Ask question

Ask question

All categories

3 years ago

7

An empty oﬃce chair is at rest on a ﬂoor. Consider the following forces:. 1. A downward force due to gravity;. 2. An upward forc

e exerted by the ﬂoor; and. 3. A net downward force exerted by the air. Which force(s) act on the oﬃce chair?. 1. None of the forces act on the chair since. it is at rest. 2. All three forces. 3. 1 only. 4. 1 and 2 only. 5. 2 and 3 only.

1 answer:

Ghella [55]3 years ago

7 0

The answer is:

<span> 4. 1 and 2 only.</span>

You might be interested in

Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec

AysviL [449]

The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
$E= \frac{\sigma}{2\epsilon_0}$
where
$\sigma$ is the charge density
$\epsilon_0$ is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
$E=30 N/C$

7 0

3 years ago

1. Define weight

andrezito [222]

1. Weight is the gravitational pull with which the earth attracts the body towards the center of the earth. The S.I unit is Newton (N)

2. The weight of an object is related to its mass with the below equation.

W = mg

Where W = weight, m = mass and g = acceleration due to gravity

3. The mass m = 50 kg and g = 9.8 m/ $s^{2}$

Substitute the parameters in the equation above.

W = 50 x 9.8

W = 490 N

4. An object is accelerating when it speeds up. If the object slows down, it means it is decelerating. The correct answer is option A

5. Newton's second law of motion is:

F = ma

Where F = force applied, m = mass and a = acceleration

Therefore, Newton's second law of motion relates an object's acceleration to its net force acting on it. The correct answer is option C

6. According to Newton's second law of motion which relates an object's acceleration to its mass, doubling the net force acting on an object a doubles its acceleration. Because mass is always constant.

7. Since the weight of an object is related to its mass with the equation.

W = mg, If the mass of an object doubles, its weight will also doubles. Option A is the correct answer.

8. If you know the mass of an object, you can calculate its weight with the formula F = mX when X = 9.8m/ $s^{2}$

9. Force is expressed in unit as Newton (N)

10. The parameters given are :

mass m = 20kg

Force F = 40N

To calculate acceleration, use the formula F = ma

Substitute all the parameters into the equation.

40 = 20a

a = 40/20

a = 2m/ $s^{2}$

The correct option is D

Learn more here : brainly.com/question/18835375

8 0

3 years ago

What is the difference between a free body diagram and a vector diagram?

Murrr4er [49]

Answer:

Free body diagrams are used to describe situations where several forces act on an object. On the other hand Vector diagrams are used to resolve (break down) a single force into two forces acting as right angles to eachother

Explanation:

Hope this helps !

4 0

2 years ago

Read 2 more answers

An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1

goblinko [34]

Answer:

$v"_{1} = v_{1} tan$ Θ

$v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

Θ = $tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )$

Explanation:

Applying the law of conservation of momentum, we have:

Δ $p_{x = 0}$

$p_{x} = p"_{x}$

$m_{1}v_{1} = m_{2}v"_{2} cos$ Θ (Equation 1)

Δ $p_{y} = 0$

$p_{y} = p"_{y}$

$0 = m_{1} v"_{1} - m_{2} v"_{2} sin$ Θ (Equation 2)

From Equation 1:

$v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

From Equation 2:

$m_{2} v"_{2}$ sinΘ = $m_{1} v_{1}$

$v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }$

Replacing Equation 3 in Equation 4:

$v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}$

$v"_{1}=v_{1}\frac{sinΘ}{cosΘ}$

$v"_{1}=v_{1}tan$ Θ (Equation 5)

And we found Θ from the Equation 5:

tanΘ= $\frac{v"_{1}}{v_{1}}$

Θ= $tan^{-1}(\frac{v"_{1}}{v_{1}})$

7 0

3 years ago

A 40.0 -kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 13

Irina18 [472]

The work done by the gravitational force = 0

Given the mass of the box = 40 kg

The box is initially at rest.

Distance moved by the applied force = 5m

Force applied = 130 N

Co-efficient of friction between the box and floor = 0.3

The box is moved only in the horizontal direction by the applied force.

Gravitational force is applied in a direction perpendicular to the applied force. hence it doesn't do any work on the box.

Therefore, the work done by the gravitational force is zero.

Learn more about the gravitational force at brainly.com/question/862529

#SPJ4

3 0

2 years ago