Answer:
its unit is Ohm
Explanation:
Resistance means material which resist the passing current through it and the value of resistance says how much the material is resisting the current and it temperature dependent and the unit is Ohm.
The average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.
The correct answer is option D.
In the given graph, we can deduce the following;
- the total time of the motion, = 1 mins + 45 s = 60 s + 45 s = 105 s
The average speed of the ant is calculated as;

The total distance from the graph is calculated as follows;
- first horizontal distance from 2 cm to 8 cm = 8 - 2 = 6 cm
- first upward distance from 3 cm to 5 cm = 5 - 3 = 2 cm
- second horizontal distance from 8 cm to 6 cm = 8 - 6 = 2 cm
- second upward distance from 5 cm to 12 cm = 12 - 5 = 7 cm
- third horizontal distance from 6 cm to 13 cm = 13 - 6 = 7 cm
- fourth downward distance from 12 cm to 9 cm = 3 cm
- final horizontal distance from 13 cm to 15 cm = 2cm
The total distance = (6 + 2 + 2 + 7 + 7 + 3 + 2) cm = 29 cm

The average velocity is calculated as the change in displacement per change in time.
The displacement is the shortest distance between the start and end positions.
- This shortest distance is the straight line connecting the start and end position. Call this line P
- From the end position at x = 15 cm, draw a vertical line from y = 9 cm, to y = 3 cm. The displacement = 9 cm - 3 cm = 6 cm
- Also, draw a horizontal line from start at x = 2 cm to x = 15 cm. The displacement = 15 cm - 2 cm = 13 cm
Notice, you have a right triangle, now calculate the length of line P.
↓end
↓
↓ 6cm
↓
start -------------13 cm------------
Use Pythagoras theorem to solve for P.

The average velocity of the ant is calculated as;

Thus, the average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.
Learn more here: brainly.com/question/589950
42.9°
Explanation:
Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:


Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at
Solving for the angle, we get

or

![\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]](https://tex.z-dn.net/?f=%5C%3B%5C%3B%5C%3B%3D%20%20%5Csin%5E%7B-1%7D%5Cleft%5B%5Cdfrac%7B34%5C%3A%5Ctext%7BN%7D%7D%7B%285.1%5C%3A%5Ctext%7Bkg%7D%29%289.8%5C%3A%5Ctext%7Bm%2Fs%7D%5E2%29%7D%5Cright%5D)

The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
The answers is A and C hope this helps :)