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3 years ago

What is the terminal velocity of a penny falls off a skyscraper

Physics

1 answer:

Naya [18.7K]3 years ago

7 0

The penny will reach terminal velocity at 50 ft. Then it will travel 25 mph till it reaches ground.

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You have a mass of 95 kg.<br> a. What is your weight on Earth?

svp [43]

Answer:

931.63N

Hope this helps :D

8 0

3 years ago

A) A spaceship passes you at a speed of 0.800c. You measure its length to be 31.2 m .How long would it be when at rest?

rosijanka [135]

Answer:

$l_o =52 \ m$

$l = 37.13 \ LY$

Explanation:

From the question we are told that

The speed of the spaceship is $v = 0.800c$

Here c is the speed of light with value $c = 3.0*10^{8} \ m/s$

The length is $l = 31.2 \ m$

The distance of the star for earth is $d = 145 \ light \ years$

The speed is $v_s = 2.90 *10^{8}$

Generally the from the length contraction equation we have that

$l = l_o \sqrt{1 -[\frac{v}{c } ]}$

Now the when at rest the length is $l_o$

$l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }$

$l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }$

$l_o=52 \ m$

Considering b

Applying above equation

$l =l_o \sqrt{1 - [\frac{v}{c } ]}$

Here $l_o =145 \ LY(light \ years )$

$l=145 * \sqrt{1 - \frac{v_s^2}{c^2 } }$

$l =145 * \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }$

$l = 37.13 \ LY$

4 0

3 years ago

If a 400-mm diameter pipe with a pipe roughness coefficient of 100 flows full of pressurized water with a head loss of 0.4 ft pe

RoseWind [281]

Answer:

Q = 913.9 gpm

Explanation:

The Hazen Williams equation can be written as follows:

$P = \frac{4.52\ Q^{1.85}}{C^{1.85}d^{4.87}}$

where,

P = Friction Loss per foot of pipe = $\frac{0.4}{1000\ ft}$ = 4 x 10⁻⁴

Q = Flow Rate in gallon/min (gpm) = ?

d = pipe diameter in inches = (400 mm)(0.0393701 in/1 mm) = 15.75 in

C = roughness coefficient = 100

Therefore,

$4\ x \ 10^{-4} = \frac{4.52\ Q^{1.85}}{(100)^{1.85}(15.75)^{4.87}}\\\\Q^{1.85} = \frac{4\ x \ 10^{-4}}{1.33\ x\ 10^{-9}} \\\\Q = (300384.75)^\frac{1}{1.85}$

5 0

3 years ago

2O POINTS!!! Easy Question. Will Mark Brainiest

wel

Answer: The Electrostatic force of attraction or repulsion between two charges shows that the Newton's third law applies to electrostatic forces.

Explanation: Consider two Oppositely charged charges separated by distance d.

The electrostatic force exerted by charge 1 on charge 2 is.

By Coulomb's Law :

F1 = k $\frac{Q1 Q2}{d2}$ .....................................(1)

The electrostatic force exerted by charge 2 on charge 1 is.

F2 = - k $\frac{Q1 Q2}{d2}$ ................................. (2)

negative sign shows that force are in opposite direction.

From Equation 1 and 2

F1 = - F2

Which implies Newton Third law.

6 0

3 years ago

A camera has a single converging lens with a fixed focal length f. (a) How far should the lens be from the film (or in a present

Varvara68 [4.7K]

Answer:

a) Due to the characteristic that a converging lens focuses light rays from infinity and parallel to its main axis. Therefore, the lens should be placed at a distance "f" from the film, in this way it will form the image of the object placed at infinity in said film.

b) Since the converging lens produces an image of an object placed at a distance of 2f, the lens must be placed at the same distance (2f), so that this object that is placed at a distance of 2f is focused.

Explanation:

4 0

3 years ago