Answer:
20m/s
Explanation:
it covers 20 metres in a second
Answer:
f=force m=mass and a=acceleration
Answer:
wavelength = v/f or wavelength equals to velocity over frequency
frequency= v/w or velocity over wavelength
frequency= 1/p or one over period or time
Answer:
n the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.
In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.
Explanation:
Velocity is a vector therefore it has magnitude and direction, a change in either of the two is the consequence of an acceleration on the system.
In the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.
= (v₂-v₁)/Δt
In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.
= v2/R
In the general case, both the module and the address change
a = Ra ( a_{t}^2 + a_{c}^2)
Answer:
4 Ohms
Explanation
(This is seriously not as hard as it looks :)
You only need two types of calculations:
- replace two resistances, say, R1 and R2, connected in a series by a single one R. In this case the new R is a sum of the two:

- replace two resistances that are connected in parallel. In that case:

I am attaching a drawing showing the process of stepwise replacement of two resistances at a time (am using rectangles to represent a resistance). The left-most image shows the starting point, just a little bit "warped" to see it better. The two resistances (6 Ohm next to each other) are in parallel and are replaced by a single resistance (3 Ohm, see formula above) in the top middle image. Next, the two resistances (9 and 3 Ohm) are nicely in series, so they can be replaced by their sum, which is what happened going to the top right image. Finally we have two resistances in parallel and they can be replaced by a single, final, resistance as shown in the bottom right image. That (4 Ohms) is the <em>equivalent resistance</em> of the original circuit.
Using these two transformations you will be able to solve step by step any problem like this, no matter how complex.