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2 years ago

Series circuit when you had one bulb and battery voltage was at 9 volts, what was current into battery?

Physics

1 answer:

KengaRu [80]2 years ago

7 0

Answer:

incomplete question, resistor must be there

Explanation:

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A 10.0-g bullet moving at 300 m/s is fired into a 1.00-kg block at rest. the bullet emerges (the bullet does not get embedded in

sesenic [268]

Momentum before the hit:
p = mv = 0.01 * 300 + 1 * 0

Momentum after the hit:
p = 0.01 * 150 + 1 * v

Momentum is conserved:
0.01 * 300 = 0.01 * 150 + v
3 = 1.5 + v
v = 1.5

The velocity of the block after the collision is 1.5 m/s.

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3 years ago

In the water cycle, which state of matter has the particles closest together?

Dmitry [639]

No, options are given but I believe the answer would be
In a water cycle Solid state of matter has the particles closest together.

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2 years ago

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How would you calculate power if work is not done?

Brilliant_brown [7]

Explanation:

This how you do it..

Calculate Watt-hours Per Day. Device Wattage (watts) x Hours Used Per Day = Watt-hours (Wh) per Day. ...

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3 years ago

Help I’ll cash app you $5 if you get it right!

Thepotemich [5.8K]

Answer: B and E

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2 years ago

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As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

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2 years ago