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Nuetrik [128]
3 years ago
6

A sample of O2 occupies 75 L at 1 atm. If the volume of the

Physics
1 answer:
Westkost [7]3 years ago
3 0

Answer:

1/2 atm

Explanation:

Givens

The temperature remains constant so the formula is

P = 1 atm

V = 75 L

P1 = ?

V1 = 75 * 2 = 150 L

Formula

P*V = P1* V1

Solution

1 * 75 = P1 * 150      Divide by 150

75/150 = P1

P1 = 1/2 atmospheres

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What is 3600Hz has in rpm?​
Setler [38]

Answer:

Explanation:

N=Rotor Speed in Revolution per minute(rpm)

for P=4 and N=3600, f comes out to be 120 Hz.

So frequency of voltage produced is 120 Hz. But this is not practical. Generally 4-Pole generator has N=1500rpm(for 50 Hz) or 1800rpm for 60 Hz. Two pole generator can have N=3600rpm(f=60Hz).

The most practical situation is generator having N=3600Hz with 2 Poles.

Hope It will be helpful!!!

8 0
3 years ago
A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit
Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M=8.7\cdot 10^{13}kg is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

m\frac{v^2}{r}=G\frac{Mm}{r^2}

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s

3 0
3 years ago
Convert the speed of light 3.0x10^8 m/s to km/day
Aleksandr [31]

Answer: 2.592 \times 10^{8}km/day

1 m = 0.001 km\\ 1 s= 1.157\times10^{-5} days\\ 1 m/s = \frac {0.001}{1.157\times10^{-5}} km/day = 86.4 km/day \\ 3.0\times 10^{8} m/s = 3.0\times 10^{8} m/s \times \frac {86.4 km/day}{1m/s} =2.592 \times 10^{8}km/day



7 0
3 years ago
Read 2 more answers
Why is understanding the concept of forces, friction and gravity important?
nydimaria [60]
Because it's the basis of how everything around you works
8 0
3 years ago
A circular 10-turn coil with a radius of 5.0 cm carries a current of 5 A. It lies in the xy plane in a uniform magnetic field =
Tju [1.3M]

Answer:

U = – 0.12J

Explanation:

Given N = 10 turns, I = 5A, r = 5×10-²m

B^ = 0.05 T iˆ+ 0.3 T kˆ

Magnitude of the magnetic field vector B = √(0.05²+0.3²) = 0.304T

Area = πr² = π(5×10-²)² = 7.85×10-³m²

Magnetic moment μ = NIA

μ = 10×5×7.85×10-³ = 0.3925Am²

U = -μ•B = –0.3925×0.304 = –0.12J

The sign is negative because the magnetic moment is aligned with the magnetic field.

3 0
3 years ago
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