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Georgia [21]
3 years ago
13

Calculate the rms speed of helium atoms near the surface of the Sun at a temperature of about 5100 K. Express your answer to two

significant figures and include the appropriate units.
Physics
1 answer:
lianna [129]3 years ago
4 0

Answer:

V_{rms}=5.6*10^3m/s

Explanation:

From the question we are told that:

Temperature T=5100K

Generally the equation for RMS Speed is mathematically given by

 V_{rms}=\sqrt{\frac{3kT}{m}}

Where

 K=Boltzman's constant

 K=1.38*10^{-23}

And

 M=molecular mass

 M=4*1.67*10^{-27}

 V_{rms}=\sqrt{\frac{3(1.38*10^{-23})5100}{4*1.67*10^{-27}}}

 V_{rms}=5.6*10^3m/s

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a car is traveling 20 m/s. it takes 4 seconds with a 7,500 n force to come to a stop. what is the mass of the car​
aev [14]

Answer:

mass equals 375 kg

Explanation:

f=ma

f=7500

a=20

m=f/a

m=7500/20

m=375

4 0
3 years ago
In a classical carnival ride, patrons stand against the wall in a cylindrically shaped room. Once the room gets spinning fast en
g100num [7]

Answer:

- the speed of a person "stuck" to the wall is 14.8 m/s

- the normal force of the wall on a rider of m=54kg is 1851 N

- the minimum coefficient of friction needed between the wall and the person is 0.29

Explanation:

Given information:

the radius of the cylindrical room, R = 6.4 m

the room spin with frequency, ω =  22.1 rev/minutes = 22.1 \frac{2\pi }{60} = 2.31 rad/s

mass of rider, m = 54 kg

the speed of a person "stuck" to the wall

v = ω R

  = 2.31 x 6.4

  = 14.8 m/s

the normal force of the wall on a rider

F = m a

a  = ω^2 R

   =  \frac{v^{2} }{R^{2} } R

   = \frac{v^{2} }{R}

F = \frac{mv^{2} }{R}

  = \frac{(54)(14.8)^{2} }{6.4}

  = 1851 N

the minimum coefficient of friction needed between the wall and the person

F(friction) = μ N

W =  μ N

m g =  μ \frac{mv^{2} }{R}

g = μ \frac{v^{2} }{R}

μ = \frac{gR}{v^{2} }

  = \frac{(9.8) (6.4)}{14.8^{2} }

  = 0.29

5 0
3 years ago
Describe how the pendulum works.
Vinil7 [7]

Answer:

A pendulum works by converting energy back and forth, a bit like a rollercoaster ride. When the bob is highest (furthest from the ground), it has maximum stored energy (potential energy). ... So as the bob swings (oscillates) back and forth, it repeatedly switches its energy back and forth between potential and kinetic.

6 0
3 years ago
A 3.00 kg stone is dropped from a 39.2 m high building. when the stone has fallen 19.6 m, the magnitude of the impulse the earth
tatiyna
<span>The formulas are, Impulse = mv-mu ....... (1) v^2 = u^2 + 2as .......... (2) We know that, u=0 a=acceleration=gravity = 9.80665 m/s^2 = 9.81 m/s^2 s=19.6 sub (2) we get, v^2 = 0+ 2*9.81*19.6 v^2 = 2*9.81*19.6 v^2 = 384.552 v = 19.6099 v = 19.61 m/s Sub v=19.61 m/s in (1) we get, Impulse = mv - mu we know that u=0; v= 19.61 m/s; m= 3.00 kg Impulse = 3(19.61) - 3(0) Impulse = 58.83-0 Impulse = 58.83 Ns. Therefore the gravitational force exerted by the stone is 58.83 Ns.</span>
4 0
4 years ago
For how long should a force of 100N act on a body of mass 20kg at rest so that it acquires a velocity of 100m/s?​
julia-pushkina [17]

Answer:

20 seconds

Explanation:

7 0
3 years ago
Read 2 more answers
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