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pickupchik [31]
3 years ago
9

tas watches as his uncle changes a flat tire on a car. his uncle raises the car using a machine called a jack. each time his unc

le pushes down on the jack handle, the car rises up. tas sketches the jack and labels it using the symbols f, to represent force, and d, to represent distance. he uses large and small letters to compare the sizes of the forces and distances. which set of labels is correct for the side of the jack that tas’s uncle pushes on? a large f and a small d a small f and a small d a large f and a large d a small f and a large d
Physics
2 answers:
Alenkinab [10]3 years ago
8 0
Small f and large L.

People needs help of machines to increase their force.

People cannot lift a car without a machine.

Using the leverage  or hydraulic principles the machines increase your force.

If you use a large leverage you execute a large movement with little force and as result the ohter side will move small distances with a greater force.

I hope this help. Please, let me know.
kakasveta [241]3 years ago
7 0

The answer is

-Small f and large D.

The explanation:

-when The car jack is an example of a machine, which is defined as anything that a person can use to make the exertion of force easier.

-So with the small force he exerts on the jack, the distance that the car is lifted up increases .

and People needs help of machines to increase their force.  People cannot lift a car without a machine. Using the leverage  or hydraulic principles the machines increase your force.

If you use a large leverage you execute a large movement with little force and as result the other side will move small distances with a greater force.

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3 years ago
A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision
Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is a=2,800m/s^{2}

Explanation:

<u>Known Data</u>

We will asume initial speed has a negative direction, v_{i}=-30m/s, final speed has a positive direction, v_{f}=26m/s, \Delta t=20ms=0.020s and mass m_{b}.

<u>Initial momentum</u>

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<u>final momentum</u>

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<u>Impulse</u>

I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s

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F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}

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