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3 years ago

13

In a hydraulic device, the surface area of the small piston is 1 cm2 and the surface area of the large piston is 8 cm2. If 10 N

of force is put on the small piston, how much force will be exerted on the large piston.

1 answer:

kotykmax [81]3 years ago

7 0

Hence the force is 80 N

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13 Henry Wadsworth Longfellow, John Greenleaf Whittier, Oliver Wendell Holmes, and James Russel Lowell were known as

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Answer:

The fireside poets also known as the schoolroom or household poets Explanation:

i helped my sis with it and got it correct

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3 years ago

A 20-ton truck collides with a 1500-lb car and causes a lot of damage to the car. During the collision:

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Answer:

B. the force of on the truck due to the collision is exactly equal to the force on the car.

Explanation:

At the time of collision between two objects of unequal or equal masses two forces emerge at the point of collision .They are called action and reaction force . According to third law of Newton , they are equal . So in the instant case of collision between car and truck , force on both of them will be equal.

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3 years ago

Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope

Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

= -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

= 26,754 (0.816)

= 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c. What is the work done by the gravitational force on the sled?

By using the formula:

Gravity work = mgdsinθ

= 97.5 kg * 9.8 m/s² * 28 m * sin 60

= 23,170 J

23,170 J the work, in joules done by the gravitational force on the sled .

D. What is the total work done?

By adding all the values

work done = -1337.3 + 21,835 + 23,170

= 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

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2 years ago

The water from a fire hose follows a path described by y equals 3.0 plus 0.8 x minus 0.40 x squared (units are in meters). If

ivanzaharov [21]

Explanation:

It is given that, the water from a fire hose follows a path described by equation :

$y=3+0.8x-0.4x^2$ ........(1)

The x component of constant velocity, $v_x=5\ m/s$

We need to find the resultant velocity at the point (2,3).

Let $\dfrac{dx}{dt}=v_x$ and $\dfrac{dy}{dt}=v_y$

Differentiating equation (1) wrt t as,

$\dfrac{dy}{dt}=0.8\times \dfrac{dx}{dt}-0.8x\times \dfrac{dx}{dt}$

$v_y=0.8\times v_x-0.8x\times v_x$

$v_y=0.8v_x(1-x)$

When x = 2 and $v_x=5\ m/s$

So,

$v_y=0.8\times 5\times (1-2)$

$v_y=-4\ m/s$

Resultant velocity, $v=\sqrt{v_x^2+v_y^2}$

$v=\sqrt{5^2+(-4)^2}$

v = 6.4 m/s

So, the resultant velocity at point (2,3) is 6.4 m/s. Hence, this is the required solution.

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3 years ago

Which one is the lunar eclipse?

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Answer:

Yes it would be a lunar eclipse here is an image that I found that might help but during a lunar eclipse when the moon is directly behind earth it will be a full moon

Explanation:

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3 years ago