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madam [21]
3 years ago
5

Find the kinetic energy of a tennis ball travelling at a speed of 46 m/s with a mass of 58 g.

Physics
1 answer:
tekilochka [14]3 years ago
6 0
It’s half the mass of the object by its velocity ^2
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As an object increases or decreases, what happens to the total energy?
kompoz [17]

Answer:As an object speeds up, the total energy increases / decreases / remains the same. As an object slows down, the total energy increases / decreases / remains the same.

Explanation:

7 0
3 years ago
What type of relationship does this graph show?
Basile [38]

Answer:

An nonlinear inverse relationship

Explanation:

7 0
3 years ago
PLEASE HELP!!!!!
Arte-miy333 [17]

Answer:

it states that energy can either be gained or lost but it only changes its form.

Explanation:

for example:as a ball is still on the table it posses a potential energy of 100j and a k.e of 0j,as it falls it gains k.e so the midpoint the p.e is equal to the k.e (50j equally) as it approches the ground it completely gains k.e (100j) and the p.e is 0j.

total energy is 100j so it has been converted from p.e to k.e.

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5 0
3 years ago
Does the size of the magnetic field change with the size of the magnet?
mestny [16]

Answer:

Not all of the time, it can also depend on the strength of the magnet.

Explanation:

say you have a small very strong magnet, that might work just as well as a large but week magnet.

4 0
3 years ago
When a small object is launched from the surface of a fictitious planet with a speed of 52.9 m/s, its final speed when it is ver
NISA [10]

Answer:

The escape speed of the planet is 41.29 m/s.

Explanation:

Given that,

Speed = 52.9 m/s

Final speed = 32.3 m/s

We need to calculate the launched with excess kinetic energy

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

K.E=\dfrac{1}{2}\times m\times(32.3)^2

We need to calculate the escape speed of the planet

Using formula of kinetic energy

\text{escape kinetic energy}=\text{launch kinetic energy}-\text{excess kinetic energy}

\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2-\dfrac{1}{2}mv^2

\dfrac{1}{2}\times v^2=\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2

v=\sqrt{2\times(\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2)}

v=41.29\ m/s

Hence, The escape speed of the planet is 41.29 m/s.

4 0
3 years ago
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