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3 years ago

Which number would be 16,000 when rounded to the nearest thousand? *

Physics

2 answers:

Dennis_Churaev [7]3 years ago

8 0

it would be for sure 15,518

yanalaym [24]3 years ago

5 0

Answer:

15,518

Explanation:

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People build a dam to create a reservoir that supplies water a nearby city needs. Describe two ways this action will likely affe

PolarNik [594]

Due to damming the water it will not flow as fast and since it wont flow as fast it will cause the water level to rise.

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4 years ago

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A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electr

Nata [24]

Answer:

- E = 11.55J

- Q = 0.17C

- E' = (1/4)E

Explanation:

- To calculate the amount of energy stored in the capacitor, you use the following formula:

$E=\frac{1}{2}CV^2$

C: capacitance = 3800.0*10^-6F

V: potential difference = 78.0V

$E=\frac{1}{2}(3800.0*10^{-6}C)(78.0V)^2=11.55J$

The energy stored in the capacitor is 11.55J

- If the electrical energy stored in the capacitor is 6.84J, the charge on the capacitor is:

$E=\frac{1}{2}QV\\\\Q=\frac{2E}{V}\\\\Q=\frac{2(6.84J)}{78.0V}=0.17C$

The charge on the capacitor is 0.17C

- If you take the capacitor as a parallel plate capacitor, you have that the energy stored on the capacitor is:

$E=\frac{1}{2}CV^2=\frac{1}{2}(\frac{\epsilon_oA}{d})V^2=\frac{1}{2}\frac{\epsilon_oAV^2}{d}\\\\$

A: area of the plates

d: distance between plates

If the distance between plates is increased by a factor of 4, you have:

$E'=\frac{1}{2}\frac{\epsilon_oAV^2}{(4d)}=\frac{1}{4}\frac{\epsilon_oAV^2}{2d}=\frac{1}{4}E$

Then, the stored energy in the capacitor is decreased by a a factor of (1/4)

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3 years ago

Why is it incorrect to say "We ran out of gas", when referring to your cars energy source?

Allushta [10]

Maybe it’s suppose to say fuel or something fossil fuel or something like that

3 0

3 years ago

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A 0.350-m-long cylindrical capacitor consists of a solid conducting core with a radius of 1.20 mm and an outer hollow conducting

Brut [27]

Answer: a. 6.52*EXP{-10}C/m. b. 2.28*EXP{-10}C. c. 38picoFarad.

d. 6.84*EXP{-10}joules

Explanation: We first calculate for the capacitance first. For a concetric cylinder with two radius R1 and R2 the capacitance C is given as

C= {2*pi*permitivity of

freespace*lenght}/In(R2/R1) from the given Question R2 is 2mm and R1 is 1.2mm, lenght is 0.35meter, permitivity of free space is 8.85*EXP {-12} and pi is 3.142.

Therefore Capacitance would be,

C = 2*3.142*8.85*EXP

{-12}*0.35/In(2/1.2)

C = 3.8*EXP {-11} Farad which is also

38*EXP {-12} Farad or 38picoFarad.

Next, we solve for our total charge Q. Charge, capacitance and voltage are related by

Q = C*V = 38*EXP {-12}*6

=2.28*EXP {-10} Coulombs

Next, we obtian charge per unit lenght, which is

Q/L = 2.28*EXP {-10}/0.35

= 6.52*EXP {-10} Coulombs/meter

Next, we obtain the energy stored in the capacitor from

Energy stored = 1/2*(C*V²)

=1/2*38*EXP {-12}*6²

=6.84*EXP {-10} Joules

Note: EXP means 10^

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3 years ago

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The electric field near the earth's surface has magnitude of about 150n/c. what is the acceleration experienced by an electron n

qaws [65]

Felectric = q*E
 Ftranslational = m*a
 Felectric = Ftranslational
 q*E = m*a
 Solve for a
 a = q/m*E 
 Our sign convention is "up is positive"
 q = 1.6*10^-19 C
 m = 1.67*10^-27 kg
 E = -150 N/C (- because it is down and up is positive)
 a = -6,4*10^5 m/s^2 (downward)
 answer
a = -6,4*10^5 m/s^2 (downward)

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3 years ago