If an object has a kinetic energy of 30 j and mass of 34kg how fast is the object moving ?
1 answer:
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Answer:
V1 =8.1 m/s
Explanation:
height at highest point (h2) = 4.1 m
height at lowest point (h1) = 0.8 m
acceleration due to gravity (g) = 9.8 m/s^{2}
from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore
mgh1 + = mgh2 +
where
- speed at highest point = V2
- speed at lowest point = V1
- mass of the girl and swing = m
- at the highest point, the speed is minimum (V1 = 0)
- at the lowest point the speed is maximum (V2 is the maximum speed)
- therefore the equation becomes mgh1 +
= mgh2
m(gh1 + ) = m(gh2)
gh1 + = gh2
V1 =
now we can substitute all required values into the equation above.
V1 =
V1 =
V1 =8.1 m/s
Answer:
A.) The arrow`s range is 624,996 m
B.) The arrow`s range is 846.887 m, when the horse is galloping
Explanation:
We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.
By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.
<u>Equations</u>
X axis:
Y axis:
A.) First, it is necessary to know t, total time.
To figure out t value, we use UAM, since time is determined by this movement.
Now, at the end of the movement, , then
Caculate the segcond degree equation to obtain the two possible values for t:
But, in physics, time it could not be negative, so we take
Caculate now:
B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.
Using the same procedure that item A, caculate X
First, we need to know the new time
And we obtain:
One more time, we take the positive time:
Finally: