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mote1985 [20]
3 years ago
7

Which weighs more a pound of feathers or a pound of bricks

Physics
1 answer:
Alecsey [184]3 years ago
5 0

Answer: neither

Explanation:

both weight exactly one pound. although the quantity of feathers may outnumber the bricks, they both weigh a pound

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What is a short circuit?
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Answer:

A short circuit is an electrical circuit that allows a current to travel along an unintended path. It has no or very low electrical impedances. The opposite of a short circuit

Explanation:

3 0
3 years ago
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A large sheet of charge has a uniform charge density of 9  μCm2. What is the electric field due to this charge at a point just
Alex73 [517]

Answer:

Explanation:

Surface charge density, σ = 9 μC/m² = 9 x 10^-6 C/m²

According to the Gauss theorem,

Electric field due to the sheet is given by

E = \frac {\sigma }{2\epsilon _{0}}

E = \frac{9\times 10^{-6}}{2\times 8.854\times 10^{-12}}

E = 5.08 x 10^5 N/C

7 0
3 years ago
A stone with a mass m is dropped from an airplane that has a horizontal velocity v at a height h above a lake. If air resistance
seropon [69]

Answer: Option B. R = (1/2)gt^2

Explanation:

S = R (horizontal distance)

V^2 = 2gS

V^2 = 2gR

R = V^2 / 2g

But V = gt

R = (gt)^2 / 2g

R = (g^2 x t^2) / 2g

R = gt^2 / 2

But t^2 = 2h/g

R = ( g x 2h/g) / 2

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R = h = (1/2)gt^2

4 0
3 years ago
If earth's axis was tilted 45 degrees at what latitudes
givi [52]
23.5 DEGREES is the answer. 
7 0
2 years ago
Two equipotential surfaces surround a +3.10 x 10-8-c point charge. how far is the 290-v surface from the 41.0-v surface?
MrMuchimi
 T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r 

where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance. 
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect, 

Point 1: 
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r

Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m 
The distance between the two points then is equal to 7.07 m.


</span>
8 0
2 years ago
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