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3 years ago

You can increase the vapor pressure of a liquid by:

Physics

1 answer:

irga5000 [103]3 years ago

4 0

Answer: Option (c) is the correct answer.

Explanation:

Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.

When we increase the temperature of a liquid substance then there will occur an increase kinetic energy of the molecules. As a result, they will move readily from one place to another.

Hence, liquid state of a substance will change into vapor state of the substance. This means that an increase in temperature will lead to an increase in vapor pressure of the substance.

Thus, we can conclude that you can increase the vapor pressure of a liquid by increasing temperature.

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What is the relationship between the masses of the objects and the gravitational force between them

irga5000 [103]

The relationship between the masses of the object and the gravitational force between them is a direct relationship

Explanation:

The gravitational force between two objects is given by the equation:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

We observe that:

- The gravitational force is proportional to the masses of the two objects, m1 and m2, so if the masses increase, the force will increase as well (so, this is a direct relationship)

- The gravitational force is inversely proportional to the square of the separation between the objects, so if the distance is increased, the force will decrease (so, this is an inverse relationship)

Learn more about gravitational force here:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0

3 years ago

When bones are moved by muscles due to

77julia77 [94]

Respiratory and circulatory

6 0

3 years ago

Read 2 more answers

24. An elevator is moving vertically up with an acceleration a. The force exerted on the floor by a passenger of mass m is

Wewaii [24]

Given :-

An elevator is moving vertically up with an acceleration a.

To Find :-

The force exerted on the floor by a passenger of mass m .

Solution :-

As the man is in a accelerated frame that is non inertial frame , we would have to think of a pseudo force .

The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .

For the FBD refer to the attachment . From that ,

$\implies Weight_{apparent}= mg + ma$

Hence option d is correct choice .

I hope this helps .

3 0

3 years ago

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$