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Nana76 [90]
3 years ago
11

4. What is the electric field strength 1.4 nm from a charge of 4.7 cC?

Physics
1 answer:
pentagon [3]3 years ago
5 0

The electric field strength is 2.16\cdot 10^{26} N/C

Explanation:

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

E=k\frac{q}{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 at which the field strength is calculated

For the charge in the problem, we have:

q=4.7 cC = 0.047 C is the charge

r=1.4 nm = 1.4\cdot 10^{-9} m

Therefore, the electric field strength is

E=(8.99\cdot 10^9)\frac{0.047}{(1.4\cdot 10^{-9})^2}=2.16\cdot 10^{26} N/C

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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antiseptic1488 [7]

Answer: 510 m/s

Explanation: specific gravity of steam is 18/29 = 0.620

It is the ratio of the density of steam over density of water

400m3/s of steam =

400m3ms * 0.620 of water

= 248m3/s of water

Total flow rate Q = 248 + 7 = 255m3/s

Using Q = AV

Where A is area and V is velocity

V = Q/A

V = 255/0.5 = 510m/s

4 0
3 years ago
After a new aircraft has passed ground and flight tests, the aircraft must receive a series of certificates before it can be mas
ollegr [7]

Answer:

A production certificate Shows that the aircraft met FAA construction and performance standards

Explanation:

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7 0
4 years ago
A woman is 1.6 m tall and has a mass of 50 kg. She moves past an observer with the direction of the motion parallel to her heigh
eduard

To solve this problem it is necessary to apply the concepts related to linear momentum, velocity and relative distance.

By definition we know that the relative velocity of an object with reference to the Light, is defined by

V_0 = \frac{V}{\sqrt{1-\frac{V^2}{c^2}}}

Where,

V = Speed from relative point

c = Speed of light

On the other hand we have that the linear momentum is defined as

P = mv

Replacing the relative velocity equation here we have to

P = \frac{mV}{\sqrt{1-\frac{V^2}{c^2}}}

P^2 = \frac{m^2V^2}{1-\frac{V^2}{c^2}}

P^2 = \frac{P^2V^2}{c^2}+m^2V^2

P^2 = V^2 (\frac{P^2}{c^2}+m^2)

V^2 = \frac{P^2}{\frac{P^2}{c^2}+m^2}

V^2 = \frac{(2.1*10^10)^2}{\frac{(2.1*10^10)^2}{(3.8*10^8)^2}+50^2}

V = 2.81784*10^8m/s

Therefore the height with respect the observer is

l = l_0*\sqrt{1-\frac{V^2}{c^2}}

l = 1.6*\sqrt{1-\frac{(2.81*10^8)^2}{(3*10^8)^2}}

l = 0.56m

Therefore the height which the observerd measure for her is 0.56m

8 0
3 years ago
How do i calculate this?
Lesechka [4]

CAR 1

Momentum = Mass/Velocity

M = 2100/20

M = 105 m/s^2

CAR 2

Momentum = Mass/Velocity

M = 2100/30

M = 70 m/s^2

8 0
3 years ago
Some waves can only travel through matter. This matter is called _________.
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This matter is called medium
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3 years ago
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