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2 years ago

A 0.15-kg ball is thrown into the air and rises to a height of 20.0 m. How much kinetic energy did the ball initially have?

2 answers:

7 0

IF the toss was straight upward, then the kinetic energy it got
from the toss is the gravitational potential energy it has at the top,
where it stops rising and starts falling.

Potential energy = (mass) x   (gravity) x (height)

   = (0.15 kg) x (9.8 m/s²) x (20 m)

   = 29.4 kg-m²/s² = 29.4 joules .

mel-nik [20]2 years ago

3 0

The conservation of energy says that energy in an isolated system remains constant, that it is not created nor destroyed but transforms from one form of energy to another.
So when the ball is thrown into the air, it has kinetic energy (KE). As it goes into the air, that kinetic energy transforms into gravitational potential energy (GPE). The higher it gets, the more GPE it has and therefore the less KE it has.
At the top, all of it is transformed into GPE and there is no KE.
But total energy at any point after the ball is thrown until it lands it has the same energy throughout.

Knowing this, you can set the beginning KE equal to the final GPE.
KE = GPE

Insert formula for GPE.

KE = mgh
KE = 0.15kg*9.8m/s^2*20.0m
KE = 29 J

The ball initially had 29 J of kinetic energy

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There is a known potential difference between two charged plates of 12000 Volts. An object with a charge of 6.5 x 10-6 C charge

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Answer:

1.97 m/s.

Explanation:

From the question,

Using the law of conservation of energy,

The energy stored in the charged plate = Kinetic energy of the mass

1/2(qV) = 1/2mv².......................... Equation 1

Where q = charge, V = voltage, m = mass, v = velocity.

make v the subject of the equation

v = √(qV/m)......................... Equation 2

Given: q = 6.5×10⁻⁶ C, V = 12000 Volts, m = 0.02 kg

Substitute these values into equation 2

v = √(6.5×10⁻⁶×12000 /0.02)

v = √3.9

v = 1.97 m/s.

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3 years ago

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2 years ago

Chegg Given that the mean radius of the Moon’s orbit is 3.84 x 108 m and its period is 2.36 x 106 sec, at what altitude above th

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Answer:

The altitude of geostationary satellite is $3.58\times10^{7}\ m$

Explanation:

Given that,

Radius of moon's orbit $r=3.84\times10^{8}\ m$

Time period $T=2.36\times10^{6}\ sec$

We need to calculate the orbital radius of geostationary satellite is

Using formula of time period

$T=\sqrt{\dfrac{4\pi^2}{GM}a^3}$

$a=((\dfrac{GM}{4\pi^2})T^2)^{\dfrac{1}{3}}$

Where, G = gravitational constant

M = Mass of earth

T = time period of geostationary satellite orbit

Put the value in to the formula

$a=((\dfrac{6.67\times10^{-11}\times5.97\times10^{24}}{4\times\pi^2})\times(86160)^2)^{\dfrac{1}{3}}$

$a=4.217\times10^{7}\ m$

We need to calculate the altitude of geostationary satellite

Using formula of altitude

$h = a-R_{e}$

Where, R = radius of earth

a = radius of geostationary satellite

Put the value into the formula

$h =4.217\times10^{7}-6.38\times10^{6}$

$h =35790000\ m$

$h=3.58\times10^{7}\ m$

Hence, The altitude of geostationary satellite is $3.58\times10^{7}\ m$

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3 years ago

The voltage V in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance R is slowly increasin

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Answer:

The current in the circuit decrease slowly .

Explanation:

Given as :

For the electrical circuit

The voltage V in the circuit is slowly decreasing

The resistance R of the resistor slowly increasing after heating

Now, From Ohm's Law

Voltage is directly proportional to the flow of current through circuit

I.e V ∝ I

Or. V = R × I

where R is the proportionate constant and this is the resistance of the resistor

whose property is to opposes the flow of current in the circuit

So, If R value more then current I reduces in the circuit

∵ Here in the circuit , The resistance is slowly increasing, so, current I is slowly decreasing .

Hence The current in the circuit decrease slowly . answer

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3 years ago