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3 years ago

5

What are the two kinds of nucleons?

1 answer:

Akimi4 [234]3 years ago

5 0

Answer:

The main types of nucleons are protons and neutrons. A proton, as its name suggests, has a positive electric charge, and a neutron has a neutral electric charge (meaning that it has no charge). The two in the nucleus of the atom make a positive charge, since the neutron has no charge at all.

Explanation:

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A battery of voltage V delivers power P to a resistor of resistance R connected to it. By what factor will the power delivered t

Anettt [7]

Answer:

Explanation:

Power P = V² / R

a ) The resistance is changed to 2.90R

Power will become 1 / 2.9 times .

b )The voltage of the battery is now 2.90V, but the resistance is R

P = (2.9V)² / R

= 8.41 x V² / R

So power becomes 8.41 times

c )The resistance is 2.90R and voltage is 2.90V

Power P = (2.9V)² / 2.9 R

= 2.9 V²/R

So power becomes 2.9 times

d ) The resistance is 2.90R and the voltage is V/2.90

Power P = ( V/2.90)² x 1 / 2.90R

1 / ( 2.9 )³ x V² / R

= 1 / 24.389 x V² / R

So power becomes 1 / 24.389 times .

4 0

3 years ago

create a poem that incorporates those ten words. Feel free to make it as silly as you like! MINIMUM of 6 lines with a MINIMUM of

Luba_88 [7]

I could make a poem for you if you actually gave the words...... what 10 words do i need to incorporate???☹︎

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3 years ago

How much work is done when you lift a 5kg bag of groceries 1.5 m?

olga55 [171]

Answer:

7.5J

Explanation:

5x1.5=7.5

8 0

3 years ago

If the released energy of one earthquake is 100 times that of another, how much greater is its magnitude on the richter scale?

Alekssandra [29.7K]

Greater by 2 on the Richter Scale :)

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3 years ago

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

6 0

3 years ago