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konstantin123 [22]

3 years ago

15

On the Moon, the acceleration due to the effect of gravity is only about 1/6 of that on Earth. An astronaut whose weight on Eart

h is 595 N travels to the lunar surface.

1 answer:

vazorg [7]3 years ago

3 0

And on the moon, the with is 595/6 N

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Which universal force acts on protons and neutrons in an atom’s nucleus

Yuliya22 [10]

The strong nuclear force overcomes the electric force of repulsion thatacts among the protons in thenucleus. B. The weak nuclear force is involved in certain types of radioactive processes. A.The strong nuclear force is a powerful force of attraction that acts only on theneutrons and protons in the nucleus.

6 0

3 years ago

When an electric stove element is hot enough, it gives off a dull red glow. When it cools to the point that it no longer glows,

DochEvi [55]

Answer:

It will have a longer wavelength

Explanation:

When an electric stove is hot and gives dull red glow. a part of the energy dissipated is emitted as visible light and part as infrared radiation in the form of heat. When the stove cools down, and no longer glows all the energy is now in the form of infrared radiation.In the electromagnetic spectrum infrared rays have a higher wavelength than visible light. Hence for the reason the radiation will have a higher wavelength since visible light is cut off.

4 0

3 years ago

A wire is joined to points X and Y in the circuit diagram shown. A diagram of a circuit with a power source on the left. Directl

Sidana [21]

The circuit change when the wire is added will see a short circuit occur and makes bulbs 1 and 2 turn off but keeps bulbs 3 and 4 lit. Option D. This is further explained below.

<h3>How does the circuit change when the wire is added?</h3>

Generally, Electronic circuits consist of a series of interconnected parts that form a closed loop through which electricity may flow.

In conclusion, If two wires are linked together, a short circuit will develop, cutting power to bulbs 1 and 2. But there is no impact on bulbs 3 and 4. There is no problem with bulbs 3 and 4.

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7 0

1 year ago

ou have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 660.0 kg and was trav

Montano1993 [528]

Answer:

vₐ₀ = 29.56 m / s

Explanation:

In this exercise the initial velocity of car A is asked, to solve it we must work in parts

* The first with the conservation of the moment

* the second using energy conservation

let's start with the second part

we must use the relationship between work and kinetic energy

W = ΔK (1)

for this part the mass is

M = mₐ + m_b

the final velocity is zero, the initial velocity is v

friction force work is

W = - fr x

the negative sign e because the friction forces always oppose the movement

we write Newton's second law for the y-axis

N -W = 0

N = W = Mg

friction forces have the expression

fr =μ N

fr = μ M g

we substitute in 1

-μ M g x = 0 - ½ M v²

v² = 2 μ g x

let's calculate

v² = 2 0.750 9.8 6.00

v = ra 88.5

v = 9.39 m / s

Now we can work on the conservation of the moment, for this part we define a system formed by the two cars, so that the forces during the collision are internal and therefore the tsunami is preserved.

Initial instant. Before the crash

p₀ = + mₐ vₐ₀ - m_b v_{bo}

instant fianl. Right after the crash, but the cars are still not moving

p_f = (mₐ + m_b) v

p₀ = p_f

+ mₐ vₐ₀ - m_b v_{bo} = (mₐ + m_b) v

mₐ vₐ₀ = (mₐ + m_b) v + m_b v_{bo}

let's reduce to the SI system

v_{bo} = 64.0 km / h (1000m / 1km) (1h / 3600s) = 17.778 m / s

let's calculate

660 vₐ₀ = (660 +490) 9.39 + 490 17.778

vₐ₀ = 19509.72 / 660

vₐ₀ = 29.56 m / s

we can see that car A goes much faster than vehicle B

5 0

3 years ago

A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. Wha

Gwar [14]

A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. The change in the electric potential energy of the proton-field system when the proton travels to x = 2.50m is -3.40 × 10⁻¹⁶ J (Option B)

<h3 /><h3>How is the change in electric potential energy of the proton-field system calculated?</h3>

Work done on the proton =Negative of the change in the electric potential energy of the proton field
In the given case, W = -qΔV
-W = qΔV
= qEcosθ
Therefore, work done on the proton = -e(8.50× $10^2$ N/C)(2.5m)(1)
= -3.40× $10^-^1^6$ J
Any change in the potential energy indicates the work done by the proton.
Therefore the positive sign shows that the potential energy increases when the proton does the work.
The negative sign shows that the potential energy decreases when the proton does the work.

To learn more about electric potential energy, refer

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3 0

1 year ago