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Vesnalui [34]
3 years ago
13

Consider a cylindrical crew module with a diameter of 5 meters, a length of 15 meters, and a crew of 6 people. The module contai

ns air that is at
standard sea-level atmospheric pressure and temperature. The sea level density of air at 20 degrees Celsius is 1.204 kg/m3.
What is the volume of the crew module? Provide your answer in m' with at least 2 decimal places.​
Physics
1 answer:
UNO [17]3 years ago
7 0

Answer:

  294.524 m^3

Explanation:

The volume of a cylinder is given by the formula ...

  V = πr^2h

For a radius of 2.5 m and a length of 15 m, the volume is ...

  V = π(2.5 m)^2(15 m) = 93.75π m^3 ≈ 294.524 m^3

The volume is about 294.52 cubic meters.

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Woman pulls a 6.87 kg suitcase,
GenaCL600 [577]

Answer:

2.24 m/s

Explanation:

resolving force of 29.2 N in x component

Fx = 29.2 cos 57.7

Fx = 15.6N

as force of friction is 12.7 N hence net force which produces acceleration is

15.6-12.7=2.9 N

by Newton 's law a=f/m

a= 2.9/6.87=0.422 m/s^2

now equation of motion is

v^2= U^2+2as

 = 0^2+2(.422)(5.93)

v^2=5.00

v=2.24 m/s

4 0
2 years ago
Dean Potter is known for slacklinging across a 915 metter deep ravine in Yosemite National Park with no safety gear. If his mass
Natasha2012 [34]

Explanation:

answer 689910

M x G x H

mass = 77

G= 9.8

Meter = 915 this is the height

77 x 9.8= 754.6

754.6 X 915 = 689910 J

8 0
3 years ago
4. How much would an object accelerate if it has a mass of 25 kg and is pushed with a force of 2 Newtons?
slega [8]

Answer:

0.4

Explanation:

f divided by m = a

3 0
3 years ago
If we decrease the amount of force, and keep all other factors the same, what will happen to the amount of work?
kozerog [31]
The answer is A. I am sure of it.
8 0
3 years ago
Read 2 more answers
Two long, parallel wires are separated by a distance of 2.2 cm. The force per unit length that each wire exerts on the other is
VikaD [51]

Answer:

i_2 = 7.6 A

Explanation:

As we know that the force per unit length of two parallel current carrying wires is given as

F = \frac{\mu_o i_1 i_2}{2\pi d}

here we know that

F = 3.6 \times 10^{-5} N/m

i_1 = 0.52 A

d = 2.2 cm

now from above equation we have

3.6 \times 10^{-5} = \frac{4\pi \times 10^{-7} (0.52)(i_2)}{2\pi (0.022)}

3.6 \times 10^{-5} = 4.73 \times 10^{-6} i_2

i_2 = 7.6 A

5 0
2 years ago
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