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2 years ago

10

If anyone knows momentum can you explain it? the khan academy problems are literally killing me

1 answer:

kupik [55]2 years ago

4 0

The body's mass and velocity are multiplied to create momentum. It happens as a result of the force used. Kg m/s² is its unit. Measurement of mass in motion is momentum.

<h3 /><h3>What is momentum?</h3>

The body's mass and velocity are multiplied to create momentum. It happens as a result of the applied force and has the unit Kg-m/s².

Equal amounts of force and momentum are applied. More quantitative than the first rule of motion, the second law of motion is usually used to describe what happens when a force is applied.

The momentum is found as;

P =mv

Where m is the mass in kilogram

v is the velocity in m/sec

P is the momentum in the kg-m/s²

The second law of motion is more quantitative, and it is frequently used to quantify what occurs when a force is applied. The second law of Newton is discussed in depth in this article.

Force and the rate of change of momentum are equal. The force is given as the product of mass and change of velocity. The impulsive force is given by the change in momentum.

I=ΔP

Hence, the body's mass and velocity are multiplied to create momentum.

To learn more about the momentum refer to the link;

brainly.com/question/4956182

#SPJ1

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<u>Explanation</u>:

We know that,

$\text { Electric power }(p)=\frac{V^{2}}{R}$

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For calculating number of turns

$\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

Given that,

$80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }$

$\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}$

$\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .$

We need to find the number of turns in the secondary winding $\left(N_{S}\right)$ to run the bulb at 120W $\left(P_{2}\right)$

Firstly find the secondary voltage in the transformer use, $\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

$\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}$

$V_{2}^{2}=\frac{120^{2} \times 120}{80}$

$V_{2}^{2}=\frac{1728000}{80}$

$V_{2}^{2}=21600$

$V_{2}=\sqrt{21600}$

$V_{2}=146.9 \mathrm{V}=V_{S}$

Now, finding the number of turns in secondary coil. Use, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

$\frac{30}{N_{S}}=\frac{65}{146.9}$

$N_{S}=\frac{30 \times 146.9}{65}$

$N_{S}=\frac{4407}{65}$ $N_{S}=67.8$

The number of turns in the secondary winding are 67.8 turns.

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