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borishaifa [10]
3 years ago
10

The escape velocity on earth is 11.2 km/s. What fraction of the escape velocity is the rms speed of H2 at a temperature of 31.0

degrees Celsius on the earth? Note that virtually all the molecules will have escaped the earth's atmosphere if this fraction exceeds 0.15.
Physics
1 answer:
mash [69]3 years ago
5 0

To solve this problem it is necessary to apply the concept related to root mean square velocity, which can be expressed as

v_{rms} = \sqrt{\frac{3RT}{n}}

Where,

T = Temperature

R = Gas ideal constant

n = Number of moles in grams.

Our values are given as

v_e =11.2km/s = 11200m/s

The temperature is

T = 30\°C = 30+273 = 303K

Therefore the root mean square velocity would be

v_{rms} = \sqrt{\frac{3(8.314)(303)}{0.002}}

v_{rms} = 1943.9m/s

The fraction of velocity then can be calculated between the escape velocity and the root mean square velocity

\alpha = \frac{v_{rms}}{v_e}

\alpha = \frac{1943.9}{11200}

\alpha = 0.1736

Therefore the fraction of the scape velocity on the earth for molecula hydrogen is 0.1736

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leonid [27]

Answer:

6.6 N

Explanation:

Let's take the direction of the force of 4.0 N as positive x-direction. This means that the force of 3.0 N is at 40 degrees above it. So the components of the two forces along the x- and y-directions are:

F_{1x} = 4.0 N\\F_{1y} = 0

F_{2x} = 3.0 N cos 40^{\circ}=2.3 N\\F_{2y} = 3.0 N sin 40^{\circ} = 1.9 N

So the resultant has components

F_x = F_{1x}+F_{2x}=4.0 N +2.3 N = 6.3 N\\F_y = F_{1y} + F_{2y} = 0 + 1.9 N = 1.9 N

So the magnitude of the resultant is

F=\sqrt{F_x^2 +F_y^2}=\sqrt{(6.3)^2+(1.9)^2}=6.6 N

And in order for the body to be balanced, the third force must be equal and opposite (in direction) to this force: so, the magnitude of the third force must be 6.6 N.

3 0
3 years ago
On Earth, a spring stretches by 5.0 cm when a mass of 3.0 kg is suspended from one end.
Neko [114]

Answer:

Mass = 18.0 kg

Explanation:

From Hooke's law,

F = ke

where: F is the force, k is the spring constant and e is the extension.

But, F = mg

So that,

mg = ke

On the Earth, let the gravitational force be 10 m/s^{2}.

3.0 x 10 = k x 5.0

30 = 5k

⇒ k = \frac{30}{5} ................ 1

On the Moon, the gravitational force is \frac{1}{6} of that on the Earth.

m x \frac{10}{6} = k x 5.0

\frac{10m}{6} = 5k

⇒ k = \frac{10m}{30} ............. 2

Equating 1 and 2, we have;

\frac{30}{5}  = \frac{10m}{30}

m = \frac{900}{50}

    = 18.0

m = 18.0 kg

The mass required to produce the same extension on the Moon is 18 kg.

8 0
3 years ago
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A car traveling with constant speed travels 150 km in 7200s what is the speed of the car
Oksana_A [137]
Velocity is distance/time 

so 150/7200=.0208km/s 

unless you have to convert it to miles or something else. but use the formula!
5 0
3 years ago
A bullet with a mass m b = 11.5 g is fired into a block of wood at velocity v b = 249 m/s. The block is attached to a spring tha
Ostrovityanka [42]

Answer:

0.358Kg

Explanation:

The potential energy in the spring at full compression = the initial kinetic energy of the bullet/block system

0.5Ke^2 = 0.5Mv^2

0.5(205)(0.35)^2 = 12.56 J = 0.5(M + 0.0115)v^2

Using conservation of momentum between the bullet and the block

0.0115(265) = (M + 0.0115)v

3.0475 = (M + 0.0115)v

v = 3.0475/(M + 0.0115)

plugging into Energy equation

12.56 = 0.5(M + 0.0115)(3.0475)^2/(M + 0.0115)^2

12.56 = 0.5 × 3.0475^2 / ( M + 0.0115 )

12.56 = 0.5 × 9.2872/ M + 0.0115

12.56 = 4.6436/ M + 0.0115

12.56 ( M + 0.0115 ) = 4.6436

12.56M + 0.1444 = 4.6436

12.56M = 4.6436 - 0.1444

12.56 M = 4.4992

M = 4.4992÷12.56

M = 0.358 Kg

4 0
3 years ago
Sics and Chemistry A (2nd Six Weeks) - SC3200SFA/01
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Force = Mass * Acceleration therefore the red ball with the higher mass will have more force and greater acceleration
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