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Blababa [14]
3 years ago
14

3. Why is static electricity not useful as a power

Physics
1 answer:
shtirl [24]3 years ago
3 0
The answer is B, because all energy is released at once in static electricity.

There’s a quizlet that mentions these questions, if you are having trouble. I’d suggest to give them a look.
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What conclusions can be drawn about the existence of carbon-12, carbon-13, and carbon-14?
SashulF [63]
<span>These are isotopes of carbon and they all contain 6 protons and 6 electrons but each contains a difference number of neutrons - 6, 7, and 8 respectively.

^ This is the answer because an isotope changes the atomic mass, NOT atomic number. That means that the neutrons are changed, not the protons. </span>
7 0
3 years ago
Read 2 more answers
A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10
Charra [1.4K]

Answer:

u(t)=1.15 \sin (8.68t)cm

0.3619sec

Explanation:

Given that

Mass,m=148 g

Length,L=13 cm

Velocity,u'(0)=10 cm/s

We have to find the position u of the mass at any time t

We know that

\omega_0=\sqrt{\frac{g}{L}}\\\\=\sqrt{\frac{980}{13}}\\\\=8.68 rad/s

Where g=980 cm/s^2

u(t)=Acos8.68 t+Bsin 8.68t

u(0)=0

Substitute the value

A=0\\u'(t)=-8.68Asin8.68t+8.68 Bcos8.86 t

Substitute u'(0)=10

8.68B=10

B=\frac{10}{8.68}=1.15

Substitute the values

u(t)=1.15 \sin (8.68t)cm

Period =T = 2π/8.68

After half period

π/8.68 it returns to equilibruim

π/8.68 = 0.3619sec

6 0
3 years ago
A toy car has a momentum of 3 kilogram meters per second south. The car has a 1-kilogram mass. Which is the velocity of the car?
antiseptic1488 [7]
Using p = mv 3 = 1× v v = 3m/s
6 0
3 years ago
What is the purpose of a free body diagram?
leonid [27]

Answer:

i think the answer is c

Explanation:

zrrxtcyvjvugugyctcyvjv

8 0
3 years ago
a typical cmall flashlight contains two batteries each having na emf of 2.0 v connected in series with a bulb havin ga resistanc
Helen [10]

Answer:

P = 0.25 W

Explanation:

Given that,

The emf of the battry, E = 2 V

The resistance of a bulb, R = 16 ohms

We need to find the power delivered to the bulb. We know that, the formula for the power delivered is given by :

P=\dfrac{V^2}{R}\\\\P=\dfrac{2^2}{16}\\\\=0.25\ W

So, 0.25 W power is delivered to the bulb.

5 0
3 years ago
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