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3 years ago

12

A runner completes the 300-meter dash in 38 seconds. What is the speed of the runner? Round your answer to the nearest tenth.The

answer would be in m/s

1 answer:

Galina-37 [17]3 years ago

7 0

The speed of the runner is 300 m /38 seconds. You can simplify this answer to be about 7.9 m/s

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n electromagnetic wave in vacuum has an electric field amplitude of 611 V/m. Calculate the amplitude of the corresponding magnet

enot [183]

Answer:

The corresponding magnetic field is

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From the question we are told that

The electric field amplitude is $E_o = 611\ V/m$

Generally the magnetic field amplitude is mathematically represented as

$B_o = \frac{E_o }{c }$

Where c is the speed of light with a constant value

$c = 3.0 *0^{8} \ m/s$

So

$B_o = \frac{611 }{3.0*10^{8}}$

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6 0

3 years ago

The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons tr

Degger [83]

Answer:

e. 1.2 x 10²³

Explanation:

According to the problem, The current equation is given by:

$I(t)=0.88e^{-t/6\times3600s}$

Here time is in seconds.

Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.

The relation between current and number of charge carriers is:

$q=\int\limits {I} \, dt$

Here the limits of integration is from 0 to infinite. So,

$q=\int\limits {0.88e^{-t/6\times3600s}}\, dt$

$q=0.88\times(-6\times3600)(0-1)$

q = 1.90 x 10⁴ C

Consider N be the total number of charge carriers. So,

q = N e

Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.

N = q/e

Substitute the suitable values in the above equation.

$N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}$

N = 1.2 x 10²³

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2 years ago