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:
cross out the false piece in blue and write the true piece in red
Answer:
E.two angles are vertical angles if, and only if they are not adjacent angles
Answer:
– 2.5 m/s²
Explanation:
We have,
• Initial velocity, u = 180 km/h = 50 m/s
• Final velocity, v = 0 m/s (it stops)
• Time taken, t = 20 seconds
We have to find acceleration, a.
a = (v ― u)/t
a = (0 – 50)/20 m/s²
a = –50/20 m/s²
a = – 5/2 m/s²
a = – 2.5 m/s² (Velocity is decreasing) [Answer]