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3 years ago

The kinetic energy of a ball with a mass of 0.5 kg and a velocity of 10 m/s is?

Physics

2 answers:

cestrela7 [59]3 years ago

8 0

<u>Answer:</u> The kinetic energy of the ball is 25 J

<u>Explanation:</u>

Kinetic energy is defined as the energy which is possessed due to its motion.

It is also defined as the half of the product of mass of the object and square of the velocity of the object.

Mathematically,

$E_K=\frac{1}{2}mv^2$

where,

$E_K$ = kinetic energy of ball

m = mass of the ball = 0.5 kg

v = velocity of ball = 10 m/s

Putting values in above equation, we get:

$E_K=\frac{1}{2}\times 0.5\times (10)^2\\\\E_K=25J$

Hence, the kinetic energy of the ball is 25 J

Vlad1618 [11]3 years ago

3 0

KE= .5*M*V^2
.5*.5*10^2
=25Joules

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Transmitters send radio waves, and receivers capture radio waves.

Explanation:

Let us look at each of the choices one by one:

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(3). Transmitters have demodulators, and receivers have modulators.

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3 years ago

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3 years ago

An asteroid orbiting the Sun has a mass of 4.00×1016 kg. At a particular instant, it experiences a gravitational force of 3.14×1

Ksivusya [100]

<h2>The asteroid is 4.11 x 10¹¹ m far from Sun</h2>

Explanation:

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M = Mass of body 1

M = Mass of body 2

r = Distance between them

Here we have

M = Mass of Sun = 1.99×10³⁰ kg

m = Mass of asteroid = 4.00×10¹⁶ kg

F = 3.14×10¹³ N

Substituting

$F=\frac{GMm}{r^2}\\\\3.14\times 10^{13}=\frac{6.67\times 10^{-11}\times 1.99\times 10^{30}\times 4\times 10^{16}}{r^2}\\\\r=4.11\times 10^{11}m$

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3 years ago

A solenoid 50-cm long with a radius of 5.0 cm has 800 turns. You find that it carries a current of 10 A. The magnetic flux throu

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Answer:

126 mWb

Explanation:

Given that:

length (L) = 50 cm = 0.5 m, radius (r) = 5 cm = 0.05 m, current (I) = 10 A, number of turns (N) = 800 turns.

We assume that the magnetic field in the solenoid is constant.

The magnetic flux is given as:

$\phi_m=NBAcos(\theta)\\Where\ B\ is\ the\ magnetic\ field\ density,A\ is \ the\ area.\\But\ B =\mu_onI.\ n \ is\ the\ number\ of\ turns\ per\ unit \ length=N/L\\Therefore,B=\frac{\mu_oNI}{L} \\substituting\ the\ value\ of\ B\ in\ the\ equation: \\\phi_m=\frac{NAcos(\theta)*\mu_oNI}{L} .\ But \ \theta=0,cos(\theta)=1\ and\ A=\pi r^2\\ \phi_m=\frac{N^2\pi r^2\mu_oI}{L} \\Substituting\ values:\\\phi_m=\frac{800^2*(\pi*0.05^2)*(4\pi*10^{-7})*10}{0.5}=0.126\ Wb=126\ mWb$