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9966 [12]
4 years ago
5

Consider two ideal gases, A & B, at the same temperature. The rms speed of the molecules of gas A is twice that of gas B. Ho

w does the molecular mass of A compare to that of B? (a) It is twice that of B. (b) It is one half that of B (c) It is four times that of B (d) It is one fourth that of B. (e) It is 1.4 times that of B.
Physics
1 answer:
uysha [10]4 years ago
7 0

Answer:

option (d)

Explanation:

The relation between the rms velocity and the molecular mass is given by

v   proportional to  \frac{1}{\sqrt{M}} keeping the temperature constant

So for two gases

\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}

\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}

{\frac{M_{B}}{M_{A}}} = 4

{\frac{M_{B}}{4}} = M_{A}

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Can anyone help me on question 13
Leona [35]

Given that in a parallel circuit:

R1 = 12 ohms

R2= 15 ohms

I = 12 A

I2 = 4 A

V=?

R=?

R3 =?

P=?

Since,

V= IR

or,

V2 = I2 * R2

V2= 4* 15

V2 = 60V

Since in a parallel circuit voltage remain same in all component of the circuit and is equal to the source voltage.

Therefore,

V= V1 = V2 = V3 = 60V

Since,

V= IR

R= V/I

R= 60/12

R= 5 ohm

That is total resistance is equal to 5 ohms.

Since for parallel circuit,

1/R= 1/R1 + 1/R2 + 1/R3

1/5= 1/12+ 1/15 + 1/R3

or

1/R3= 1/5- 1/12- 1/15

1/R3= 1/20

or

R3= 20 ohms

Since,

V=IR

I= V/R

I1= V1/ R1

I1= 60/12

I1= 5 A

I3= V3/R3

I3= 60/20

I3= 3A

Since,

P=VI

P= 60*12

P=  720 watt

P1= V1* I1

P1= 60* 5

P1= 300 watt

P2= V2* I2

P2= 60* 4

P2= 240watt

P3= V3*I3

P3= 60*3

P3= 180 watt

Hence we have,

R1= 12 ohms , R2= 15 ohms, R3= 20 ohms, R= 5 ohms

I1= 5A, I2= 4A, I3= 3A, I= 12 A

V1= V2= V3= V= 60V

P1= 300 watt, P2= 240 watt, P3 = 180 watt, P= 720 watt



6 0
3 years ago
In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of
aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation 

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

(c) the angular speed of the merry-go-round and child after the child has jumped on.

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

7 0
3 years ago
Please help me! :( <br> I will give brainliest
-BARSIC- [3]
It’s 8 babe good luck
8 0
3 years ago
If a person (weighs 70kg) jumped of a moving car (at 100km/h) and fell on asphalt what is the amount of force applyied to his bo
jok3333 [9.3K]

Answer:

The amount of force applied to his body is 1944.44 N

<em>The chances of the person dying is very high owing to the high impact force with which the person would experience when he or she lands on the asphalt road due to the jump out of the moving car.</em>

Explanation:

We all know that,

F = Ma where,

F = Force

M = weight of the person

a = acceleration or velocity of the moving car

Therefore;

F = { 70 x (100 x 1000) } / [3600]

= [7 000 000] / 3600

= <u>1944.44 N</u>

6 0
3 years ago
What is the electrical force between q1 and q2? Recall that k = 8.99 × 109 N•meters squared over Coulombs squared.. 4.3 × 10 N 3
Rudiy27

Answer:

Explanation:

Incomplete question but for understanding.

We want to find the electrical force between two charges, then you can use the coulombs law which states that the force of attraction or repulsion between two charges is directly proportional to the product of the two charges and inversely proportional to the square of their distance apart,

So,

F = kq1•q2 / r²

Where k is a constant and it is given as

K = 8.99 × 10^9 Nm²/C²

q1 and q2 are the charges and in this question it is not given, so the question is incomplete. Let assume that,

q1 = - 1.609 × 10^-19 C electron

q2 = 1.609 × 10^-19 C proton

Since unlike charges attract, then it is force of attraction

Also, r is the distance apart and it is not given, let assume the distance between the two charges is 2 × 10^-5m

Then,

F = kq1•q2 / r²

F = 8.99 × 10^9 × 1.609 × 10^-19 × 1.609 × 10^-19 / (2 × 10^-5)²

F = 5.82 × 10^-19 N

7 0
3 years ago
Read 2 more answers
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