Which statements describe the factors affecting the strength of an electrical force? Check all that apply.

The potential difference between a pair of oppositely charged parallel plates is 398 V. If the spacing between the plates is dou

LiRa [457]

Answer:

Explanation:

capacitance of parallel plate capacitor

c = ε A / d , d is distance between plates , A is surface area , ε is constant

As d becomes two times , Capacitance c = 1/ 2 times ie c / 2

potential V = Q / C

Q is constant , potential

v = Q / c /2

= 2 . Q / C

= 2 V

So potential difference becomes 2 times.

NEW P D = 398 X 2

= 796 V.

6 0

3 years ago

What is center of mass?

yuradex [85]

“a point representing the mean position of the matter in a body or system.”

3 0

2 years ago

Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit.

Effectus [21]

Answer:

The width of the slit is 0.167 mm

Explanation:

Wavelength of light, $\lambda=608\ nm=608\times 10^{-9}\ m$

Distance from screen to slit, D = 88.5 cm = 0.885 m

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m

We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

$y=\dfrac{mD\lambda}{a}$

where

a = width of the slit

$a=\dfrac{mD\lambda}{y}$

$a=\dfrac{5\times 0.885\ m\times 608\times 10^{-9}\ m}{0.0161\ m}$

a = 0.000167 m

$a=1.67\times 10^{-4}\ m$

a = 0.167 mm

So, the width of the slit is 0.167 mm. Hence, this is the required solution.

7 0

2 years ago

Light waves reflect off the objects that you see. Where do those light waves originate?

xxMikexx [17]

Answer:

The Sun is a natural source for visible light waves and our eyes see the reflection of this sunlight off the objects around us.

3 0

2 years ago

An automobile of mass 2000 kg moving at 30 m/s is braked suddenly with a constant braking force of 10000 N. How far does the car

saveliy_v [14]

Answer:

The car traveled the distance before stopping is 90 m.

Explanation:

Given that,

Mass of automobile = 2000 kg

speed = 30 m/s

Braking force = 10000 N

For, The acceleration is

Using newton's formula

$F = ma$

Where, f = force

m= mass

a = acceleration

Put the value of F and m into the formula

$-10000 =2000\times a$

Negative sing shows the braking force.

It shows the direction of force is opposite of the motion.

$a = -\dfrac{10000}{2000}$

$a=-5\ m/s^2$

For the distance,

Using third equation of motion

$v^2-u^2=2as$

Where, v= final velocity

u = initial velocity

a = acceleration

s = stopping distance of car

Put the value in the equation

$0-30^2=2\times(-5)\times s$

$s = 90\ m$

Hence, The car traveled the distance before stopping is 90 m.

6 0

2 years ago