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2 years ago

A nuclear particle with no charge

Physics

1 answer:

pishuonlain [190]2 years ago

7 0

A nuclear particle with no charge is a neutron

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Now explore friction force. Set the piece of plastic or wood on the table and push it steadily across the tabletop using your fi

ehidna [41]

Answer:

The tabletop is smooth so my finger is down it fast and easy. The fabric however slowed my finger down considerably, and it was harder for me to move my finger across it.

Explanation:

Hope this helps.

3 0

2 years ago

A 2 kg ball is moving 3 m/s when it starts rolling up a hill.

AURORKA [14]

Answer:

the height reached is = 0.458 [m]

Explanation:

We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:

$Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]$

Replacing the values on the equation we have:

$Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\$

This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.

$Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\$

Now we have:

$h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]$

In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]

6 0

3 years ago

Find the angle for the third-order maximum for 591 nm wavelength light falling on a diffraction grating having 1460 lines per ce

Marat540 [252]

Answer:

15.32°

Explanation:

We have given the wavelength $\lambda =591nm=591\times 10^{-9}m$

Diffraction grating is 1460 lines per cm

So $d=\frac{10^{-2}}{1460}=6.71\times 10^{-6}m$ (as 1 m=100 cm )

For maximum diffraction

$dsin\Theta =m\lambda$ here m is order of diffraction

So $6.71\times 10^{-6}sin\Theta =3\times 591\times 10^{-9}$

$sin\Theta =0.264$

$\Theta =15.32^{\circ}$

6 0

2 years ago

What would happen to the moon if there was no sun and earth ?

iragen [17]

Answer:

High tides would be much smaller than they are now, and low tides would be even lower. This is because the sun would be influencing the tides, not the moon; however, the sun has a weaker pull, which would decrease the tides. ... Winds could become much faster and much stronger without the moon.

Explanation:

google

4 0

3 years ago

PLEASE HELP ME WITH MY HOMEWORK!!!!!

IgorC [24]

Known :

l = 7 cm

w = 4 cm

Asked :

h = ...?

Answer :

V = B triangle × h (long)

35 = ½ × 4 × h × 7

35 = ½ × 28 × h

35 = 14 h

h = 35 ÷ 14

h = 2,5 cm

Sorry if I am wrong, I only study

5 0

3 years ago