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3 years ago

2. What is the kinetic energy of a 7.26 kg bowling ball that is rolling at a speed of 2 m/s?

2 answers:

6 0

Kinetic energy is the energy possessed by an object when that object is moving in space. The higher the mass of an object or higher the speed of an object the higher the kinetic energy will be.

So to calculate the Kinetic Energy we can use the following formula

K.E=(1/2)*m*v^2

Inserting the values in formula gives:

K.E=1/2*7.26*2^2

14.52J

This is the final answer which gives the kinetic energy of the ball.

Dimas [21]3 years ago

3 0

Answer:

The kinetic energy of bowling ball is 14.52J

Explanation:

Given that the formula of kinetic energy is (1/2)mv². Then substitute the following value into the formula:

m = 7.26kg

v = 2m/s

k.e = (1/2)×7.26×2²

= 14.52J

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The average velocity is -4.17 m/s

Explanation:

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For the student in this problem, we have:

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So the displacement is

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Learn more about average speed and velocity:

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4 years ago

Outside a spherically symmetric charge distribution of net charge Q, Gauss's law can be used to show that the electric field at

Sergio039 [100]

Answer:

Q at the center of the distribution.

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3 years ago

What is the kinetic energy k of an electron with momentum 1.05×10−24 kilogram meters per second?

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Momentum = mv
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4 years ago

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Particle q₁ has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10⁻³ N, what is the s

Ne4ueva [31]

The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
$F=qvB \sin \theta$
where $\theta$ is the angle between the directions of v and B.

1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is $q=2.7 \mu C=2.7 \cdot 10^{-6}C$ . In this case, v and B are perpendicular, so $\theta=90^{\circ}$ , therefore we have:
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2) In this second case, the angle between v and B is $\theta=55^{\circ}$ . The charge is now $q=42.0 \mu C=42.0 \cdot 10^{-6}C$ , and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
$F=qvB \sin \theta=(42\cdot 10^{-6}C)(1.21 \cdot 10^3 m/s)(2.8 T)(\sin 55^{\circ})=0.12 N$

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3 years ago

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