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Pepsi [2]
3 years ago
10

(I WILL GIVE OUT BRAINLIEST) Two objects were lifted by a machine. One object had a mass of 2 kilograms, and was lifted at a spe

ed of 2 m/sec. The other had a mass of 4 kilograms and was lifted at a rate of 3 m/sec. Which object had more potential energy when it was lifted to a distance of 10 meters?
Physics
1 answer:
LekaFEV [45]3 years ago
4 0

Answer:

Consinder the first body

potential \: energy = mgh \\  = 2 \times 10 \times 10 \\  = 200 \: joules

Consider the second body

potential \: energy = mgh \\  = 4 \times 10 \times 10 \\  = 400 \: joules

The body of mass 4 kg had the greater potential energy than the body of 2 kg

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Jamal plugged his radio into the wall. The radio's plug had copper wires surrounded by rubber. The rubber protects Jamal from___
Mariana [72]
The rubber protects him from being electrocuted by the flow of current going through the plug.
Hope this helped!!
3 0
4 years ago
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you read a primary source and a secondary source that discuss the same experiment. There is a difference in the conclusion made
Nuetrik [128]
You should trust the primary source more.

This is because the primary source is make its conclusion from direct observation, while the secondary source is possibly making reference to another secondary source or to another primary.

The primary source should be trusted more because it is from direct observation.     
4 0
3 years ago
Two equal point charges QQQ are separated by a distance ddd. One of the charges is released and moves away from the other due on
lys-0071 [83]

Answer:

The kinetic energy K of the moving charge is K = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

Explanation:

The potential energy due to two charges q₁ and q₂ at a distance d from each other is given by U = kq₁q₂/r.

Now, for the two charges q₁ = q₂ = Q separated by a distance d, the initial potential energy is U₁ = kQ²/d. The initial kinetic energy of the system K₁ = 0 since there is no motion of the charges initially. When the moving charge is at a distance of r = 3d, the potential energy of the system is U₂ = kQ²/3d and the kinetic energy is K₂.

From the law of conservation of energy, U₁ + K₁ = U₂ + K₂

So, kQ²/d + 0 = kQ²/3d + K

K₂ = kQ²/d - kQ²/3d = 2kQ²/3d

So, the kinetic energy K₂ of the moving charge is K₂ = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

4 0
4 years ago
~~~NEED HELP ASAP~~~<br>Please solve each section and show all work for each section.
anastassius [24]

Explanation:

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass m_A as

x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)

y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)

Substituting (2) into (1), we get

\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)

where f_N= \mu_kN, the frictional force on m_A. Set this aside for now and let's look at the forces on m_B

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>

Let the x-axis be (+) up along the inclined plane. We can write the forces on m_B as

x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)

y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)

From (5), we can solve for <em>N</em> as

N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)

Substituting (7) into (4) we get

m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba

Collecting similar terms together, we get

(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag

or

a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)

Putting in the numbers, we find that a = 1.4\:\text{m/s}. To find the tension <em>T</em>, put the value for the acceleration into (7) and we'll get T = 21.3\:\text{N}. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get N = 50.9\:\text{N}

8 0
3 years ago
A graph shows how the temperature of a substance changes as energy is added steadily over time.
OlgaM077 [116]

Answer:

A flat, horizontal line  

Explanation:

A flat, horizontal line indicates a phase change.

The temperature does not increase because the added heat goes into converting one phase into another.

A is wrong. A downward-sloping line indicates that the temperature is decreasing with time.

C is wrong. An upward-sloping line indicates that the temperature is increasing with time.

8 0
3 years ago
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