When a car suddenly stops,the objects in the backseat are thrown forward.this is explained by-

DochEvi [55]

We have that Gravity a force acting down ward against any object and frictional force is a force acting a against movement, Hence

True
True

<h3>Gravity and frictional force</h3>

Generally, Gravity a force acting down ward against any object on earth at an acceleration of about 9.8m/s

and frictional force is a force acting a against the movement based on contact with another surface

Therefore

Gravity is a field force acting downward on the couch:

This is True

Friction from the carpeted floor is a contact force acting against your push.

This is True

For more information on Force visit

brainly.com/question/26115859

6 0

3 years ago

An electron is released from rest at a distance of 0.570 m from a large insulating sheet of charge that has uniform surface char

Lelu [443]

Explanation:

Formula to calculate the electric field of the sheet is as follows.

E = $\frac{\sigma}{2 \epsilon_{o}}$

And, expression for magnitude of force exerted on the electron is as follows.

F = Eq

So, work done by the force on electron is as follows.

W = Fs

where, s = distance of electron from its initial position

= (0.570 - 0.06) m

= 0.51 m

First, we will calculate the electric field as follows.

E = $\frac{\sigma}{2 \epsilon_{o}}$

= $\frac{4.60 \times 10^{-12}C/m^{2}}{2 \times 8.854 \times 10^{-12}C^{2}/N m^{2}}$

= 0.259 N/C

Now, force will be calculated as follows.

F = Eq

= $0.259 N/C \times 1.6 \times 10^{-19} C$

= $0.415 \times 10^{-19} N$

Now, work done will be as follows.

W = Fs

= $0.415 \times 10^{-19} N \times 0.51 m$

= $2.12 \times 10^{-20} J$

Thus, we can conclude that work done on the electron by the electric field of the sheet is $2.12 \times 10^{-20} J$ .

3 0

4 years ago

Read 2 more answers

A superball with a mass m = 61.6 g is dropped from a height h = [02]____________________ m. It hits the floor and then rebounds

aleksklad [387]

<em>There is not enough data to solve the problem, but I'm assuming the initial height as h = 10 m for the question to have a valid answer and the student can have a reference to solve their own problem</em>

Answer:

(a) $\Delta P=1.67 \ kg.m/s$

(b) $\Delta P=0.86\ m/s$

Explanation:

<u>Change of Momentum</u>

The momentum of a given particle of mass m traveling at a speed v is given by

$P=m.v$

When this particle changes its speed to a value v', the new momentum is

$P'=m.v'$

The change of momentum is

$\Delta P=m.v'-m.v$

$\Delta P=m.(v'-v)$

Defining upward as the positive direction, we'll compute the change of momentum in two separate cases.

(a) The initial height of the superball of m=61.6 gr = 0.0616 Kg is set to h= 10 m. This information leads us to have the initial potential energy of the ball just after it's dropped to the floor:

$U=m.g.h=0.0616\cdot 9.8\cdot 10 =6.0368\ J$

This potential energy is transformed into kinetic energy just before the collision occurs, thus

$\displaystyle \frac{1}{2}mv^2=6.0368$

Solving for v

$\displaystyle v=\sqrt{\frac{6.0368\cdot 2}{0.0616}}$

$v=-14\ m/s$

This is the speed of the ball just before the collision with the floor. It's negative because it goes downward. Now we'll compute the speed it has after the collision. We'll use the new height and proceed similarly as above. The new height is

$h'=88.5\% (10)=8.85\ m$