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3 years ago

What formula relstes work n power

Physics

1 answer:

777dan777 [17]3 years ago

8 0

<span>Raj is trying to make a diagram to show what he has learned about nuclear fusion.
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Find the required answer to the following by substituting the numbers. (Bonus points for knowing the units). PLEASE ANSWER

Alex73 [517]

Answer:

a) F=ma

F=13.2×10

F=132

b) I=P/V

I=75/15

I=5

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3 years ago

Ts to

Leni [432]

The charge not experience a net magnetic force if it moves to the right.

<h3>What is the magnetic force?</h3>

The magnetic force is a force that is felt by an object owing to the presence of a magnet. We should note that when a charge is placed in a magnetic field, it experiences a magnetic force.

The direction in which the charge will experience no net magnetic force is when the charge moves towards the right.

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2 years ago

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3 years ago

a 3.00 kg block moving 2.09 m/s right hits a 2.22 kg block moving 3.92 m/s left. afterward, the 3.00 kg block moves 1.11 m/s lef

LuckyWell [14K]

The momentum of the 2.22 kg afterwards is 0.91 kg m/s to the right

Explanation:

We can solve this problem by applying the law of conservation of momentum.

In fact, the total momentum before and after the collision must be conserved. So, we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 3.00 kg$ is the mass of the first block

$u_1 = 2.09 m/s$ is the initial velocity of the first block (we take the right as positive direction)

$v_1 = -1.11 m/s$ is the final velocity of the first block (to the left)

$m_2 = 2.22 kg$ is the mass of the second block

$u_2 = -3.92$ is the initial velocity of the second block

$v_2$ is the final velocity of the second block

Re-arranging the equation and substituting the values, we find: the final velocity of the second block:

$v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(3.00)(2.09)+(2.22)(-3.92)-(3.00)(-1.11)}{2.22}=0.41 m/s$

(to the right, since it is positive)

And so, the momentum of the 2.22 kg block afterwards is:

$p_2 = m_2 v_2 = (2.22)(0.41)=0.91 kg m/s$ (to the right)

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3 years ago

Monochromatic light is incident on a grating that is 75 mmwide

statuscvo [17]

Answer:

Explanation:

Given

Grating Width $w=75 mm$

no of lines $N=50,000 lines$

inclination of second order maxima $\theta =32.5^{\circ}$

Condition of maxima For Diffraction grating

$d\sin \theta =m\lambda$

where $d=Grating\ constant$

$\lambda =wavelength$

$m=order\ of\ maxima$

$d=\frac{width}{N}$

$d=\frac{75}{50,000}$

$d=0.0015\ mm$

therefore for second order maxima wavelength of light is given by

$\lambda =\frac{d\sin \theta }{m}$

$\lambda =\frac{0.0015\times \sin (32.5)}{2}$

$\lambda =403\times 10^{-6}\ m$

$\lambda =403\ nm$

3 0

3 years ago