Answer:
2065.005 m/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²
Now H-h = 42.3 - 0.2 = 42.1 cm = 0.421 m
The final velocity will be the initial velocity
Acceleration of the frog is 2065.005 m/s²
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.