A pivoted beam supports a load P at one end and a load Q at the other end.

A mole of gas at 0.0oC and 760. torr occupies 22.4 L. What is the volume of this gas at 25.oC and 760. torr

daser333 [38]

Answer:

Explanation:

Given

Temperature of gas at First stage $T_1=0^{\circ}C\approx 273\ K$

Pressure of gas at First stage $P_1=760\ torr$

Volume Occupies $V_1=22.4\ L$

If the Pressure and Temperature at second stage is

$P_2=760\ torr$

$T_2=25^{\circ}C\approx 298\ K$

Using ideal gas Equation

$PV=nRT$

where P=Pressure

V=volume

R=Universal Gas constant

T=Temperature

n=no of moles

as n and R is constant therefore

$\frac{PV}{T}=constant$

thus $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=\frac{760}{760}\times \frac{298}{273}\times 22.4$

$V_2=24.45\ L$

8 0

3 years ago

A ball thrown at 4 meters/second exerts 200 Newtons of force. The mass of the ball is:

Harlamova29_29 [7]

Answer:

50

Explanation:

200 divided by 4 gives you the mass of 50

5 0

3 years ago

the aeroplane in fig 3.1 flies an outward journey from Budapest (Hungary) to palermo (Italy) in 2.75 the distance is 2200 KM (i)

azamat

Answer: what is the cause

Explanation:

5 0

3 years ago

If we compare the force of gravity too strong nuclear force we would conclude that

Yuri [45]

The force of gravity is much weaker than the strong nuclear force. But the strong nuclear force only acts over short distances, such as within the nuclues. The gravitational force can act over infinite distance.

6 0

4 years ago

Read 2 more answers

Starting from rest, a small block of mass m slides frictionlessly down a circular wedge of mass M and radius R which is placed o

guapka [62]

Answer:

Part a)

$V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

$v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

$y_{cm} = \frac{mR}{m + M}$

Explanation:

PART A)

As we know that there is no external force on the system of two masses in horizontal direction

So here the two masses will have its momentum conserved in horizontal direction

So we have

$mv + MV = 0$

Also we know that here no friction force on the system so total energy will always remains conserved

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}MV^2 = mgR$

now we have

$\frac{1}{2}m(\frac{MV}{m})^2 + \frac{1}{2}MV^2 = mgR$

$\frac{1}{2}MV^2(\frac{M}{m} + 1) = mgR$

so we have

$V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

and another block has speed

$v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

$y_{cm} = \frac{mR}{m + M}$

7 0

3 years ago