Answer:
A. 456 seconds
Explanation:
We are given that two students walk in the same direction along a straight path at a constant speed.
One student walks with a speed=0.90 m/s
second student walks with speed=1.9 m/s
Total distance covered by each students=780 meter
We have to find who is faster and how much time extra taken by slower student than the faster student.
Time taken by one student who travel with speed 0.90 m/s=
Time=
Time taken by one student who travel with speed 0.90 m/s
=
Time taken by one student who travel with speed 0.90 m/s
=866.6 seconds
Time taken by second student who travel with speed 1.9 m/s=
=410.5 seconds
The second student who travels with speed 1.9 m/s is faster than the student travels with speed 0.90 m/s .
Extra time taken by the student travels with speed 0.90 m/s=866.6-410.5=456.1 seconds
Extra time taken by the student travels with speed 0.90 m/s=456 seconds
Hence, option A is true.
Answer:
1275J
Explanation:
Given parameters:
Force on box = 85N
Distance moved = 15m
Unknown:
Work done = ?
Solution:
Work done is the amount of force applied on a body to move it through a specific distance.
Work done = Force x distance
Now insert the parameters and solve;
Work done = 85 x 15 = 1275J
Answer:
The potential energy of the more massive one is twice that of the other.
Explanation:
Potential energy is given by
<em>PE</em> = <em>mgh</em>
where <em>m</em> = mass of body, <em>g</em> = acceleration of gravity and <em>h</em> = height or elevation.
For the less massive car, let the mass be
. Then its <em>PE</em> is

For the massive car, let the mass be
. Its <em>PE</em> is

But 

Hence, the potential energy of the more massive one is twice that of the other.
Answer:
I thought "Born Without a Heart" was pretty good.
Hi
Explanation:
Im sure the answer is letter B