What is transferred by a radio wave?

A tow truck drags a stalled car along a road. The chain makes an angle of 30° with the road and the tension in the chain is 1400

AURORKA [14]

Answer:

The work is done by the truck pulling the car 1 km is 1,212,436 J

Explanation:

Work is defined as the force that is applied on a body to move it from one point to another. When a force is applied, an energy transfer occurs. Then it can be said that work is energy in motion.

So work is one of the forms of energy transmission between bodies. To perform a job, you must exert a force on a body and it moves.

In the International System of Units, work is measured in Joule. A Joule is the work that a constant force of 1 Newton does on a body that moves 1 meter in the same direction and direction as the force. Then, Joule is equivalent to Newton per meter.

The work is equal to the product of the force by the distance and by the cosine of the angle that exists between the direction of the force and the direction that travels the point or the object that moves:

Work= Force*Distante* cos (θ)

In this case:

Force= 1,400 N= 1,400 $kg*\frac{m}{s^{2} }$

Distance= 1 km= 1,000 m

θ= Angle that exists between the direction of the force and the direction= 30°

Replacing:

Work= 1,400 N* 1,000 m* cos (30°)

Work= 1,212,435. 565 Joule≅ 1,212,436 J

<u><em> The work is done by the truck pulling the car 1 km is 1,212,436 J</em></u>

7 0

3 years ago

A 1250 kg car is pulling a 325 kb trailer. The car and trailer have an acceleration of 2,15m/sec2 together. Determine the net fo

xenn [34]

To solve this problem we will apply the definition of Newton's second law, which says that force is equivalent to body mass by acceleration. In this case the mass of the trailer is 325Kg and its acceleration is $2.15m / s ^ 2$ , so we will proceed to replace and multiply these values to find the net force on this object.

$F= ma$

$F= (325 kg)(2.15 m/s^2)$

F= 698.75N

Under the reference system in which the direction of travel is the positive direction, the direction of the force will be positive.

7 0

3 years ago

Jose’s lab instructor gives him a solution of sodium phosphate that is buffered to a pH of 4. Because of an error that he made w

Kazeer [188]

I am thinking that maybe the problem is not with the calibration. It might be that the buffered solution is already expired since at this point the solution is already not stable and will give a different pH reading than what is expected.

7 0

3 years ago

Lexy throws a dart with an initial velocity of 25 m/s at an angle of 60° relative to the ground. what is the approximate vertica

Serjik [45]

An initial velocity is:
v o = 25 m/s
The vertical component of the initial velocity:
v o y = v o * sin 60° =
= v o * √3 / 2 = 25 m/s * √3 / 2 = 21.65 m/s
Answer:
The approximate vertical component of the initial velocity is 21.65 m/s.

7 0

3 years ago

Read 2 more answers

1) A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/

mojhsa [17]

Answer: a) 49.560 and 21.13 b) i) 50 N, ii) 196 N iii) 196 N iv) 47.685 N

c) i) 594.72 ii) 0 iii) 0 iv) 0

d) 594.72

Explanation: question a)

The force is inclined at an angle of 25° to the horizontal

The horizontal component of force = 50 cos 25° = 49.560 N

The vertical component of force = 50 sin 30°= 21.130N

Question b)

i) according to the question applied force is 50 N

ii) if g = 9.8m/s², w=mg where m = mass of object = 20kg hence weight = 20* 9.8 = 196 N

iii) the normal force is the force the floor exerts on the body as a result of the weight of the object.

Normal reaction R = W = mg, we already deduced that w = mg, hence R = 196 N.

iv) according to newton's laws of motion

F - Fr = ma

F = applied force = horizontal component of force = 49.560 N.

We need to get the acceleration (a) by using Newton laws of motion before we can be able to compute the frictional force..

The body started from rest hence initial velocity u = 0

Final velocity v = 1.5m/s distance covered (s) = 12m

v ² = u² + 2as

But u = 0

v² = 2as

1.5² = 2(a) * 12

2.25 = 24a

a = 2.25/24 = 0.09735m/s²

From F - Fr = ma

49.560 - Fr = 20 * 0.09735

49.560 - Fr = 1.875

Fr = 49.560 - 1.875

Fr = 47.685 N

Question c)

i) The applied force = 49.560 N, distance covered = 12m

Work done = force * distance

Work done = 49.560 * 12

Work done = 594.72 J

ii) the weight of the object does not make the object move a distance, hence work done = 0 ( since distance covered is 0)

iii) the normal force is the same thing as the weight and they did not cover any distance hence work done is zero.

iv) the frictional force does not cover any distance, hence work done is zero.

Question d)

The total work done = work done by applied force + work done by weight + work done by normal reaction + work done by frictional force.

Total work done = 594.72 + 0 + 0 + 0 = 594.72 J

8 0

3 years ago