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3 years ago

5

Suppose two equal charges of 0.65 C each are separated by a distance of 2.5 km in air. What is the magnitude of the force acting

between them, in newtons?

1 answer:

mafiozo [28]3 years ago

7 0

Answer:

Force between two equal charges will be 608.4 N

Explanation:

We have given charges $q_1=0.65C\ and\ q_2=0.65C$

Distance between the charges = 2.5 km = 2500 m

According to coulombs law force between two charges is given by

$F=\frac{1}{4\pi \varepsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}$ , here K is constant which value is $9\times 10^9Nm^2/C^2$

So force $F=\frac{9\times 10^9\times 0.65\times 0.65}{2500^2}=608.4N$

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1. A 3.0 kg mass is tied to a rope and swung in a horizontal circle. If the velocity of the mass is 4.0 ms and

saul85 [17]

10.67m/s²

32N

Explanation:

Given parameters:

Mass of the body = 3kg

velocity of the mass = 4m/s

radius of circle = 0.75m

Unknown:

centripetal acceleration = ?

centripetal force = ?

Solution:

The centripetal force is the force that keeps a radial body in its circular motion. It is directed inward:

Centripetal acceleration = $\frac{v^{2} }{r}$

v is the velocity of the body

r is the radius of the circle

putting in the parameters:

Centripetal acceleration = $\frac{4^{2} }{0.75}$

Centripetal acceleration = 10.67m/s²

Centripetal force = m $\frac{v^{2} }{r}$

m is the mass

Centripetal force = mass x centripetal acceleration

= 3 x 10.67

= 32N

learn more:

Acceleration brainly.com/question/3820012

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3 years ago

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 21.0 cm and carries a

Burka [1]

Solution :

a). B at the center :

$=\frac{u\times I}{2R}$

Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.

Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE

b). Also, the sum of the fields must be zero.

Therefore,

$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$

So,

$\frac{I_1}{d_1}= \frac{I_2}{d_2}$

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Therefore, the current in the outer wire is 24.38 ampere.

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3 years ago

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A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward

AleksandrR [38]

Answer:

(a) 104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.

The component of F perpendicular to the ramp is Fy.

(a)

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.

Resolve F into its x-component from Pythagorean theorem:

Fx=Fcos30°

Solve for F:

F= Fx/cos30°

Substitute for Fx from given data:

Fx=90 N/cos30°

=104 N

(b) Resolve r into its y-component from Pythagorean theorem:

Fy = Fsin 30°

Substitute for F from part (a):

Fy = (104 N) (sin 30°)

= 52 N

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3 years ago

Match the situation with the energy transformation ITEMBANK: Move to Top A boy shooting a rubber band across the classroom A chi

Sonbull [250]

A boy shooting a rubber band across the classroom -->
Elastic potential energy transformed into kinetic energy
<span>The initial energy is the energy stored in the muscles of the boy's arm, which is elastic potential energy. This is converted into motion of the rubber, therefore kinetic energy

A child going down a slide on a playground --> </span>Gravitational potential energy transformed into kinetic energy
On top of the slide, all the energy of the child is gravitational potential energy due to its height with respect to the ground (E=mgh). when it moves down the slide, this is converted into kinetic energy, because the child acquires a speed v (E=1/2 mv^2)
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Rubbing your hands together to warm them on a cold day --> </span>Kinetic energy being transformed into thermal energy <span>
When rubbing hands, we are moving them (kinetic energy), and this energy raises the temperature of the hand's surface (thermal energy)

Turning on a battery operated light --> </span>
Chemical potential energy transformed into radiant energy <span>
A battery works by mean of chemical reactions (chemical potential energy), producing light (so, emitting energy by radiation, i.e. radiant energy)

Using a dc electric motor --> </span> Electrical energy transformed into kinetic energy<span>
A dc electric motor works using currents (so, electrical energy), and the energy produced can be used for example to accelerate a car (kinetic energy)

Using a gas power heater to warm a room --> </span>Chemical potential energy transformed into thermal energy
<span>A gas power heater burns gases (so, chemical reaction, i.e. chemical potential energy) to raise the temperature of the room (thermal energy)

Using a hand crank generator to produce electric current --> Kinetic energy transformed into electrical energy
In a hand-crank generator, the handle is being rotated (kinetic energy) in order to produce an electric current (electrical energy)

Using the light in your room that is plugged into the wall --> </span>Electrical energy transformed into radiant energy
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3 0

3 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

3 years ago