Given :
Initial speed, u = 0 m/s.
Final speed, v = 9 m/s.
Time taken to reach that speed, t = 2 sec.
To Find :
The acceleration of the athlete.
Solution :
We know, in uniform acceleration motion :
Therefore, acceleration of the athlete is 4.5 m/s².
Answer:
W = 2038400 J
Explanation:
Let
water be Δx meters thick at the depth of x.
So it has volume(V) of 6x.Δx
Now we know that weight of the water is given by
W = ρgV
=1000 x 9.8 x (6x.Δx)
It is also know that the water is lifted to a distance of (7-x)
Therefore, work done is
W =
=
=
=
= 58800 x 34.667
= 2038400 J
Answer:
The weight of the water above a diver exerts pressure on their body. The deeper a diver descends, the more water they have above them, and the more pressure it exerts on their body
Answer:
a) 2.85 m/s
b) -24.01 m/s
c) 1
Explanation:
From the question, we can attest that the motion of the fish that was dropped by the seagul is that of a projectile motion. This motion is made up of two other motions which are, a horizontal uniform motion and a vertical motion, at constant acceleration. From the question, we are asked to find the horizontal motion. And then, the horizontal component of the fish's velocity does not change, therefore its the horizontal component of the fish's velocity is 2.85 m/s
To find the vertical component of the fish's velocity, we use the equation
v(y) = u(y) + gt
Where u(y) is the initial velocity which is zero, and g is the acceleration due to gravity which is -9.8 m/s. We are also told from the question that it took 2.45 s, and that's our time t. Applying this into the equation, we have
v(y) = 0 + -9.8 * 2.45
v(y) = -24.01 m/s
For the 3rd part, the horizontal component of the fish's velocity would decrease