The momentum of an object is proportional to its weight and speed.<br> a. true<br> b. false

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

Two technicians are discussing electromagnetic induction. Technician A says that the induced voltage can be increased if the spe

Inga [223]

Answer:Both

Explanation:

There are three ways to increase the induced voltage in electromagnetic induction:

1) increase the speed at which the conductor moves through the magnetic field. This means that the lines of flux are cut more quickly and more emf is induced.

2) use stronger magnets which provides a stronger magnetic field and more densely packed lines of flux.

3) use a coil of multiple loops.

Hence both technicians were correct.

3 0

3 years ago

Hello i need help with this!!

mr Goodwill [35]

Answer:

Yes

Explanation:

If lamp A burnt out there would still be a wire above it that connects lamp B and C to the power source

3 0

2 years ago

Hydrogen is the second most abundant gas in the atmosphere? True or false?

Alenkinab [10]

False as oxygen is the second most abundant and nitrogen is the most abundant at 78%.

7 0

3 years ago

An object of mass m is hung from a spring and set into oscillation. The period of the oscillation is measured and recorded as T.

sergejj [24]

Answer: $\sqrt{2}T$

Explanation:

Given

object of mass m is suspended from spring and set in oscillation with time Period T

We know Time period of a mass in oscillation is given by

$T=2\pi \sqrt{\frac{m}{k}}$

where k=spring constant

When mass m is replaced by a mass of 2 m time period is given by

$T'=2\pi \sqrt{\frac{2m}{k}}$

$T'=\sqrt{2}\times 2\pi \sqrt{\frac{m}{k}}$

$T'=\sqrt{2}T$

i.e. New time period becomes $\sqrt{2}$ times of previous one

7 0

3 years ago

The momentum of an object is proportional to its weight and speed. a. true b. false