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2 years ago

49. In what direction does the force of friction act?

Physics

1 answer:

liubo4ka [24]2 years ago

6 0

Frictional force offers the resistance to the applied force opposing its motion. Thus, it always acts in the direction opposite to that of the object in motion. If force is applied to the left, then friction acts in the right.

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A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i

nydimaria [60]

1. -23.2 J

The gravitational potential energy of the ball is given by

$U=mgh$

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J$

2. 41.1 J

Again, the gravitational potential energy of the ball is given by

$U=mgh$

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

$U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J$

3. 0 J

As before, the gravitational potential energy of the ball is given by

$U=mgh$

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(0 m)=0 J$

4 0

3 years ago

A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld

Triss [41]

The magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

Speed of the proton = 5.02 × 10 ⁶ m /a

Angel of between the velocity and the magnetic force = 60 °

The magnitude of magnetic field B = 0.180 T

The magnitude of the magnetic force on the proton is,

$F = q(v \times B)$

$F = qvB \: sin \: θ$

$F = 1.6 \times 10 ^{ - 19} \times 5.02 \times 10 ^{6} \times 0.180 \times \: sin \: 60°$

$= 1.25 \times 10 ^{ - 13} \: N$

Therefore, the magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

To know more about magnetic force, refer to the below link:

brainly.com/question/23096032

#SPJ4

4 0

1 year ago

How long does it take a glass of salt water to freeze than a glass of fresh water

max2010maxim [7]

A glass of salt water will take a slightly longer time & slightly lower temperature (28 F as compared to 32 F for fresh water) to freeze than a glass of fresh water.

Hope this helps!

6 0

3 years ago

You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh

lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

$F=\frac{kQq}{r^2}$

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

$\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}$

And the force on the charge due to the charge on the north is,

$\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}$

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of $45^\circ$ from the north or from the north direction.