Formula of kinetic energy
[tex]e = \frac{1}{2} m {v}^{2} [/tex]
Therefore Kinetic energy of ball B is 4 times more than ball A.
ans is KB=4KA
A spaceship travels toward the Earth at a speed of 0.97c. The occupants of the ship are standing with their torsos parallel to the direction of travel. According to Earth observers, they are about 0.50 m tall and 0.50 m wide. Calculate what the occupants’ height and width according to the others on the spaceship?
Answer:
Explanation:
We shall apply length contraction einstein's relativistic formula to calculate the length observed by observer on the earth . For the observer , increased length will be observed for an observer on the earth
[tex]L=\frac{.5}{\sqrt{1-(\frac{.97c}{c})^2 } }[/tex]
[tex]L=\frac{.5}{.24}[/tex]
L= 2.05
The length will appear to be 2.05 m . and width will appear to be .5 m to the observer on the spaceship. . It is so because it is length which is moving parallel to the direction of travel. Width will remain unchanged.
(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on the street, (a) what was its initial speed, (b) what altitude does it reach, (c) when was it thrown, and (d) when does it reach the street again? Giancoli, Douglas C.. Physics (p. 45). Pearson Education. Kindle Edition.
Answer:
Assuming that the vertical speed of the ball is 14 m/s we found the given values:
a) V₀ = 23.4 m/s
b) h = 27.9 m
c) t = 0.96 s
d) t = 4.8 s
Explanation:
a) Assuming that the vertical speed is 14 m/s (founded in the book) the initial speed of the ball can be calculated as follows:
[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]
Where:
[tex]V_{f}[/tex]: is the final speed = 14 m/s
[tex]V_{0}[/tex]: is the initial speed =?
g: is the gravity = 9.81 m/s²
h: is the height = 18 m
[tex] V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s [/tex]
b) The maximum height is:
[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]
[tex] h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m [/tex]
c) The time can be found using the following equation:
[tex] V_{f} = V_{0} - gt [/tex]
[tex] t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s [/tex]
d) The flight time is given by:
[tex] t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s [/tex]
I hope it helps you!
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There is a lever with 5 m long. The fulcrum is 2 m from the right end. Each end hangs a box. The whole system is in balance. If the box hung to the right end is 12 kg, then what is the mass of the box hung to the left end?
Answer:
8kg
Explanation:
For the box to be in equilibrium. The clockwise moment ensued by the box on the right should be same as that ensued by the one on the right. Hence :
M ×3 = 12 ×2
M = 24/3 = 8kg
Note mass is used because trying to compute the weight by multiplying by the acceleration of free fall due to gravity on both sides will cancel out.
How many times can a three-dimensional object that has a radius of 1,000 units fit something with a radius of 10 units inside of it? How many times can something with a radius of 2,000 units fit something with a radius of 1 unit?
Answer:
# _units = 1000
Explanation:
This exercise we can use a direct proportion rule.
If a volume of radius r = 1 is one unit, how many units can fit in a volume of radius 10?
# _units = V₁₀ / V₁
The volume of a body of radius 1 is
V₁ = 4/3 π r₁³
V₁ = 4/3π
the volume of a body of radius r = 10
V₁₀ = 4/3 π r₂³
V10 = 4/3 π 10³
the number of times this content is
#_units = 4/3 π 1000 / (4/3 π 1)
# _units = 1000