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2 years ago

14

If a ball of mass 0.30kg is dropped onto a flat surface, which it hits at 9m/s and rebounds to 3m. Find the rebound speed just a

fter the impact, if g=9.8m/s^2. Round your answer to the nearest tenth

1 answer:

VikaD [51]2 years ago

4 0

Answer:

vinegar without v and i

Explanation:

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A flat loop of wire consisting of a single turn of cross-sectional area 8.20 cm2 is perpendicular to a magnetic field that incre

olga nikolaevna [1]

Answer:

The induced current is $I = 6.25*10^{-4} \ A$

Explanation:

From the question we are told that

The number of turns is $N = 1$

The cross-sectional area is $A = 8.20 cm^2 = 8.20 * 10^{-4} \ m^2$

The initial magnetic field is $B_i = 0.500 \ T$

The magnetic field at time = 1.02 s is $B_t = 2.60 \ T$

The resistance is $R = 2.70\ \Omega$

The induced emf is mathematically represented as

$\epsilon = - N * \frac{ d\phi }{dt}$

The negative sign tells us that the induced emf is moving opposite to the change in magnetic flux

Here $d\phi$ is the change in magnetic flux which is mathematically represented as

$d \phi = dB * A$

Where dB is the change in magnetic field which is mathematically represented as

$dB = B_t - B_i$

substituting values

$dB = 2.60 - 0.500$

$dB = 2.1 \ T$

Thus

$d \phi = 2.1 * 8.20 *10^{-4}$

$d \phi = 1.722*10^{-3} \ weber$

So

$|\epsilon| = 1 * \frac{ 1.722*10^{-3}}{1.02}$

$|\epsilon| = 1.69 *10^{-3} \ V$

The induced current i mathematically represented as

$I = \frac{\epsilon}{ R }$

substituting values

$I = \frac{1.69*10^{-3}}{ 2.70 }$

$I = 6.25*10^{-4} \ A$

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4 years ago