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3 years ago

When doing a squat, how do you do it without getting hurt?

Physics

2 answers:

Aleksandr [31]3 years ago

7 0

I do weight lifting so you need feet flat on floor but out back straight and also a spotter

padilas [110]3 years ago

6 0

Make sure your feet and shoulder width apart, and parallel to each other also when you do the squat make sure your knees dont pass your toes and keep your back straight!:) have fun

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A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.3 m/s rel

Sonja [21]

consider east-west direction along X-axis and north-south direction along Y-axis

$V_{ra}$ = velocity of migrating robin relative to air = 12 j m/s

(where "j" is unit vector in Y-direction)

$V_{ag}$ = velocity of air relative to ground = 6.3 i m/s

(where "i" is unit vector in X-direction)

$V_{rg}$ = velocity of migrating robin relative to ground = ?

using the equation

$V_{rg}$ = $V_{ra}$ + $V_{ag}$

$V_{rg}$ = 12 j + 6.3 i

$V_{rg}$ = 6.3 i + 12 j

magnitude : sqrt((6.3)² + (12)²) = 13.6 m/s

direction : tan⁻¹(12/6.3) = 62.3 deg north of east

4 0

3 years ago

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

3 years ago

Which of the following statements apply to electric charges?

Gre4nikov [31]

Answer:

The statement "If a positively charged rod is brought close to a positively charged object, the two objects will repel " applies to electric charges.

Explanation:

There are only two types of electric charges. Both having own magnitude but different charge.

1. Positive charge

2. Negative charge

Like charges repel each other and opposite charges always attract each other.

When a positively charged rod is brought close to a positively charged object, the rod and the object will repel.

6 0

3 years ago

A circular loop with radius r is rotating with constant angular velocity ω in a uniform electric field with magnitude E. The axi

inn [45]

Answer:

$\Phi_{E} = E\pi r^2 \omega t$

Explanation:

The electric flux is defined as the multiple of electric field and the area that the electric field passes through, such that

$\Phi_{E} = \vec{E}\vec{A}$

When calculating the electric flux, the angle between the directions of electric field and the area becomes important, especially if the angle is changing with time.

The above formula can be rewritten as follows

$\Phi_{E} = EA\cos(\theta)$

where θ is the angle between the electric field and the area of the loop. Note that, the direction of the area of the loop is perpendicular to the plane of the loop.

If the loop is rotating with constant angular velocity ω, then the angle can be written as follows

$\theta = \omega t$

At t = 0, cos(0) = 1 and the electric flux through the loop is at its maximum value.

Therefore the electric flux can be written as a function of time

$\Phi_{E} = E\pi r^2 \omega t$

3 0

3 years ago

The alternative to nuclear fission reactors in a nuclear fusion reactor. Explain why it is much more difficult to get a fusion r

Sliva [168]

Answer: The major challenges are as

1) understanding of the plasma: Plasma is a soup like mixture of subatomic particles of different atoms nuclei and electrons that are shattered apart by the temperature at which plasma is formed. further research is needed to understand the behavior of plasma so that it can be put to a proper use.

2) Confinement of plasma: Once we get the plasma we need to hold it so that we can obtain heat from it to drive a steam turbine but the sheer temperature of plasma is in millions of Celsius thus currently making it impossible to confine conventionally. Scientists use a loop of electric and magnetic fields to keep it in circulatory like manner so that it can be studied.

3) finally to obtain electricity from the plasma it should be stable to produce electricity. But currently to obtain pressure, temperature so that we have a sustained supply is highly difficult in technical and economical aspects.

Inertial confinement: In order to get the nuclei of atoms close enough for fusion this type of method used compression of the nuclei into highly small volumes.This is accomplished by use of lasers which are directed towards the fuel pellets that implode and travel towards other nuclei making fusion possible. It's main advantage is that it requires lesser time to initiate fusion but the disadvantage being that a large power is used to fire the lasers and the lasers should all hit the small target.

Magnetic Confinement: In this method we use a magnetic and electric fields in a properly designed space to keep the plasma in motion. In motion the nuclei of the atoms come close enough to initiate fusion.It's advantage being less power is required to start the process as compared to inertial confinement and the disadvantage being that plasma confinement is currently not properly understood.

5 0

3 years ago