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faltersainse [42]
3 years ago
9

25-2y plzzz help I'm stuck on this

Physics
2 answers:
Likurg_2 [28]3 years ago
8 0
There's nothing here to factor, distribute, simplify, or anything else.

You can't do anything with it, except figure out what number it
amounts to ... and you can't do that until you know what ' y ' is.
romanna [79]3 years ago
7 0
I think you cant subtract this because you have no like terms. So the answer is the question.
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Mrs. Stark weighs 70kg. How much Potential Energy does she have if she stands on the roof of an 80m tall building?
sesenic [268]

70kg

1 earth gravity

80 M of height

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54917.2 Joules of energy

Mrs. Stark has 54,917.2 Joules of energy.

3 0
3 years ago
g Drop the object again and carefully observe its motion after it hits the ground (it should bounce). (Consider only the first b
Anestetic [448]

Answer:

a) quantity to be measured is the height to which the body rises

b) weighing the body , rule or fixed tape measure

c)   Em₁ = m g h

d) deformation of the body or it is transformed into heat during the crash

Explanation:

In this exercise of falling and rebounding a body, we must know the speed of the body when it reaches the ground, which can be calculated using the conservation of energy, since the height where it was released is known.

a) What quantities must you know to calculate the energy after the bounce?

The quantity to be measured is the height to which the body rises, we assume negligible air resistance.

So let's use the conservation of energy

starting point. Soil

          Em₀ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₀ = Em_f

         Em₀ = m g h₀

b) to have the measurements, we begin by weighing the body and calculating its mass, the height was measured with a rule or fixed tape measure and seeing how far the body rises.

c) We use conservation of energy

starting point. Soil

          Em₁ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₁ = Em_f

         Em₁ = m g h

d) to determine if the energy is conserved, the arrival energy and the output energy must be compared.

There are two possibilities.

* that have been equal therefore energy is conserved

* that have been different (most likely) therefore the energy of the rebound is less than the initial energy, it cannot be stored in the possible deformation of the body or it is transformed into heat during the crash

7 0
3 years ago
WAVES AND SOUND
lawyer [7]

Answer:

i think it is B

Explanation:

6 0
3 years ago
When a bat hits a baseball, what is transferred from the bat to the ball?​
GenaCL600 [577]

Answer:

kinetic energy

Explanation:

4 0
3 years ago
Suppose that a comet that was seen in 563 A.D. by Chinese astronomers was spotted again in year 1951. Assume the time between ob
Mars2501 [29]

Answer:

a=2.77*10^{13}m

R_a=5.49*10^{13}m

Explanation:

The period of the comet is the time it takes to do a complete orbit:

T=1951-(-563)=2514 years

writen in seconds:

2514years*\frac{3,154*10^7s}{1year}=7.93 *10^{10}s

Since the eccentricity is greater than 0 but lower than 1 you can know that the trajectory is an ellipse.

Therefore, if the mass of the sun is aprox. 1.99e30 kg, and you assume it to be much larger than the mass of the comet, you can use Kepler's law of periods to calculate the semimajor axis:

T^2=\frac{4\pi^2}{Gm_{sun}}a^3\\ a=\sqrt[3]{\frac{Gm_{sun}T^2}{4\pi^2} } \\a=1.50*10^{6}m

Then, using the law of orbits, you can calculate the greatest distance from the sun, which is called aphelion:

R_a=a(1+e)\\R_a=2.77*10^{13}(1.986)\\R_a=5.49*10^{13}m

8 0
3 years ago
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