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3 years ago

14

3. Why is static electricity not useful as a power

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