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3 years ago

The mass of an object with 500 J of kinetic energy moving with a velocity of 5 m/s is

Physics

1 answer:

Ann [662]3 years ago

4 0

40kg

Explanation:

Given parameters:

Kinetic energy = 500J

Velocity = 5m/s

Unknown:

Mass of the object = ?

Solution:

Kinetic energy is the energy due to motion of a body. It is expressed as:

K.E = $\frac{1}{2}$ mv²

m is the mass of the body

v is the velocity

To find the mass, make it the subject of the expression:

m = $\frac{2 K.E}{v[tex]^{2}$ }[/tex]

m = $\frac{2 x 500}{5[tex]^{2}$ }[/tex] = 40kg

Learn more:

Kinetic energy brainly.com/question/6536722

#learnwithBrainly

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What is the definition of energy

natita [175]

Energy (in Physics) is the ability to do work.

5 0

3 years ago

A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?

Pie

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.

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2 years ago

Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini

amid [387]

Answer:

Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

$E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}$

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

$E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}$

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

$V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r$