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boyakko [2]
3 years ago
10

An empty parallel plate capacitor is connected between the terminals of a 16.7-v battery and charges up. the capacitor is then d

isconnected from the battery, and the spacing between the capacitor plates is doubled. as a result of this change, what is the new voltage between the plates of the capacitor?
Physics
1 answer:
katen-ka-za [31]3 years ago
4 0
16.7V * 2 = 33.4V
_____
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Where is the potential energy equal to zero?
s2008m [1.1K]

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im sure your already past this but it's E.

Explanation:

This is because in this case potential energy is linear to height, which means that the higher the more potential energy.

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Your tired 10 lb. baby cousin is cradled in your arms as you're walking back and forth across a 15m room to get her to fall asle
Mashcka [7]

Work done = Force x Distance

Force = 10 lb = 44.5 N

Work Done = 44.5 N x 15 m

= 667.5 N-m

6 0
3 years ago
Sunspots are _______ and _________ than the gases surrounding them.
Scrat [10]
The answer is B. darker; cooler
the dark spots give off less energy then the rest of the sun.

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3 0
4 years ago
Read 2 more answers
A car with a mass of 1,500 kg is accelerating at a rate of
AURORKA [14]

Answer:

The net force acting on the car is

3

×

10

3

Newtons.

Hope this helps you

Explanation:

Force is defined as the product of the mass of the body and its aaceleration,

⇒

F

=

m

a

Substituting the above given values we get,

F

=

(

1500

k

g

)

(

2.0

m

/

s

2

)

=

3000

N

=

3

×

10

3

N

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4 0
3 years ago
Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father
Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity \omega_o = 0

angular acceleration \alpha = 4.44 rad/s²

Using the formula:

\omega = \omega_o+ \alpha t

Making t the subject of the formula:

t= \dfrac{\omega- \omega_o}{ \alpha }

where;

\omega = 1.53 \ rad/s^2

∴

t= \dfrac{1.53-0}{4.44 }

t = 0.345 s

b)

Using the formula:

\omega ^2 = \omega _o^2 + 2 \alpha \theta

here;

\theta = angular displacement

∴

\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }

\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }

\theta =0.264 \ rad

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

x = \dfrac{0.264 \times 1}{2 \pi}

x = 0.042 revolutions

c)

Here; force = 270 N

radius = 1.20 m

The torque = F * r

\tau = 270 \times 1.20 \\ \\  \tau = 324 \ Nm

However;

From the moment of inertia;

Torque( \tau) = I \alpha \\ \\  Since( I \alpha) = 324 \ Nm. \\ \\  Then; \\ \\  \alpha= \dfrac{324}{I}

given that;

I = 84.4 kg.m²

\alpha= \dfrac{324}{84.4} \\ \\  \alpha=3.84 \ rad/s^2

For re-tardation; \alpha=-3.84 \ rad/s^2

Using the equation

t= \dfrac{\omega- \omega_o}{ \alpha }

t= \dfrac{0-1.53}{ -3.84 }

t= \dfrac{1.53}{ 3.84 }

t = 0.398s

The required time it takes= 0.398s

5 0
3 years ago
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