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3 years ago

5

What must be the distance in meters between point charge q1 = 23.5 µc and point charge q2 = -64.2 µc for the electrostatic force

between them to have a magnitude of 4.67 n?

1 answer:

svlad2 [7]3 years ago

7 0

The answer for your problem is shown on the picture.

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PLEASE ANSWER ASAP!!!!!!!

Viefleur [7K]

Answer:

B. Mechanical energy= 50J+30J=80J

4 0

3 years ago

Free runners jump long distances and land on the ground or a wall. How do they apply Newton’s second law to lessen the force of

Veseljchak [2.6K]

As we know that as per Newton's II law we have

$F = \frac{dP}{dt}$

here we will have

$dP$ = change in momentum

$dt$ = time interval in which momentum is changed

now in order to have least injury during jumping we need to have least force on the jumper

so in order to have least force we can say that the momentum must have to change in maximum time so that amount of force must be least

So we need to increase the time in which momentum of the system is changed

5 0

3 years ago

Kelli weighs 440 N, and she is sitting on a playground swing that hangs 0.41 m above the ground. Her mom pulls the swing back an

kifflom [539]

Answer:

Explanation:

Initial height from the ground = .41 m

Final height = 1m

Height by which Kelli was raised ( h )= .59 m

When she passes through the lowest point , she loses P E

= mgh

= 440 x .59

= 259.6 J

kinetic energy possessed by her

= 1/2 mv²

= .5 x (440/9.8) x 2²

= 89.8 J

Difference of energy is lost due to work by air friction

work done by friction = 89.8 - 259.6

= - 169.8 J

4 0

3 years ago

16. Two capacitors have an equivalent

Gennadij [26K]

Answer:

C1 + C2 = 30 parallel connection

C1 * C2 / (C1 + C2) = 7.2 series connection

C1 * C2 = 7.2 * (C1 + C2) = 216

C2 + 216 / C2 = 30 using first equation

C2^2 + 216 = 30 C2

C2^2 - 30 C2 + 216 = 0

C2 = 12 or 18 solving the quadratic

Then C1 = 18 or 12

5 0

3 years ago

A light bulb emits light uniformly in all directions. The average emitted power is 150.0 W. At a distance of 5 m from the bulb,

beks73 [17]

Answer:

a) 0.477 W/m²

b) 13.407 N/C

c) 18.96 N/C

Explanation:

P = Power = 150 W

r = Distance = 5 m

ε₀ = Permittivity of space = 8.854×10⁻¹² F/m

a) Average intensity

$\bar{S}=\frac{P}{A}\\\Rightarrow \bar{S}=\frac{150}{4\pi r^2}\\\Rightarrow \bar{S}=\frac{150}{4\pi 5^2}\\\Rightarrow \bar{S}=0.477\ W/m^2$

∴ Average intensity is 0.477 W/m²

b) Rms value

$\bar{S}=c\epsilon_0E_{rms}^2\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{S}}{c\epsilon_0}}\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{0.477}}{3\times 10^8\times 8.854\times 10^{-12}}}\\\Rightarrow E_{rms}=13.407\ N/C$

∴ Rms value of the electric field is 13.407 N/C

c) Peak value

$E_0=\sqrt 2E_{rms}\\\Rightarrow E_0=\sqrt 2\times 13.407\\\Rightarrow E_0=18.96\ N/C$

∴ Peak value of the electric field is 18.96 N/C

8 0

3 years ago