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PSYCHO15rus [73]
3 years ago
15

Calculate the electrical energy expended in a device across which the circuit voltage drops by 50 volts in moving a charge of 6

coulombs.
Physics
2 answers:
solmaris [256]3 years ago
5 0
1 volt = 1 joule per coulomb 50 volts = 50 joules per coulomb 50 joules/coulomb times 6 coulombs = 300 joules
olasank [31]3 years ago
4 0
I=\dfrac{Q}{t} \\  \\  \\ W=VIt=V\dfrac{Q}{t}t=VQ=50\times 6=300
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Explanation: lol I just born that that sign ig
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the human nervous system can propagate nerve impulses at about 10 squared metered per second. estimate the time it takes to trav
rusak2 [61]
Use the formular d = v x t
d = 2m
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t= d / v
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3 years ago
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The radius of Earth is about 6450 km. A 7070 N spacecraft travels away from Earth. What is the weight of the spacecraft at a hei
Triss [41]

Answer:

(a) 1767.43 N

(b) 182.45 N

Explanation:

Radius of earth, R = 6450 km

Weight of person, W = 7070 N

mass of person, m = W / g = 7070 / 9.8 = 721.4 kg

(a) h = 6450 km

The value of acceleration due to gravity on height is given by

g' = g\left ( \frac{R}{R+h} \right )^2

g' = g\left ( \frac{6450}{6450+6450} \right )^2

g' = g / 4 = 9.8 / 4 = 2.45 m/s^2

The weight of the person at such height is

W' = m x g' = 721.4 x 2.45

W' = 1767.43 N

(b) h = 33700 km

The value of acceleration due to gravity on height is given by

g' = g\left ( \frac{R}{R+h} \right )^2

g' = g\left ( \frac{6450}{6450+33700} \right )^2

g' = g x 0.0258 = 9.8 x 0.0258 = 0.253 m/s^2

The weight of the person at such height is

W' = m x g'

W' = 721.4 x 0.253

W' = 182.45 N

3 0
3 years ago
what english word does the following- the first two signify a male, the 1st 3 letters support a female, the first four support a
Zarrin [17]

Answer:

lol i know - i was rushing -_-

he= male

her=female

One word with both is <u>heroin</u> but Im not 100% sure

8 0
2 years ago
Suppose that a comet has a very eccentric orbit that brings it quite close to the Sun at closest approach (perihelion) and beyon
Nutka1998 [239]

Answer:

16.63min

Explanation:

The question is about the period of the comet in its orbit.

To find the period you can use one of the Kepler's law:

T^2=\frac{4\pi}{GM}r^3

T: period

G: Cavendish constant = 6.67*10^-11 Nm^2 kg^2

r: average distance = 1UA = 1.5*10^11m

M: mass of the sun = 1.99*10^30 kg

By replacing you obtain:

T=\sqrt{\frac{4\pi}{GM}r^3}=\sqrt{\frac{4\pi^2}{(6.67*10^{-11}Nm^2/kg^2)(1.99*10^{30}kg)}(1.496*10^8m)^3}\\\\T=997.9s\approx16.63min

the comet takes around 16.63min

8 0
3 years ago
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