Which describes the motion of the media for a surface wave?

If we add 50 Joules of thermal energy to a heat engine, and that heat engine does 30 Joules of work, how much thermal energy is

Natalka [10]

Answer:

The correct answer should be

A. 20 Joules

Explanation:

I'm taking the K12 Unit Test: Energy - Part 1 right now

7 0

2 years ago

There is a seasaw that's holding two men. The seesaw has a length of 18m that can pivot from a point at its center. Man 1 has a

Alex

Answer:

Distance= 2.3864m

Explanation:

So that the balance is in equilibrium parallel to the floor, we must match the moment each man makes with respect to the pivot point.

In many cases the point of application of force does not coincide with the point of application in the body. In this case the force acts on the object and its structure at a certain distance, by means of an element that transfers that action of this force to the object.

This combination of force applied by the distance to the point of the structure where it is applied is called the moment of force F with respect to the point. The moment will attempt a rotation shift or rotation of the object. The distance from the force to the point of application is called the arm.

Mathematically it is calculated by expression:

M= F×d

The moment caused by the first man is:

M1= 75kg × (9.81m/s²) × 1.75m= 1287.5625 N×m

The moment caused by the second man must be equal to that caused by the first by which:

M2= 1287.5625 N×m= 55kg × (9.81m/s²) × distance ⇒

⇒distance= (1287.5625 N×m)/( (55kg × (9.81m/s²) )= 2.3864m

At this distance from the pivot point, the second should sit down so that the balance is balanced parallel to the ground.

3 0

3 years ago

Which phrase describes speed?

Anna11 [10]

Answer:

the answer is B

Explanation:

speed is the rate at which the distance covered changes or the distance divided by the time taken.

scalar is always positive.

6 0

3 years ago

Read 2 more answers

A 25kg chair initially at rest on a horizontal floor requires a 165 N horizontal force to set it in motion. Once the chair is in

SCORPION-xisa [38]

Answer:

$\mu_k=0.51$

Explanation:

Given that

Mass , m = 25 kg

We know that when body is in rest condition then static friction force act on the body and when body is in motion the kinetic friction force act on the body .That is why these two forces are given as follows

Static friction force ,fs= 165 N

Kinetic friction force ,fk = 127 N

If the body is moving with constant velocity ,it means that acceleration of that body is zero and all the forces are balanced.

Lets take coefficient of kinetic friction = μk

The kinetic friction is given as follows

fk = μk m g

Now by putting the values

127 = μk x 25 x 9.81

$\mu_k=\dfrac{127}{25\times 9.81}$

$\mu_k=0.51$

Therefore the value of coefficient of kinetic friction will be 0.51

4 0

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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4 0

2 years ago