This is something u are going to have to do
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
Answer:
17 °C
Explanation:
From specific Heat capacity.
Q = cm(t₂-t₁)................. Equation 1
Where Q = Heat absorb by the metal block, c = specific heat capacity of the metal block, m = mass of the metal block, t₂ = final temperature, t₁ = Initial temperature.
make t₁ the subject of the equation
t₁ = t₂-(Q/cm)............... Equation 2
Given: t₂ = 22 °C, Q = 5000 J, m = 4 kg, c = 250 J/kg.°c
Substitute into equation 2
t₁ = 22-[5000/(4×250)
t₁ = 22-(5000/1000)
t₁ = 22-5
t₁ = 17 °C
Given data:
* The mass of the ball is 2 kg.
* The gravitational field strength at the surface of planet X is 5 N/kg.
Solution:
The weight of the ball on the planet X is,

where m is the mass of ball, a is the gravitational field strength,
Substituting the known values,

Thus, the weight of the ball on the surface of planet X is 10 N.
Answer:
a = F/m
Explanation:
So we have to isolate a, in order to do this we need to move m to the other side, and we do that by diving both sides by m, resulting in a = F/m