The given problem can be exemplified in the following diagram:
Since there is no friction or any other external force, the only force acting in the direction of the movement is the component of the weight of the block, therefore, applying Newton's second law:

Replacing the values:

We may cancel out the mass:

Using the gravity constant as 9.8 meters per square second:

Solving the operations:

Therefore, the acceleration is 6.3 meters per square second.
Answer:
Frequency, 
Explanation:
Visible red light has a wavelength of 680 nanometers (6.8 x 10⁻⁷ m). The speed of light is 3.0 x 10 ⁸ m / s. What is the frequency of visible red light?
It is given that,
Wavelength of a visible red light is, 
Speed of light is, 
We need to find the frequency of visible red light. It can be calculated using below relation.

So, the frequency of visible red light is
.
Answer:
Explanation:
In a conductor, electric current can flow freely, in an insulator it cannot.
Metals such as copper typify conductors, while most non-metallic solids are said to be good insulators, having extremely high resistance to the flow of charge through them.
Most atoms hold on to their electrons tightly and are insulators.
The correct answer of this question is : A) Change alternating current into direct current.
EXPLANATION :
As per the question, we are given vacuum tube. Vacuum tube can be of various types. Normally it contains two electrodes called cathode and anode which are enclosed in an evacuated glass chamber . There are also other types of vacuum tubes which contain extra electrodes like control grid .
The vacuum tube can be used as a rectifier. It means that it can be used as an electronic device which will convert alternating current into direct current. It may be a half wave rectifier or a full wave rectifier. Actually the direct current obtained during the rectification of alternating current is pulsating in nature.
Hence, the correct answer is that a vacuum tube can be used to change alternating current into direct current.
Since momentum is a vector quantity, take any direction as positive and other as negative. Answer won't change.