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3 years ago

A 2.0 kg mass is lifted 4.0 m above the ground. find the change in gravitational energy

Physics

1 answer:

Makovka662 [10]3 years ago

6 0

Answer:

78.4 J

Explanation:

Given:

Mass of the object (m) = 2.0 kg

Height or displacement of the mass (h) = 4.0 m

Now, change in gravitational potential energy (ΔU) = ?

Change in gravitational potential energy of an object is the energy stored by the object when it is raised to some height above the reference ground level.

Change in gravitational potential energy of an object of mass 'm' raised to a height 'h' above the ground is given as:

$\Delta U=mgh$

Where, 'g' is the acceleration due to gravity and has a value of 9.8 m/s².

Now, plug in all the given values and solve for ΔU. This gives,

$\Delta U=(2.0\ kg)(9.8\ m/s^2)(4.0\ m)\\\\\Delta U=78.4\ J$

Therefore, the change in gravitational potential energy is 78.4 J.

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If an atom has twelve electrons, how many electrons are in the n=1 energy level? how many electrons are in the n=2 energy level?

Sav [38]

The shells correspond with the principal quantum numbers (n = 1, 2, 3, 4 ...) Each shell can contain only a fixed number of electrons. The first shell can hold up to two electrons, the second shell can hold up to eight electrons, the third shell can hold up to 18 and so on.

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3 years ago

the ocean floor is, on average, 4267 m below sea level. What is the pressure in the atmosphere at this depth?

hichkok12 [17]

The pressure at the depth h in the ocean is given by (Stevin's law)
$p= p_0 + \rho g h$
where
$p_0 = 1.0 \cdot 10^5 Pa$ is the atmospheric pressure
and $\rho g h$ is the pressure exerted by the column of water of height h=4267 m, with $\rho = 1000 kg/m^3$ being the water density and $g=9.81 m/s^2$ .
Substituting, we find
$p=1.0 \cdot 10^5 Pa + (1000 kg/m^3)(9.81 m/s^2)(4267 m)=4.20 \cdot 10^7 Pa$
We want to convert this into atmospheres: we know that 1 atm corresponds to the atmospheric pressure at sea level, so $1 atm=1.0\cdot 10^5 Pa$ , therefore we just need to divide by this number:
$p= \frac{4.20 \cdot 10^7 Pa}{1.0 \cdot 10^5 Pa/atm} =420 atm$

7 0

3 years ago

A small metal ball with a mass of m = 62.0 g is attached to a string of length l = 1.85 m. It is held at an angle of θ = 48.5° w

notka56 [123]

The distance x will the ball land after flies off with a horizontal initial velocity is 3.0635 m.

<h3>What is mechanical energy?</h3>

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

A small metal ball with a mass of m = 62.0 g is attached to a string of length l = 1.85 m. It is held at an angle of θ = 48.5° with respect to the vertical.

The ball is then released. When the rope is vertical, the ball collides head-on and perfectly elastically with an identical ball originally at rest. This second ball flies off with a horizontal initial velocity from a height of h = 3.76 m, and then later it hits the ground.

The conservation of energy principle states that total mechanical energy remains conserved in all situations where there is no external force acting on the system.

Kinetic energy = Potential energy

1/2 mv² =mgh₁

The velocity at the bottom, when the height h = 5m, is

v= √2gh₁...................(1)

The vertical height h₁ = l- lcosθ

h₁ = l- lcosθ

h₁ = 1.85 - 1.85cos48.5°

h₁ =0.6241 m

Putting the values in equation (1), we get

v = √2x 9.81 x0.6241

v = 3.499 m/s

The horizontal distance traveled is

x = vt

x = v x √2h/g

Plug the values, we get

x = 3.499 x √2x3.76 / 9.81

x = 3.0635 m

Thus, the horizontal distance ball travels is 3.0635 m.

Learn more about mechanical energy.

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6 0

2 years ago

A person lowers a bucket into a well by turning the hand crank, as the drawing illustrates. The crank handle moves with a consta

dimulka [17.4K]

Answer:

0.453 m/s

Explanation:

Assuming the handle has diameter of 0.4 m while inner part diameter is 0.1 m then the circumference of outer part is $\pi d_h$ where d is diameter and subscript h denote handle. By substituting 0.4 for the handle's diameter then cirxumference of outer part is $\pi\times 0.4\approx 1.256 m$

The rate of rotation will then be 1.81/1.256=1.441 rev/s

Similarly, circumference of inner part will be $\pi d_i$ where subscript i represent inner. Substituting 0.1 for inner diameter then

$\pi\times 0.1\approx 0.3142 m$

The rate of rotation found for outer handle applies for inner hence speed will be 0.3142*1.441=0.453 m/s

7 0

3 years ago

One speaker generates sound waves with amplitude A.

raketka [301]

Answer:

iv) It is 9x bigger than before

Explanation:

As the amplitudes of the new speakers add directly with the original one, taking into account the phase that they have, the composed amplitude of the sound wave is as follows:

At = A + 4A -2A = 3 A

The intensity of the wave, assuming it propagates evenly in all directions, is constant at a given distance from the source, and can be expressed as follows:

I = P/A

where P= Power of the wave source, A= Area (for a point source, is equal to the surface area of a sphere of radius r, where is r is the distance to the source along a straight line)

For a sinusoidal wave, the power is proportional to the square of the amplitude, so the intensity is proportional to the square of the amplitude also.

If the amplitude changes increasing three times, the change in intensity will be proportional to the square of the change in amplitude, i.e., it will be 9 times bigger.

So, the statement iv) is the right one.

7 0

3 years ago