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Shalnov [3]
3 years ago
10

Allison wants to determine the density of a bouncing ball. which metric measurements must she use?

Physics
1 answer:
lubasha [3.4K]3 years ago
3 0
Density depends on mass and volume so option D is correct answer. Hope this helps!
You might be interested in
If the loop is removed from the field region in a time interval of 2.8 ms , find the average emf that will be induced in the wir
nekit [7.7K]

The question is incomplete. The complete question is :

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magnetic field. A field of 1.2 T is directed along the positive z-direction, which is upward. (a)If the loop is removed from the field region in a time interval of 2.8 ms, find the average emf that will be induced in the wire loop during the extraction process.

Solution :

Let us consider a $\text{circular loo}p \text{ of wire}$ which has a \text{radius} of r = 15 cm.

It is oriented horizontally along the xy-plane and is located in the region of an $\text{uniform magnetic field}$, such that it points in the positive z direction and having a magnitude of B = 1.2 T.

Now if the loop $\text{is removed from the field region}$ in a time interval of Δt = 2.8 ms. Initially the magnetic field and the area points is in the same direction, so that the angle between them is Ф = 0°, thus the initial and the final fluxes are :

$\phi_{B,i}=BA \cos (\phi) = BA  $    and   $\phi_{B,f} = 0$

Area A = $\pi r^2.$ The induced emf equals to the change in the flux, and is divided by the time that it takes to go from the initial flux, Δt and multiplied by the number of turns N = 1, i.e. ,

$\epsilon = -\frac{\Delta \phi_{B}}{\Delta t}$

  $=-\frac{0-(1.2 T)\pi(0.15^2)}{2.8 \times 10^{-3}}$

  = 30.27 V

Therefore, the emf generated is 30.27 V.

 

7 0
2 years ago
the temperature of a body fell from 100°c to 50°c in 10 minutes. the surrounding temperature was 20°c. what is the temperature a
Anna007 [38]

Answer:

The temperature is 10 c

4 0
3 years ago
What is the speed of light ?​
Brrunno [24]

Answer:

299792458 metres per second

Explanation:

3 0
3 years ago
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
Express the distance between the moon and earth in meters with a meter prefix (km).
tankabanditka [31]

Answer;

= 3.86 × 10^8 Meters

Explanation;

-The distance between the Earth and the moon is 386000 km

But; 1 km = 1000 m

Therefore; 386000 km will be equivalent to;

= 386000 × 1000

= 386000000 m

= 3.86 × 10^8 meters

3 0
3 years ago
Read 2 more answers
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