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3 years ago

9

Lien uses a spring scale to pull a block toward the right across the lab table. The scale reads 8 N. Which force should Lien con

clude is also 8 N?

1 answer:

lana66690 [7]3 years ago

6 0

Answer:

<h2>Frictional force</h2>

Explanation:

For the block placed on the table, there are several force acting on the body along the horizontal and vertical axis. All this forces tends to keep the body in a state of equilibrium.The forces acting along the horizontal are the moving force (Fm) and the frictional force (Ff).

Frictional force are forces that acts opposite to the force that causes the body to move (moving force).

If Lien uses a spring scale to pull a block toward the right across the lab table and the scale reads 8 N, this means that the force that causes the body to move is the 8N force (moving force).

Taking the sum of force along the horizontal;

$\sum fx = ma_x$

Since the body is static, max = 0

$\sum fx = 0\\fm+(-Ff)= 0$

Note that the frictional force acts is the force of opposition acting in the negative x direction.

$fm = 0+Ff\\fm = Ff$

Since Fm = 8N, Ff will also be equal to 8N.

Based on the above proof, Lien can also conclude that 8 N is a frictional force

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It is possible for two electric field lines to cross each other.
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8 0

3 years ago

antsy andy tries to push his 6kg desk away from him and finds that it takes him 12 N of force to start the desk from rest and 8

Rama09 [41]

Answer:

static coefficient = 0,203 & kinetic coefficient = 0,14

Explanation:

There are two (2) conditions, when the desk is about to move and when the desk is moving. In the attachements you can see the two free body diagram for each condition.

In the first condition, there is no movement and the force is 12 N, in the image we can see the total forces are equal to 0 and by the definition of the friction force we can get the static friction coefficient.

In the second condition there is movement in the direction of the force which is equal to 8 N, again by the definition of the friction force we can get the kinetic friction coefficient. Since the desk is moving with constant velocity there is not acceleration.

4 0

3 years ago

A planetâs moon revolves around the planet with a period of 39 Earth days in an approximately circular orbit of radius of 4.8Ã10

Alchen [17]

Answer:

v = 895 m/s

Explanation:

Time period is given as 39 Earth Days

$T = 39 days \times 24 hr \times 3600 s$

$T = 3369600 s$

now the radius of the orbit is given as

$r = 4.8 \times 10^8 m$

so the total path length is given as

$L = 2 \pi r$

$L = 2\pi (4.8 \times 10^8)$

$L = 3.015 \times 10^9$

now the speed will be given as

$v = \frac{L}{T}$

$v = \frac{3.015 \times 10^9}{3369600}$

$v = 895 m/s$

7 0

3 years ago

A playground merry-go-round has a radius of 4.6 m and a moment of inertia of 200 kg-m2 and turns with negligible friction about

tankabanditka [31]

Answer:

8050 J

Explanation:

Given:

r = 4.6 m

I = 200 kg m²

F = 26.0 N

t = 15.0 s

First, find the angular acceleration.

∑τ = Iα

Fr = Iα

α = Fr / I

α = (26.0 N) (4.6 m) / (200 kg m²)

α = 0.598 rad/s²

Now you can find the final angular velocity, then use that to find the rotational energy:

ω = αt

ω = (0.598 rad/s²) (15.0 s)

ω = 8.97 rad/s

W = ½ I ω²

W = ½ (200 kg m²) (8.97 rad/s)²

W = 8050 J

Or you can find the angular displacement and find the work done that way:

θ = θ₀ + ω₀ t + ½ αt²

θ = ½ (0.598 rad/s²) (15.0 s)²

θ = 67.3 rad

W = τθ

W = Frθ

W = (26.0 N) (4.6 m) (67.3 rad)

W = 8050 J

6 0

2 years ago

A force, F1, of magnitude 2.0 N and directed due east is exerted on an object. A second force exerted on the object is F2 = 2.0

IgorLugansk [536]

Answer:

Magnitude of the force is

$F_3 = 2.83$

direction of the force is given as

$\theta = 45 degree$ West of South

Explanation:

As we know that force is a vector quantity and in order to find the resultant of two or more forces we need to add them vectorialy

So here we have

$\vec F_1 + \vec F_2 + \vec F_3 = 0$

here we know that first force is of magnitude 2 N towards east

$\vec F_1 = 2 \hat i N$

second force is also of 2.0 N due North

$\vec F_2 = 2 \hat j$

now from above equation

$2\hat i + 2\hat j + \vec F_3 = 0$

$\vec F_3 = -2\hat i - 2\hat j$

so magnitude of the force is given as

$F_3 = \sqrt{2^2 + 2^2}$

$F_3 = 2.83$

direction of the force is given as

$\theta = tan^{-1}\frac{F_y}{F_x}$

$\theta = tan^{-1}\frac{-2}{-2}$

$\theta = 45 degree$ West of South

3 0

2 years ago

Read 2 more answers