A worker pushes horizontally on a large crate with a force of 265.0 N and the crate moved 4.3 m. How much work was done?
1 answer:
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Answer:
The speed of the stone when it is 4.66 m higher is 236.057 m/s.
Explanation:
Given the initial velocity and vertical distance, we can use the fourth kinematic equation () to find v final, or the v to the left of the equal sign. We know
(initial velocity) is 24.7 m/s, y (change in vertical distance) is 4.66 m, and a is another way to write g (acceleration due to gravity), or 9.8
.
From here you could plug in the values and solve for v final, but to make the solving process simpler, we can simplify the given equation, <em>then </em>plug in the known values.
To isolate v final, we can take the square root of and do the same to the right side of the equation. Therefore, we can find v final with:
, where v initial is outside of the square root because it squared...
If we plug in the known values to the simplified equation, we get:
The final answer is 236.057 m/s.
Answer:
12 cm
Explanation:
We shall use Lens makers formula here which is as follows
Put μ = 1.5 , R₁ = 10 cm ,R₂ = - 15 cm ( according to sign convention )
= .5 x ( 15 + 10 ) / 15 x 10
=
F = 12 cm
A. 103.9 m/s
The motion of the rocket until it runs out of fuel is an accelerated motion with constant acceleration a = 3.6 m/s^2, so we can use the following equation
where
v is the velocity of the rocket when it runs out of fuel
u = 0 is the initial velocity of the rocket
a = 3.6 m/s^2 is the acceleration
d = 1500 m is the distance covered during this first part
Solving for v, we find
B. 28.9 s
We can calculate the time taken for the rocket to reach this altitude with the formula
where
d = 1500 m
a = 3.6 m/s^2
Solving for t, we find
C. 2050.8 m
We can calculate the maximum altitude reached by the rocket by using the law of conservation of energy. In fact, from the point it runs out of fuel (1500 m above the ground), the rocket experiences the acceleration due to gravity only, so all its kinetic energy at that point is then converted into gravitational potential energy at the point of maximum altitude:
where h is distance covered by the rocket after it runs out of fuel, and v=103.9 m/s is the velocity of the rocket when it starts to decelerate due to gravity. Solving for h,
So the maximum altitude reached by the rocket is
D. 39.5 s
The time needed for the second part of the trip (after the rocket has run out of fuel) can be calculated by
where
h = 550.8 m is the distance covered in the second part of the trip
g = 9.8 m/s^2
Solving for t,
So the total time of the trip is
E. 200.5 m/s
When the rocket starts moving downward, it is affected by gravity only. So the gravitational potential energy at the point of maximum altitude is all converted into kinetic energy at the instant the rocket hits the ground:
where
v is the final velocity of the rocket
h = 2050.8 m is the initial altitude of the rocket
Solving for v,
F. 60 s
We need to calculate the time the rocket takes to fall down to the ground from the point of maximum altitude, and that is given by
where
h' = 2050.8 m
g = 9.8 m/s^2
Solving for t,
So the total time of the trip is