Answer:
The average velocity of a train moving along a straight track if its displacement is 192 m was during a time period of 8.0 s is 24 
.
Explanation:
Velocity is a physical quantity that expresses the relationship between the space traveled by an object and the time used for it. Then, the average velocity relates the change in position to the time taken to effect that change.
Velocity considers the direction in which an object moves, so it is considered a vector magnitude.
In this case, the displacement is 192 m and the time period is 8 s. Replacing:
Solving:
velocity= 24
<em><u>The average velocity of a train moving along a straight track if its displacement is 192 m was during a time period of 8.0 s is 24 </u></em><em><u>.</u></em>
Answer:False
Explanation:
Anything with evidence it can test.
Answer:
Magnitude of net force will be 432.758 N
Explanation:
We have given x component of acceleration 
And vertical component of acceleration 
Mass of the ball m = 0.40 kg
So net acceleration 
Now according to second law of motion
Force = mass × acceleration
So F = 0.40×1081.896 = 432.758 N
It decreases because it gave its momentum to the other car.
Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s
