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antoniya [11.8K]
3 years ago
14

A power source of 2.0 V is attached to the ends of a capacitor. The capacitance is 4.0 μF. What is the amount of charge stored i

n this capacitor?
Physics
1 answer:
ioda3 years ago
8 0
C = 4 \ \mu F = 4 \cdot 10^{-6} \ F. \newline
q = Cu = 4 \cdot 10^{-6} \cdot 2 = 8 \cdot 10^{-6} = 0.000008 \ C.
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Calculate the power provided by a 12 volt car battery with a rating of 350 amperes.
mamaluj [8]

Answer:

Option D

4200 W

Explanation:

Power, P is also given as the product of voltage and current, expressed as P=VI

Here, P is power, V is voltage in the xircuit and I is current theough voltage.

Taking 12 V for voltage across and 350A for current across circuit then power will be

P=350*12=4200 W

Therefore, option D is correct.

8 0
3 years ago
What's an easy way to create an interference pattern of waves?
igor_vitrenko [27]

Answer:

B

Explanation:

5 0
3 years ago
Early humans used rocks as tools to make other things as well as to construct buildings. Why was it better for them to use rocks
babymother [125]
Wood isn't as tough as rock. Wood also breaks down in weather and cracks under pressure. Plus rocks were more accessible.
6 0
3 years ago
Read 2 more answers
The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m
Sonbull [250]

Answer:

density is 10^{6} Mg/µL

Explanation:

given data

density of nuclear = 10^{18} kg/m³

1 ml = 1 cm³

to find out

density of nuclear matter in Mg/µL

solution

we know here

1 Mg = 1000 kg

so

1 m³ is equal to 10^{6} cm³

and here 1 cm³ is equal to  1 mL

so we can say 1 mL is equal to 10³ µL

so by these we can convert density

density = 10^{18} kg/m³

density = 10^{18} kg/m³ × \frac{10^{-3} }{10^{6} }  Mg/µL

density =  10^{6} Mg/µL

8 0
3 years ago
Read 2 more answers
To demonstrate the tremendous acceleration of a top fuel dragracer, you attempt to run your car into the back of a dragster that
noname [10]

Answer:

a. 2v₀/a   b. 2v₀/a  

Explanation:

a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.

Since the dragster starts from rest with an acceleration, a, using

s' = ut + 1/2at² where u = 0 and s' = distance moved by dragster

s' = 0t + 1/2at²

s' = 1/2at²

Since the distance moved by me and the dragster must be the same,

s = s'

v₀t. =  1/2at²

v₀t. - 1/2at² = 0

t(v₀ - 1/2at) = 0

t= 0 or v₀ - 1/2at = 0

t= 0 or v₀ = 1/2at

t= 0 or t = 2v₀/a  

So the maximum time tmax = 2v₀/a

b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s =  v₀(2v₀/a)

= 2v₀/a  

4 0
3 years ago
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