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trasher [3.6K]
3 years ago
10

The image shows the positions of a car on a roller coaster track. Arrange the cars in order based on their gravitational potenti

al energy. Begin with the lowest potential energy and end with the highest.
A B C D E

Physics
2 answers:
RideAnS [48]3 years ago
5 0

Correct order, from lowest potential energy to highest potential energy:

E - C - D - B - A

Explanation:

The gravitational potential energy of the car is given by:

U=mgh

where

m is the car's mass

g is the gravitational acceleration

h is the height of the car relative to the ground

In the formula, we see that m and g are constant, so the potential energy of the car depends only on its height above the ground, h. The higher the car from the ground, the larger its potential energy. Therefore, the position with least potential energy will be E, since the height is the minimum. Then, C will have more potential energy, because the car is at higher position, and so on: the position with greatest potential energy is A, because the height of the car is maximum.

larisa [96]3 years ago
3 0

Answer:

E,C,D,B,A

Explanation:

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A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w
amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁  + m₂u₂ = v(m₁  +  m₂)

10 x 3   + 25 x 0   = v(10  +  25)

30  = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\

the final kinetic energy of the package;

K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J

the fraction of the initial energy lost;

= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71

7 0
3 years ago
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sergeinik [125]

Answer:

1. geothermal

2. nuclear

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4 0
3 years ago
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A particle moves in a circular path of radius 0.10 m with a constant angular speed of 5 rev/s. The acceleration of the particle
Darya [45]

Answer:

2.5 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s²

For circular motion, the expression for acceleration is given as,

a = ω²r ................ Equation 1

Where a = acceleration of the particle, ω = angular speed of the particle, r = radius of the circular path.

Given: ω = 5 rev/s = 31.42 rad/s, r = 0.10 m.

Substitute into equation 1

a = 5²(0.10)

a = 25(0.10)

a = 2.5 m/s²

Hence the acceleration of the particle = 2.5 m/s²

Hence, none of the option  is correct

7 0
4 years ago
In the image, not to scale, which phase of the moon would you observe from earth?​
timofeeve [1]

Answer:

The "new" moon phase is being shown because the moon shown is directly between the sun and earth and is not visible from the earth because of the brightess of the sun.

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3 years ago
a tennis ball is hit upward with a tennis racket an initial velocity of 12 m/s. What will the ball's speed be when it returns to
zaharov [31]

Answer: 12 m/s

Explanation:

If we ignore air resistance, gravity alone will reduce the upward velocity to zero at the top of the flight. As gravity is a conservative force, it will return exactly the same amount of energy to the tennis ball when it returns to the original elevation.

7 0
3 years ago
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