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Bumek [7]
3 years ago
6

What is gravity?what is one solar day?​

Physics
1 answer:
mafiozo [28]3 years ago
6 0

Answer:

<h2>Gravity :</h2><h3>the force that attracts a body towards the centre of the earth, or towards any other physical body having mass.</h3>

<h2>Solar day</h2><h3>A solar day is the time it takes for the Earth to rotate about its axis so that the Sun appears in the same position in the sky.</h3><h2> or</h2><h3>It is the time between successive meridian transits of the sun at a particular place.</h3>
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3 years ago
A 3) How far will 20 N of force stretch a spring with a spring constant of 140 N/m?​
anyanavicka [17]

Answer:

I think the answer is

Explanation:

140N/m/20N =

7m is the answer

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2 years ago
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3 years ago
The speed of light in a vacuum is 2.99x10^8 m/s. calculate its speed in miles per hour.
34kurt
You'll never get the correct answer without the correct conversion factor. Note carefully that you have no decimal. It should be 
<span>1 km = 0.6214 miles </span>
<span>1000 m = 1 km </span>
<span>60 seconds = 1 minute </span>
<span>60 minutes = 1 hour. </span>
<span>2.998E8 m/s x (1 km/1000m) x (0.6214 miles/km) x (60 sec/min) x (60 min/hr) = ?</span>
6 0
3 years ago
A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^
xeze [42]

Answer:

v_{max}=52.38\frac{m}{s}

v_{100}=33.81

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

\sum{F}=0=F_d-W

F_d=W

kv_{max}^2=m*g

v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

\sum{F}=ma=W-F_d

ma=W-F_d

ma=mg-kv_{100}^2

a=g-\frac{kv_{100}^2}{m} (1)

consider the next equation of motion:

a = \frac{(v_{x}-v_0)^2}{2x}

If assuming initial velocity=0:

a = \frac{v_{100}^2}{2x} (2)

joining (1) and (2):

\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}

\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g

v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g

v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}

v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}} (3)

v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}

v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}

v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}

v_{100}=\sqrt{1,143.3}

v_{100}=33.81

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

v = v_0 +at

as stated before, the initial velocity is 0:

v =at (4)

joining (1) and (4) and reducing you will get:

\frac{kt}{m}v^2+v-gt=0

solving for v:

v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }

Plots:

5 0
3 years ago
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