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3 years ago

7

As the earth rotates through one revolution, a person standing on the equator traces out a circular path whose radius is equal t

o the radius of the earth (6.38×10^6). What is the average speed of this person in meters per second? If we are standing in latitude 43° north what is our average speed?

1 answer:

mrs_skeptik [129]3 years ago

6 0

Answer:

The average speed will be "1038 mph".

Explanation:

Period for one revolution is:

T = 24 hours

= 86,400 sec

x = c = 2πr

The given values is:

r = 6.38×10⁶ m

Now,

⇒ $T=\frac{2 \pi r}{v}$

Or,

⇒ $v=\frac{2 \pi r}{T}$

On substituting the values, we get

$=\frac{2 \pi (6.38\times 10^6)}{86,400}$

$=464 \ min/sec$

$=(464 \frac{min}{sec}) (\frac{3600 \ sec}{1 \ hour} )(\frac{1 \ mi}{1609 \ m} )$

$=1038 \ mph$

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2 years ago

By what potential difference must a proton [m_0 = 1.67E-27 kg) be accelerated to have a wavelength lambda = 4.23E-12 m? By what

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Explanation:

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We need to find the potential difference. The relationship between potential difference and wavelength is given by :

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2 years ago

A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.665 m. W

Liula [17]

Answer:

Impulse is 1.239 kg.m/s in upward direction

Explanation:

Taking upward motion as positive and downward motion as negative.

Downward motion:

Given:

Mass of ball (m) = 0.150 kg

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Initial velocity (u) = 0 m/s

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Using equation of motion, we have:

$v_d^2=u^2+2aS\\\\v=\pm\sqrt{u^2+2aS}\\\\v_d=\pm\sqrt{0+2\times -9.8\times -1.25}\\\\v_d=\pm\sqrt{24.5}=\pm4.95\ m/s$

Since, the motion is downward, final velocity must be negative. So,

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Upward motion:

Given:

Displacement of ball (S) = 0.665 m

Initial velocity ( $v_d$ ) = 4.95 m/s(Upward direction)

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

$v_{up}^2=v_d^2+2aS\\\\v_{up}=\pm\sqrt{v_d^2+2aS}\\\\v_{up}=\pm\sqrt{24.5+2\times -9.8\times 0.665}\\\\v_{up}=\pm\sqrt{10.966}=\pm3.31\ m/s$

Since, the motion is upward, final velocity must be positive. So,

$v_{up}=3.31\ m/s$

Now, impulse is equal to change in momentum. So,

Impulse = Final momentum - Initial momentum

$J=m(v_{up}-v_d)\\\\J=(0.150\ kg)(3.31-(-4.95))\ m/s\\\\J=0.150\ kg\times 8.26\ m/s\\\\J=1.239\ Ns$

Therefore, the impulse given to the ball by the floor is 1.239 kg.m/s in upward direction.

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