 meters

Hope, this helps you.

8 0
3 years ago
This glass of lemonade is sitting in the hot summer sun. As time passes, in which direction will heat transfer take place?
KiRa [710]

Answer:

c

Explanation:

8 0
3 years ago
Read 2 more answers
Bradley drops a rock in a well. It falls for 12 seconds. How deep is the well?
erik [133]
It’s 12 seconds long
3 0
3 years ago
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