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3 years ago

I need help this is for economics, can anyone help me ?

Physics

1 answer:

PolarNik [594]3 years ago

4 0

i can <u>when do u need it by</u> working on it now!!!

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What are the 3 Newton's law of motions?

irinina [24]

Answer:

laws of motion relate an object’s motion to the forces acting on it. In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

4 0

2 years ago

Two rigid tanks of equal size and shape are filled with different gases. The tank on the left contains oxygen, and the tank on t

fredd [130]

Answer:

The number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.

Explanation:

Given:

Molar mass of oxygen, $M_O=32$

Molar mass of hydrogen, $M_H=2$

We know ideal gas law as:

$PV=nRT$

where:

P = pressure of the gas

V = volume of the gas

n= no. of moles of the gas molecules

R = universal gs constant

T = temperature of the gas

∵ $n=\frac{m}{M}$

where:

m = mass of gas in grams

M = molecular mass of the gas

∴Eq. (1) can be written as:

$PV=\frac{m}{M}.RT$

$P=\frac{m}{V}.\frac{RT}{M}$

as: $\frac{m}{V}=\rho\ (\rm density)$

So,

$P=\rho.\frac{RT}{M}$

Now, according to given we have T,P,R same for both the gases.

$P_O=P_H$

$\rho_O.\frac{RT}{M_O}=\rho_H.\frac{RT}{M_H}$

$\Rightarrow \frac{\rho_O}{32}=\frac{\rho_H}{2}$

$\rho_O=16\rho_H$

∴The molecules of oxygen are more densely packed than the molecules of hydrogen in the same volume at the same temperature and pressure. So, <em>the number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.</em>

5 0

3 years ago

A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p

777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

$v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s$