Answer:
10.4 m/s
Explanation:
The problem can be solved by using the following SUVAT equation:

where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
For the diver in the problem, we have:
is the initial velocity (positive because it is upward)
is the acceleration of gravity (negative because it is downward)
By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.
784 Newtons or 176.37 lbs
Work is force times distance. If there's no distance, there's no work being done.
If you put a penny in each light spot the penny that the light is shining on will recive the most energy.
1.96s and 1.86s. The time it takes to a spaceship hovering the surface of Venus to drop an object from a height of 17m is 1.96s, and the time it takes to the same spaceship hovering the surface of the Earth to drop and object from the same height is 1.86s.
In order to solve this problem, we are going to use the motion equation to calculate the time of flight of an object on Venus surface and the Earth. There is an equation of motion that relates the height as follow:

The initial velocity of the object before the dropping is 0, so we can reduce the equation to:

We know the height h of the spaceship hovering, and the gravity of Venus is
. Substituting this values in the equation
:

To calculate the time it takes to an object to reach the surface of Venus dropped by a spaceship hovering from a height of 17m, we have to clear t from the equation above, resulting:
Similarly, to calculate the time it takes to an object to reach the surface of the Earth dropped by a spaceship hovering from a height of 17m, and the gravity of the Earth
.
