Answer:
The Kinetic energy of Sphere is higher than the cylinder.
( KS > KC )
Explanation:
Given - A sphere with the same mass and radius as the original cylinder, but a smaller rotational inertia, is released from rest from the top of the ramp. KS and KC are the sphere's and cylinder's total kinetic energy at the bottom of the ramp, respectively.
To find - How do KS and KC compare, and why ?
Proof -
We know that,
The total energy of an object = Potential energy + linear kinetic energy + rotational kinetic energy.
⇒E = mgh + +
Now,
Mass of sphere = m
Radius of sphere = r
So,
The moment of inertia of a uniform solid sphere =
Also,
Mass of cylinder = m
Radius of cylinder = r
So,
The moment of inertia of a uniform solid cylinder =
Now,
Total energy for the sphere , Es = mgh +
Total energy for the cylinder, Ec = mgh +
As they always have the same total energy,
So, for height h of the sphere's velocity has to be higher.
Therefore,
The Kinetic energy of Sphere is higher than the cylinder.
Answer:a.Dry cell
Explanation: A dry cell is a type of electric battery, commonly used for portable electrical devices like a flashlight
The gravitational acceleration on the surface of planet A is:
where G is the gravitational constant,
is the mass of planet A and
its radius.
Similarly, the gravitational acceleration on the surface of planet B is:
The ratio between the gravitational acceleration on planet A and B becomes:
The problem says that the two masses are equal:
while planet A has 3 times the radius of planet B:
. Substituting into the ratio, we get:
so, gravity on planet B is 9 times stronger than planet A.
Answer:
<em>Option D: It iwill actually warm the room</em>
<em></em>
Explanation:
<u>To complete your given question the available options are:</u>
A. It will cool the room very effectively
B. It will cool the room, but inefficiently
C. It will not change anything
D. It will actually warm the room.
This is a fun and somewhat tricky case. So, let us first understand a few principles in order to answer our question. To begin with<u><em>, the basic operating principle of the fridge is to take the hot air from the surrounding environment and cool it to the desired temperature in order to sustain all products inside the fridge</em></u>. It can also be thought as 'transferring' heat from the interior (i.e. inside the fridge) to the exterior (i.e. outside the fridge and into the surrounding).<em> In fact if you check the back of a fridge during operation, you will noticed a much higher temperature in that area. Which is due to the heat removed by the 'fridge operation system' in order to cool that interior air.</em> Therefore, this heat must transfer somewhere else, which typically ends up on the little fan located on the back of the fridge. We can also think of it in terms of the 2nd Law of Thermodynamics, which essentially tells us that the system (in this case the fridge and its surrounding)<em> MUST reach an Equillibrium.</em>
Therefore, when you do open the door of the fridge, you might initially (and for an instant almost) feel this 'cool' air coming out; thinking the surrounding air should soon cool down as well. But, due to our discussion above along with the principles of the 2nd Law of Thermodynamics, and considering the fridge operation over time,<em><u> the more cool air the fridge looses, the more the fridge system works to cool the air, thus the more the fans of the fridge work, which results to increasing heat getting 'dumped' by the fridge system and thus to the surrounding. </u></em>
<em></em>
<em>Consequently when you open the fridge door you will actually warm the room. (i.e. Option D). </em>
To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.
PART A)
Here,
M = Mass of Earth
R = Distance from center to the satellite
Replacing with our values we have,
PART B) The period of satellite is given as,
PART C) The gravitational force on the satellite is given by,