A bird flying straight upward at 5 m/s drops a berry when it is 300 m above the ground. How fast is the berry going when it hits
the ground (neglect air resistance) ___ m/s.
1 answer:
Answer:
v=77.62 m/s
Explanation:
Given that
h= - 300 m
speed of the bird ,u= 5 m/s
Lets take Speed of the berry when it hit the ground = v m/s
we know that ,if object is moving upward
v² = u² - 2 g h
u=Initial speed
v=Final speed
h=Height
Now by putting the values
v² = u² - 2 g h
v² = 5² - 2 x 10 x (-300) ( take g = 10 m/s²)
v² =25 + 20 x 300
v² ==25 + 6000
v² =6025
v=77.62 m/s
Therefore the final speed of the berry will be 77.62 m/s.
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<u>Explanation :</u>
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m = 46 kg
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Answer:
(a) Acceleration of female astronaut = 9.33*10^-12 m/s^2; Acceleration of male astronaut = 7.56*10^-12 m/s^2
(b) 27.013 days
(c) No, their accelerations would not be constant.
Explanation:
In the given question, we have:
Mass of female astronaut = 60.0 kg
Mass of male astronaut = 74.0 kg
Distance between them (S) = 23.0 m
(a) The free-body diagram is shown in the attached figure.
Using the equations below, the initial accelerations of the two astronauts can be calculated:
Force of gravity (F) =
G = 6.67*10^-11
F = (6.67*10^-11 *60*74)/23^2 = 5.598*10^-10 N
For the female astronaut, her initial acceleration = F/ = 5.598*10^-10/60 = 9.33*10^-12 m/s^2
For the male astronaut, his acceleration = F/ = 5.598*10^-10/74 = 7.56*10^-12 m/s^2
(b) Since the different between their mass is not much, we can deduce that:
= (9.33*10^-12 + 7.56*10^-12)/2 = 8.445*10^-12 m/s^2
Using the equation below, we can calculate the the time:
S = ut + 1/2 (at^2) where u = 0
23 = 1/2 (8.445*10^-12)*t^2
t^2 = 5.447*10^12
t = 2333883.044 s = 27.013 days
(c) No, their accelerations will not be constant. It will increase because their radii would be decreasing.
