Answer:
Explanation:
Brownian motion is a random (irregular) motion of particles e.g smoke particle. The set up in the diagram can be used to observe the motion of smoke.
1. The apparatus used are:
A is a source of light
B is a converging lens
C is a glass smoke cell
D is a microscope
2. The uses of the apparatus are:
A - produces the light required to so as to see clearly the movement of the particles.
B - converges the rays of light from the source to the smoke cell.
C - is made of glass and used for encamping the smoke particles so as not to mix with air.
D - is used for the clear view or observation or study of the motion of the smoke particles in the cell.
Nuclear energy is released during: fission. radioactive decay. man-induced splitting of atoms. Match the basic components of a nuclear reactor with their descriptions. 1. slows down neutrons -> moderator. 2. absorb emitted neutrons -> control rods.
Answer:
The work done by a particle from x = 0 to x = 2 m is 20 J.
Explanation:
A force on a particle depends on position constrained to move along the x-axis, is given by,
We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.
We know that the work done by a particle is given by the formula as follows :
So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.
Answer:
4 A
Explanation:
We are given that
I=12 A
We have to find the current flowing through each resistor.
We know that in parallel combination current flowing through different resistors are different and potential difference across each resistor is same.
Formula :
Using the formula
Substitute the values
Hence, current flows through any one of the resistors is 4 A.
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
