Please help question in photo

How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33Ã—Ã—10âˆ

AlekseyPX

Answer:

There are 756.25 electrons present on each sphere.

Explanation:

Given that,

The force of repression between electrons, $F=3.33\times 10^{-21}\ N$

Let the distance between charges, d = 0.2 m

The electric force of repulsion between the electrons is given by :

$F=k\dfrac{q^2}{r^2}$

$q=\sqrt{\dfrac{Fr^2}{k}}$

$q=\sqrt{\dfrac{3.33\times 10^{-21}\times (0.2)^2}{9\times 10^9}}$

$q=1.21\times 10^{-16}\ C$

Let n are the number of excess electrons present on each sphere. It can be calculated using quantization of charges. It is given by :

q = ne

$n=\dfrac{q}{e}$

$n=\dfrac{1.21\times 10^{-16}}{1.6\times 10^{-19}}$

n = 756.25 electrons

So, there are 756.25 electrons present on each sphere. Hence, this is the required solution.

8 0

3 years ago

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

$I=\frac{1}{3}mL^2$

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0

3 years ago

You and a friend each carry a 15 kg suitcase up two flights of stairs, walking at a constant speed. Take each suitcase to be the

AlekseyPX

Answer:

Both of you did the same work but you expended more power.

Explanation:

<em>Work done</em> by an object is calculated by force applied multiplied by the distance.

W=F*d

From the figure given below let us calculate force applied bith you and yopur friend.

Let us take the stairs in positive x direction,

Work done by you W₁ ,

The force applied Fₓ = F - mgsinθ =maₓ

here aₓ = 0, because both of you move with constant speed

F - mgsinθ = 0

F= mgsinθ

The work done by you on the suitcase is

W = F L cos0° , where L is he length of the staircase.

W = FL = mgsinθL , by substituting value of F

Work done by you is W₁ = mgLsinθ

Similarly work done by your friend is W₂ = mgLsinθ.

Because both of you carry suitcase of same weight and in staircase is in same angle the force applied is same .

Therefore <em>work done by both of you is same</em> . Both of you did equal work.

The power , is defined as amount of energy converted or transfered per second or rate at which work is done .

P = $\frac{W}{t}$ = $\frac{FL}{t}$

Power spend by you P₁ = mgLsinθ/t

P₁ = 15*9.8*Lsinθ/30

P₁ = 4.9L sinθ eqn 1

Power spend by your friend is P₂ = mgLsinθ/t

P₂ =15*9.8*Lsinθ/60

P₂ = 2.45Lsinθ eqn 2

Dividing eqn 1 and eqn 2

P₁ = 2P₂

You have spend more power than your friend .

Hence Both of you did equal work but you spend more power.

7 0

3 years ago

A professional baseball player can throw the ball around 45 m/s if the distance between the pitcher and the batter is 18.39 m. H

svetlana [45]

Answer:

The time taken for the ball to get to the batter is 0.41 s.

Explanation:

Given;

initial velocity of the baseball, u = 45 m/s

horizontal distance between the pitcher and the batter, X = 18.39 m

The horizontal distance or range of a projectile is given as;

X = ut

where;

t is the time of flight

u is the initial velocity

t = X / u

t = 18.39 / 45

t = 0.41 s

Therefore, the time taken for the ball to get to the batter is 0.41 s.

6 0

2 years ago

A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the

Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

$K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2$ = 85.75 J

The speed of the ball after leaving the bat is:

$K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}$