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Arisa [49]
1 year ago
13

An empty container is filled with helium to a pressure P at a temperature T. Neon, which has atoms that are 5 times more massive

than those of helium, is then added to the container until the new pressure in the container is 2P at the same temperature T. The ratio of the average speed of the helium atoms to the average speed of the neon atoms is:_________
a. 25 to 1
b. 5 to 1
c. 5to 1
d. 2to 1 E
e. 1 to 1
Physics
2 answers:
Law Incorporation [45]1 year ago
4 0

Answer:

I got the same test question I picked c

ahrayia [7]1 year ago
3 0

The ratio of the average speed of the helium atoms to the average speed of the neon atoms is 5 to 1 C.

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0.6295 A

Explanation:

I=mg/BL put values in this formula.  

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A motorcycle of mass 100 kilograms travels around a flat, circular track of radius 10 meters with a constant speed of 20 meters
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100/10 = 10 , 10 × 10 = 100÷20 = 5

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Caelyn wanted to find out what shampoo made her hair the shiniest . Everyday she washed her hair with different shampoos and the
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Answer:

IV: type of shampoo used

DV: what shampoo made her hair the skinniest

control: water

Constant: rated from scale from 1-10

Explanation:

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2 years ago
A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0° below the horizontal. It is held in position by a fri
liubo4ka [24]

Answer:

T = 7.64 kN

F_y = 0.52 kN(Downwards)

F_x = 3.23 kN (Towards Left)

Explanation:

As we know that beam is in equilibrium

So here we can use torque balance as well as force balance for the beam

Now by torque balance equation at the pivot we can say

F(4.50 cos\theta) + mg(2cos\theta) = T \times 3

As we know that

mg = 1.40 kN

F = 5 kN

so we will have

5 kN(4.50 cos25) + 1.40 kN(2 cos25) = 3 T

T = 7.64 kN

Now force balance in vertical direction

F + mg = Tsin65 + F_y

5 + 1.40 = 7.64 sin65 + F_y

F_y = 0.52 kN(Downwards)

Force balance in horizontal direction

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7 0
2 years ago
a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w
AysviL [449]

Answer:

His resulting velocity will be 0.187 m/s backwards.

Explanation:

Given:

Mass of the man is, M=75\ kg

Mass of the ball is, m=4\ kg

Initial velocity of the man is, u_m=0\ m/s(rest)

Initial velocity of the ball is, u_b=0\ m/s(rest)

Final velocity of the ball is, v_b=3.50\ m/s

Final velocity of the man is, v_m=?\ m/s

In order to solve this problem, we apply law of conservation of momentum.

It states that sum of initial momentum is equal to the sum of final momentum.

Momentum is the product of mass and velocity.

Initial momentum = Initial momentum of man and ball

Initial momentum = Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm

Final momentum = Final momentum of man and ball

Final momentum = Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14

Now, initial momentum = final momentum

0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s

The negative sign implies backward motion of the man.

Therefore, his resulting velocity is 0.187 m/s backwards.

3 0
2 years ago
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