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vladimir2022 [97]
3 years ago
7

How does an inclined plane affect the effort needed to move a load vertically?

Physics
1 answer:
lys-0071 [83]3 years ago
4 0

Answer:

Decreases

Explanation:

"Effort" usually refers to the applied force.  An inclined plane decreases the force required while increasing the distance that the force is exerted over.  So even though there's less force needed, the amount of work stays the same.

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Gamma rays are used to _____.
Ipatiy [6.2K]
<span>d. check pipes for cracks or rust

The answer above is correct.

Gamma rays are primarily used in the medical field to kill cancer cells and examine the body's internal structures.
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How long does it take for an 8 kg pumpkin to hit the ground if dropped from a height of 55 m.
Sati [7]

Answer:

t = 3.35 s

Explanation:

It is given that,

Mass of a pumpkin, m = 8 kg

It is dropped from a height of 55 m

We need to find the time taken by it to hit the ground.

Initial velocity of the pumpkin, u = 0

Using second equation of motion to find it as follows :

h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2(55)}{9.8}} \\\\t=3.35\ s

So, it will take 3.35 seconds to hit the ground.

7 0
3 years ago
Isaac walks 4 blocks north. Then he turns around and walks 1 block south.
kramer

Answer:

Isaac walked a distance of 5 blocks, and his displacement was 3 blocks north.

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Distance is what he covered from the beginning, while displacement was what he covered in a specific direction

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3 years ago
Atoms of the same element that differ only in the number of neutrons are known as.
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3 years ago
a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while th
anzhelika [568]

The maximum mass of a load that can be lifted by the jack and the distance covered are:

m = 160.2 Kg

h = 25 cm

Given that a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.

The parameters given are

F_{1} = 250

A_{1} = Area of the small piston = πr^{2}

A_{1} = 22/7 x 0.4^{2}

A_{1} = 0.5 m^{2}

F_{2} = ?

A_{2} = Area of the large piston = πr^{2}

A_{2} = π x 1

A_{2} = 3.14 m^{2}

To calculate the force on the large piston, we will use the below formula

F_{1}/ A_{1} = F_{2} / A_{2}

Substitute all the parameters into the equation

250/0.5 =  F_{2}/3.14

F_{2} = 1570 N

To calculate the maximum mass of a load that can be lifted by the jack, let us apply Newton second law

F = mg

1570 = 9.8m

m = 1570/9.8

m = 160.2 Kg

.(take g=9.81ms^-2)​

If the applied force moves through a distance of 25cm, the distance through which the load is lifted will be

F_{1}/ 0.25A_{1} = F_{2} / A_{2}h

250/0.125 = 1570/3.14h

make h the subject of the formula

6280h = 1570

h = 1570/6280

h = 0.25 m

Therefore, the distance through which the load is lifted is 25 cm

Learn more here: brainly.com/question/13596980

5 0
2 years ago
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