All of the following are physical stimuli for touch except _________

A book is sitting on a table. Jackie is pushing the book with a force of 9 N to the right. If the force of friction is 4 N to th

Free_Kalibri [48]

Answer:

5 N right

Explanation:

Fx = 9N-4N

Fx = 5N

Since we can define the x and y axis. We have x to the right as positive.

8 0

3 years ago

Refer to the first diagram. What is the weight of the person hanging on the end of the seesaw in Newtons?

irina1246 [14]

Due to equilibrium of moments:

1) The weight of the person hanging on the left is 250 N

2) The 400 N person is 3 m from the fulcrum

3) The weight of the board is 200 N

Explanation:

1)

To solve the problem, we use the principle of equilibrium of moments.

In fact, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

The moment of a force is defined as:

$M=Fd$

where

F is the magnitude of the force

d is the perpendicular distance of the force from the fulcrum

In the first diagram:

- The clockwise moment is due to the person on the right is

$M_c = W_2 d_2$

where $W_2 = 500 N$ is the weight of the person and $d_2 = 2 m$ is its distance from the fulcrum

- The anticlockwise moment due to the person hanging on the left is

$M_a = W_1 d_1$

where $W_1$ is his weight and $d_1 = 4 m$ is the distance from the fulcrum

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the weight of the person on the left:

$W_1 d_1 = W_2 d_2\\W_1 = \frac{W_2 d_2}{d_1}=\frac{(500)(2)}{4}=250 N$

2)

Again, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment due to the person on the right is

$M_c = W_2 d_2$

where $W_2 = 400 N$ is the weight of the person and $d_2$ is its distance from the fulcrum

- The anticlockwise moment due to the person on the left is

$M_a = W_1 d_1$

where $W_1 = 300 N$ is his weight and $d_1 = 4 m$ is the distance from the fulcrum.

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the distance of the person on the right:

$W_1 d_1 = W_2 d_2\\d_2 = \frac{W_1 d_1}{W_2}=\frac{(300)(4)}{400}=3 m$

3)

As before, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment around the fulcrum this time is due to the weight of the seesaw:

$M_c = W_2 d_2$

where $W_2$ is the weight of the seesaw and $d_2 = 3 m$ is the distance of its centre of mass from the fulcrum

- The anticlockwise moment due to the person on the left is

$M_a = W_1 d_1$

where $W_1 = 600 N$ is his weight and $d_1 = 1 m$ is the distance from the fulcrum

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the weight of the seesaw:

$W_1 d_1 = W_2 d_2\\W_2 =\frac{W_1 d_1}{d_2}= \frac{(600)(1)}{3}=200 N$

#LearnwithBrainly

8 0

3 years ago

When asked to name all the forces on a marker sitting in equilibrium on a desk, a student responds: “Since the marker on the des

guapka [62]

Answer:

The forces acting on the pen which is still on the table can have two forces acting on them. The forces are gravitational force and the equal and opposite force to the gravitational forces.

The equal and opposite forces that is applied on the pen keeps the pen still on the table.

So, the statement that no force is applied on the pen which is kept still on the table is wrong as two forces are applied on the pen.

As both the forces are equal and opposite so it is cancelled and is still.

6 0

3 years ago

What must the charge (sign and magnitude) of a particle of mass 1.41 gg be for it to remain stationary when placed in a downward

Yuri [45]

Answer:

$q = 2.067 \times 10^{-5}\ C$

Explanation:

Given,

mass = 1.41 g = 0.00141 Kg

Electric field,E = 670 N/C.

We know,

Force in charge due to Electric field.

F = E q

And also we know

F = m g

Equating both the equation of motion

m g = E q

$q =\dfrac{mg}{E}$

$q =\dfrac{0.00141 \times 9.81}{670}$

$q = 2.067 \times 10^{-5}\ C$

Charge of the particle is equal to $q = 2.067 \times 10^{-5}\ C$

6 0

4 years ago

A quelle distance le soletl se trouve-t-il de la terre?

Mnenie [13.5K]

The explanation for the following answer is explained below.

Explanation:

The sun is at an average distance of about 93,000,000 miles(150 million kilometers) away from the earth.It is so far away that light from the Sun,travelling at a speed of 186,000 miles (300,000 kilometers) per second, takes about 8 minutes to reach the earth.Earth does not travel around the Sun in a perfect circle.Instead its orbit is elliptical,like a stretched circle,with the sun just off the center of the orbit. At its closest,the Sun is 91.4 million miles (147.1 million kilometers ) away us.At its farthest ,the Sun is 94.5 million miles (152.1 million km) away.The Earth is closest to the Sun during winter in the northern hemisphere

5 0

3 years ago

All of the following are physical stimuli for touch except __________ energy.