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Zarrin [17]
3 years ago
8

An object of mass m moving at a speed of v1 possesses kinetic energy that is equal to KE1. When the object's speed is doubled, t

he kinetic energy becomes KE2. How can you relate KE2, to KE1?
Answer and I will give you brainiliest

​

Physics
1 answer:
Vera_Pavlovna [14]3 years ago
7 0

Answer:

KE₂ = 4KE₁

Explanation:

KE₁ = ½mv₁²

KE₂ = ½mv₂²

KE₂ = ½m(2v₁)²

KE₂ = 4(½mv₁²)

KE₂ = 4KE₁

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2 decaliters + 800 deciliters = __________ liters?
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Answer is 100 liters
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Describe the Compton Effect​
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2 years ago
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A satellite of mass m is in a circular orbit of radius R2 around a spherical planet of radius R1 made of a material with density
Amiraneli [1.4K]

Answer:

a)      K = 2/3 π G m ρ R₁³ / R₂ ,  b) U = - G m M / r

Explanation:

The law of universal gravitation is

     F = G m M / r²

Part A

Let's use Newton's second law

     F = m a

The acceleration is centripetal

     a = v² / R₂

     

      G m M / R₂² = m v² / R₂

      v² = G M / R₂

They give us the density of the planet

    ρ = M / V

    V = 4/3 π R₁³

    M =   ρ V

    M =   ρ 4/3 π R₁³

    v² = 4/3 π G  ρ R₁³ / R₂

    K = ½ m v²

    K = ½ m (4/3 π G ρ R₁³ / R₂)

    K = 2/3 π G m ρ R₁³ / R₂

Part B

Potential energy and strength are related

     F = - dU / dr

     ∫ dU = - ∫ F. dr

The force was directed towards the center and the vector r outwards therefore there is an angle of 180º between the two cos 180 = -1

    U- U₀ = G m M ∫ dr / r²

    U - U₀ = G m M (- r⁻¹)

We evaluate for

    U - U₀ = -G m M (1 / r_{f} -  1 /r_{i})

They indicate that for ri = ∞     U₀ = 0

    U = - G m M / r

6 0
3 years ago
Which statements did both Aristotle and Ptolemy assume? Check all that apply.
tensa zangetsu [6.8K]

Bodies in space traveled in circles.

The planets revolved around the Earth.

6 0
3 years ago
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Point charges q1 and q2 are separated by a distance of 60 cm along a horizontal axis.
amm1812

Answer:

38 cm from q1(right)

Explanation:

Given, q1 = 3q2 , r = 60cm = 0.6 m

Let that point be situated at a distance of 'x' m from q1.

Electric field must be same from both sides to be in equilibrium(where EF is 0).

=> k q1/x² = k q2/(0.6 - x)²

=> q1(0.6 - x)² = q2(x)²

=> 3q2(0.6 - x)² = q2(x)²

=> 3(0.6 - x)² = x²

=> √3(0.6 - x) = ± x

=> 0.6√3 = x(1 + √3)

=> 1.03/2.73 = x

≈ 0.38 m = 38 cm = x

8 0
3 years ago
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