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3 years ago

What is the gravitational potential energy of a 3 kg ball that is 1 meter above the floor?

Physics

1 answer:

Sveta_85 [38]3 years ago

7 0

Answer:

PE=mgh

M= Mass (kg)

G= Gravitational field strength (N/kg)

H= Hight (m)

PE= Gravitational Potential Energy (J)

Explanation:

Gravitational Potential Energy is the energy stored in a object due to its position above the Earth's surface.

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Which choice correctly ranks these items from smallest to largest?

Vera_Pavlovna [14]

Answer:

your answer is: electron → carbon atom → quantum dot → E. coli bacteria cell → comma

Explanation:

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4 years ago

A ball with a mass of 5.0 g is moving at a speed of 2.0 m/s. Would doubling the mass or doubling the speed have a greater effect

Arte-miy333 [17]

Answer:

Yes

Explanation:

Kinetic energy is K.E1/2mv2 so that means it is directly proportional to mass and velocity.

5 0

3 years ago

Read 2 more answers

An inelastic collision of two objects is characterized by the following.

serious [3.7K]

Options:

(a) Total kinetic energy of the system remains constant.

(b) Total momentum of the system is conserved.

(d) Neither A nor B are true.

Answer:

(b) Total momentum of the system is conserved.

Explanation:

An inelastic collision is a type of collision in which momentum is conserved and kinetic energy is not conserved. That is, there is loss of kinetic energy.

In an inelastic collision:

Total momentum before collision = Total momentum after collision

An example of inelastic collision is seen in the ballistic pendulum, The ballistic pendulum is a device in which a projectile such as a bullet is fired into a suspended heavy wooden stationary block.

8 0

3 years ago

A 1-kilogram mass is attached to a spring whose constant is 14 N/m, and the entire system is then submerged in a liquid that imp

ch4aika [34]

Answer:

Part(a): The equation of motion is $\bf{x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}}$ .

Part(b): The equation of motion is $\bf{x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}}$ .

Explanation:

If ' $m$ ' be the mass of the object, ' $k$ ' be the force constant and ' $\beta$ ' be the damping constant, then the equation of motion of the particle can be written as

$\dfrac{d^{2}x}{dt^{2}} + \dfrac{\beta}{m} \dfrac{dx}{dt} + \dfrac{k}{m}x= 0.........................................(I)$

Given $m$ = 1 Kg, $k$ = 14 $N~m^{-1}$ , $\beta$ = 9. Substituting these values in equation (I),

$\dfrac{d^{2}x}{dt^{2}} + 9~\dfrac{dx}{dt} + 14~x= 0$

Taking a trial solution $x(t) = e^{mt}$ , the auxiliary equation can be written as

$m^{2} + 9m + 14 = 0............................................................(II)$

and its solutions are $m_{1} = -2~and~m_{2} = -7$ , resulting the general solution

$x(t) = C_{1}~e^{-2t} + C_{2}~e^{-7t}....................................................................(III)$

The velocity at any instant of time of the mass is

$v(t) = -2C_{1}~e^{-2t} _7~C_{2}~e^{-7t}..............................................................(IV)$

Part(a):

Given $x(t=0) = 1 m,~and~v(t=0) = 0~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(V)\\&and,& 0 = -2C_{1} - 7C_{2}.......................(VI)$

Solving equations (V) and (VI), we have

$C_{1} = \dfrac{7}{5}~and~C_{2} = \dfrac{-2}{5}$

So the equation of motion is

$x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}$

Part(b):

Given $x(t=0) = 1 m,~and~v(t=0) = - 12~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(VII)\\&and,& -12 = -2C_{1} -7C_{2}.......................(VIII)$

Solving equations (V) and (VI), we have

$C_{1} = -1~and~C_{2} = \dfrac{11}{5}$

So the equation of motion is

$x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}$

3 0

4 years ago

Two identical masses are moving down a slop. One mass is traveling twice as fast as the other mass. How does the kinetic energy

svp [43]

I believe it is D because the more weight the more total speed it will end with

3 0

3 years ago

Read 2 more answers