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salantis [7]
3 years ago
8

How much force is required to move a electron through an electric field with strength of 1.375 x 10^19 N/C?

Physics
1 answer:
Luda [366]3 years ago
4 0

Answer:

The magnitude of the force required to move the electron through the given field is 2.203 N

Explanation:

Given;

The field strength of the electron, E = 1.375  x 10¹⁹ N/C

charge of electron, q = 1.602 x 10⁻¹⁹ C

The magnitude of the force required to move the electron through the given field is calculated as follows;

F = Eq

F = (1.375  x 10¹⁹ N/C) (1.602 x 10⁻¹⁹ C)

F = 2.203 N

Therefore, the magnitude of the force required to move the electron through the given field is 2.203 N

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A ball is swung in a horizontal circle at a constant speed. Each circle takes 0.85 seconds to complete and the rope is 0.40 m lo
BaLLatris [955]

Answer:

The centripetal acceleration will be "21.785 m/s²".

Explanation:

The given values are:

Time,

t = 0.85 seconds

Length of rope,

r = 0.40 m

Mass of ball,

m = 0.80 kg

As we know,

⇒ w=\frac{2 \pi}{t}

On substituting the values, we get

⇒      =\frac{2\times 3.14}{0.85}

⇒      = \frac{6.28}{0.85}

⇒      =7.38 \ rad/s^2

The centripetal acceleration will be:

⇒  a=r\times w^2

⇒     =0.40\times (7.38)^2

⇒     =0.40\times 54.46

⇒     =21.785 \ m/s^2

7 0
3 years ago
A flywheel of mass 182 kg has an effective radius of 0.62 m (assume the mass is concentrated along a circumference located at th
musickatia [10]

Answer:

A)5524J,

B) 29.2Nm

Explanation:

This question can be treated using work- energy theorem

Work= change in Kinectic energy

W= Δ KE

Work= difference between the final Kinectic energy and intial Kinectic energy.

We know that

Kinectic energy= 1/2 mv^2 .............eqn(1)

This can be written in term of angular velocity, as

KE= 1/2 I

4 0
3 years ago
While standing atop a building 49.6 m tall, you spot a friend standing on a street corner. Using a protractor and a dangling plu
olga nikolaevna [1]

Answer:

75degree don't forget wind and gravity force pulling down

6 0
3 years ago
Light of wavelength 559 nm is used to illuminate normally two glass plates 22.1 cm in length that touch at one end and are separ
umka21 [38]

Answer:

M = 222 fringes

Explanation:

given

λ = 559 n m = 559 × 10⁻⁹ m

radius = 0.026 mm = 0.026 ×10⁻³ m

length of the glass plate = 22.1 ×10⁻² m

using relation

2t=(m+\dfrac{1}{2})\lambda\ \ (m=0,1,2,3...)\\where\ 0\leq t\leq 2r\\m = \dfrac{2t}{\lambda}-\dfrac{1}{2}

m_{max} = \dfrac{2\times 2r}{\lambda}-\dfrac{1}{2}\\m_{max} = \dfrac{2\times 2\times 0.026\times 10^{-3}}{559\times 10^{-9}}-\dfrac{1}{2}

 = 221.79  

 = 221 (approx.)

hence no of bright fringe

M = m + 1

   = 221 +1

M = 222 fringes

6 0
3 years ago
A small box slides down a ramp on a friction with surface. If the total energy of the system is conserved, which computational m
spayn [35]

Answer:

WWWEEEEEEEEEEEEEEEEEE

Explanation:

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