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GuDViN [60]
3 years ago
12

1. Looking at the planet vs. eccentricity table, which two planets have the greatest eccentricity?

Physics
1 answer:
AVprozaik [17]3 years ago
6 0

Answer:

Pluto & Mercury

Explanation:

Pluto's eccentricity is 0.248

Mercury's eccentricity is 0.206

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An 8.2 kg object accelerates at 8.0 m/s^2. What is the acceleration?
ZanzabumX [31]

Answer:

8.0m/s²

Explanation:

meters per second squared is a unit of acceleration so 8.0m/s² is the answer

6 0
2 years ago
The force of friction always opposes the _______.
Ierofanga [76]

Answer:

motion

Explanation:

uh look it up?

8 0
3 years ago
Read 2 more answers
The amount of energy that must be absorbed or lost to raise or lower the temperature of 1 g of liquid water by 1°c _____.
Fiesta28 [93]

Answer:

4.2 J

Explanation:

Specific heat capacity: This is defined as the amount of a heat required to rise a unit mass of a substance through a temperature of 1 K

From specific heat capacity,

Q = cmΔt.............................. Equation 1

Where Q = amount of energy absorbed or lost, c = specific heat capacity of water, m = mass of water, Δt = Temperature rise.

Given: m = 1 g = 0.001 kg, Δt = 1 °C

Constant : c = 4200 J/kg.°C

Substitute into equation 1

Q = 0.001×4200(1)

Q = 4.2 J.

Hence the energy absorbed or lost = 4.2 J

6 0
3 years ago
When is your kinetic energy the least when swinging on a park swing?
Afina-wow [57]

Answer:

An active pendulum has the most kinetic energy at the lowest point of its swing when the weight is moving fastest.

Explanation:

SO YOU HAVE THE LEAST KINETIC ENERGY AT THE HIGHEST POINT OF THE SWING WHEN IT'S NOT ACTIVE

7 0
3 years ago
Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the s
Nadusha1986 [10]

Answer:

Explanation:

Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .

velocity of approach = 1.5 - 0 = 1.5

velocity of separation = v₁ + v₂

coefficient of restitution = velocity of separation / velocity of approach

.8 = v₁ + v₂ / 1.5

v₁ + v₂ = 1.2

applying law of conservation of momentum

m x 1.5 + 0 = mv₂ - mv₁

1.5 = v₂ - v₁

adding two equation

2 v ₂= 2.7

v₂ = 1.35 m /s

v₁ = - .15 m / s

During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.

For second collision ,

coefficient of restitution = velocity of separation / velocity of approach

.5 = v₃ + v₄ / 1.35

v₃ + v₄ = .675

applying law of conservation of momentum

m x 1.35 + 0 = mv₄ - mv₃

1.35 = v₄ - v₃

adding two equation

2 v ₄= 2.025

v₄ = 1.0125 m /s

v₃ = - 0 .3375  m / s

3 0
4 years ago
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