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san4es73 [151]
3 years ago
13

According to the work-energy theorem, if work is done on an object, its potential and/or kinetic energy changes. Consider a car

that accelerates from rest on a flat road. What force did the work that increased the car’s kinetic energy?
1. the force of the car engine

2. air resistance

3. the friction between the road and the tires

4. gravity
Physics
1 answer:
Leni [432]3 years ago
6 0

Answer:

The force of the car engine.

Explanation:

The work- energy theorem states that the work done on an object is equal to the change in its kinetic energy. Its expression is given by :

W=\dfrac{1}{2}m(v^2-u^2)

Also, W = F.d

Fd=\dfrac{1}{2}m(v^2-u^2)

Where

F is the force applied by the engine of car

d is the displacement

m is the mass of an object

u is the initial speed

v is the final speed

So, the force of the car engine increased the car’s kinetic energy. Hence, this is the required solution.

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If a 5.0 kg box is pulled simultaneously by a 10.0 N force and a 5.0 N force, then its acceleration must be Group of answer choi
vivado [14]

Answer:

We cannot tell from the information given

Explanation:

Given;

mass of the box, m = 5 kg

first force, F₁ = 10 N

second force, F₂ = 5 N

(I) Assuming the two forces are acting horizontally in opposite direction, the resultant force on the box is calculated as;

∑Fx = 10 N - 5 N

      = 5 N

Apply Newton's second law of motion;

∑Fx = ma

a = ∑Fx/m

a = 5 / 5

a = 1 m/s² in the direction of the 10 N force.

(II) Also, if the two forces are acting in the same direction, the resultant force is calculated as;

∑Fx = 10 N + 5 N

∑Fx = 15 N

a = 15 / 5

a = 3 m/s²

Therefore, the information given is not enough to determine the acceleration of the box.

8 0
3 years ago
Why did she use an red infra lamp
mixas84 [53]
To act as the Sun' was accepted but if you put 'sunlight' alone it was not accepted. The examiner wanted you to state that the infra red radiation was needed to warm up the water.
4 0
3 years ago
A force of 1 N is the only horizontal force exerted on a block, and the horizontal acceleration of the block is
SVEN [57.7K]

The mass of the first block will be three times the mass of the second block.

According to Newton's second law of motion, the force acting on a body is directly proportional to the acceleration as shown;

F\ \alpha \ a

F = ma

F is the acting force

m is the mass

a is the acceleration of the body

Given the following parameters

Constant force F =  1N

For the first block with the acceleration of "a"

1 = m₁a

a = m₁/1

m₁ = a .................1

For the second block, acceleration is thrice that of the first. This means;

F = m(3a)

1 = 3ma

m_2=\frac{1}{3a} ..........................2

Divide both equations

\frac{m_1}{m_2} =\frac{a}{(\frac{1}{3a} )}\\\frac{m_1}{m_2} = 3\\m_1 = 3m_2

From the calculation, we can conclude that the mass of the first block will be three times the mass of the second block.

Learn more here: brainly.com/question/19030143

4 0
3 years ago
An object is suspended by a string from the ceiling of an elevator. If the tension in the string is equal to 25 N at an instant
Phantasy [73]

By Newton's second law, the net force on the object is

∑ <em>F</em> = <em>T</em> - <em>mg</em> = - <em>ma</em>

where

• <em>T</em> = 25 N, the tension in the string

• <em>m</em> is the mass of the object

• <em>g</em> = 9.8 m/s², the acceleration due to gravity

• <em>a</em> = 2.0 m/s², the acceleration of the elevator-object system

Solve for <em>m</em> :

25 N - <em>m</em> (9.8 m/s²) = - <em>m</em> (2.0 m/s²)

==>   <em>m</em> = (25 N) / (9.8 m/s² - 2.0 m/s²) ≈ 3.2 kg

4 0
3 years ago
The normal force of a parked car is 15,000 Newtons. The coefficient of static friction between the rubber of the tires and the a
Shtirlitz [24]

Answer:

11250 N

Explanation:

From the question given above, the following data were obtained:

Normal force (R) = 15000 N

Coefficient of static friction (μ) = 0.75

Frictional force (F) =?

Friction and normal force are related by the following equation:

F = μR

Where:

F is the frictional force.

μ is the coefficient of static friction.

R is the normal force.

With the above formula, we can calculate the frictional force acting on the car as follow:

Normal force (R) = 15000 N

Coefficient of static friction (μ) = 0.75

Frictional force (F) =?

F = μR

F = 0.75 × 15000

F = 11250 N

Therefore, the frictional force acting on the car is 11250 N

3 0
3 years ago
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