All categories

3 years ago

A point charge is placed 100 m from a 6 uC charge generating an electric field. What is the

Physics

1 answer:

kvv77 [185]3 years ago

3 0

The strength of the electric field is 5 N/C

Explanation:

The magnitude of the electric field produced by a single-point charge is given by:

$E=\frac{kQ}{r^2}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

Q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

$Q=6 \mu C = 6\cdot 10^{-6}C$ is the charge producing the field

r = 100 m is the distance from the charge at which we want to calculate the field

Substituting into the equation, we find the s trength of the electric field:

$E=\frac{(8.99\cdot 10^9)(6\cdot 10^{-6})}{(100)^2}=5.4 N/C \sim 5 N/C$

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

You might be interested in

A football player kicks a 0.94 kg football with a force of 2.4 N. Calculate the acceleration of the football as the player kicks

Anon25 [30]

Hello, I hope this helps :)
So the equation to figure these kinds of questions is F=MA
F refers to force, which in this situation F would be 2.4
M refers to mass, mass would be 0.94
A refers to acceleration, which we are trying to figure out
So if we put in the information we know into the equation, it is now 2.4=0.94*A
So with that equation we can figure that 2.4 divided by 0.94 equals A
So 2.4/0.94= 2.553191489362 :'D
Don't worry, the rounded and correct answer is 2.6
Have a nice day =)

7 0

3 years ago

Read 2 more answers

A 26.0 kg beam is attached to a wall with a hinge while its far end is supported by a cable such that the beam is horizontal. If

Nuetrik [128]

Answer:

Force exerted by the hinge on the beam = 109.24N

Explanation:

Weight = mg = 26 x 9.81 = 255.06 N

Vertical component = T sin θ

Horizontal component = Tcos θ

Now, there are 3 vertical forces acting on the beam. These are;

- The downward force which is the weight of the beam.

- The vertical components of the tension in the cable.

-The force that hinge exerts on the beam are the upward forces.

Hence, for the beam to remain horizontal, the sum of the upward forces must be equal to the weight of the beam.

For us to determine the vertical component of the tension in the cable, we will do a torque problem. Let the pivot point be at the hinge. Let’s assume that the length of the beam is L. The vertical component of the tension in the cable will produce clockwise torque while the weight of the beam will produce counter clockwise torque.

Tbus;

Clockwise torque = TL sin 61

Since the center of mass of beam is at the middle of the beam, the distance from the hinge to the weight of the beam is L/2.

Counter clockwise torque = WL/2

Thus;

TL sin 61 = WL/2

L will cancel out.

T sin 61 = 255.06/2

T x 0.8746 = 127.53

T = 127.53/0.8746 = 145.82 N

Now, the equation to determine the vertical component of the force that the hinge exerts on the beam is given as;

T + F = W

Thus;

145.82 + F = 255.06

F = 255.06 - 145.82 = 109.24 N

8 0

3 years ago

What task requires the most work, lifting a 12-kg sack 2 meters or lifting a 25-kg sack 1 meter?

MrMuchimi

Multiply the masses by the respective distances:

(12 kg) (2 m) = 24 J

(25 kg) (1 m) = 25 J

so the heavier bag takes more work to lift, and (b) is the answer.

(d) is technically correct if the sacks are carrying different contents whose masses are not equal, but since we don't know what's inside each sack, assume 12 kg and 25 kg are the masses of each sack *and* their contents.

5 0

2 years ago

a coin press creates a pressure of 3.20*10^8 Pa on a nickel of radius 0.0106 m. how much force does the press exert on the coin?

Naya [18.7K]

Answer:

1.13 x 10⁵N

Explanation:

Given parameters:

Pressure of the coin press = 3.2 x 10⁸ Pa

radius of the nickel coin = 0.0106m

Unknown:

Force of the press on coil = ?

Solution:

Our knowledge of pressure will help us solve this problem.

Pressure is defined as the force applied per unit area on a body.

Pressure = $\frac{force}{area}$

Force = Pressure x Area

Since the pressure is known;

Area of the coin = Area of a circle = π r²

where r is the radius of the coin;

Area of the coin = π x 0.0106² = 3.53 x 10⁻⁴m²

Force = 3.2 x 10⁸ x 3.53 x 10⁻⁴ = 1.13 x 10⁵N

3 0

3 years ago

A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t

kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

$\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W$

where we have used the values given by the information of the problem for N B and A.

the emf is given by:

$emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV$

hope this helps!!

5 0

3 years ago