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SCORPION-xisa [38]
3 years ago
9

What effect does dropping the sandbag out of the cart at the equilibrium position have on the amplitude of your oscillation? Vie

w Available Hint(s) What effect does dropping the sandbag out of the cart at the equilibrium position have on the amplitude of your oscillation? It increases the amplitude. It decreases the amplitude. It has no effect on the amplitude.
Physics
2 answers:
ozzi3 years ago
6 0

Answer:

It has no effect on the amplitude.

Explanation:

When the sandbag is dropped, then the cart is at its maximum speed. Dropping the sand bag does not affect the speed instantly, this is because the energy remains within the system after the bag as been dropped. The cart will always return to its equilibrium point with the same amount of kinetic energy, as a result the same maximum speed is maintained.

amm18123 years ago
5 0

Answer:

It decreases the amplitude.

Explanation:

If it is located at the end, its kinetic energy will be equal to zero, so the system as a whole would only have potential energy, which is defined as:

Ep = (1/2) * k * A²

Where A is the amplitude of the oscillation.

According to the energy principle, said energy must remain conserved, therefore, the total energy is equal to:

Etot = (1/2) * k * A²

Since the system is in equilibrium, according to the expression of conservation of energy, the energy in the position located at the end would be equal to the energy in equilibrium. As the sandbag falls, its kinetic energy decreases, which is why the total energy of the system also decreases.

According to the total energy expression, it is directly proportional to the square of the oscillation amplitude. If the total energy is decreased, the amplitude of the oscillation would also decrease.

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Twenty grams of a solid at 70°C is place in 100 grams of a fluid at 20°C. Thermal equilibrium is reached at 30°C.
zaharov [31]

Answer:

c. is more than that of the fluid.

Explanation:

This problem is based on the conservation of energy and the concept of thermal equilibrium

heat= m s \Delta T


m= mass

s= specific heat

\DeltaT=change in temperature

let s1= specific heat of solid and s2= specific heat of liquid

then

Heat lost by solid= 20(s_1)(70-30)=800s_1


Heat gained by fluid=100(s_2)(30-20)=1000s_2


Now heat gained = heat lost

therefore,

1000 S_2=800 S_1

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8 0
3 years ago
A car travels due east with a speed of 52.0 km/h. Raindrops are falling at a constant speed vertically with respect to the Earth
serg [7]

Answer:

a) v = 6.43 m/s

b) v = 15.8 m/s

Explanation:

Speed of car = 56 km/h

56 km/h = 14.4 m/s

Angle rain makes on the glass to the vertical = 66°

Thus knowing that the opposite side of the angle is the distance moved by the car, and the adjacent side is the distance traveled by the rain in the same time

both of which are directly proportional to their velocities

Then

tan(66°) = 14.44m/s ÷ x

or x = 14.44/tan(66°)

Which is the vertical raindrop velocity of the relative to earth

v = 6.43 m/s vertically towards earth

For v relative to the car is we have vector sum of both velocities

v = √(14.44^2 + 6.43^2) = 15.8 m/s which is the velocity relative to car

= 15.8 m/s

6 0
3 years ago
A race car moving along a circular track has a centripetal acceleration of 15.4 m/s? If the car has
Helen [10]

Answer:

r = 58.44 [m]

Explanation:

To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.

a = v²/r

where:

a = centripetal acceleration = 15.4 [m/s²]

v = tangential speed = 30 [m/s]

r = radius or distance [m]

r = v²/a

r = 30²/15.4

r = 58.44 [m]

3 0
3 years ago
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