Answer:
fgfgfdgdg
Explanation:
gfdgfgfgd
Answer:
Approximately , assuming that .
Explanation:
Under the assumptions, the package would start with an initial upward velocity of and accelerate downward at a constant (negative because acceleration points downward.)
Right before landing, the package would be below where it was released. Hence, the displacement of the package at that moment would be (negative since this position is below the initial position.)
The duration of the motion can be found in the following steps:
Rearrange the SUVAT equation to find , the velocity of the package right before reaching the ground. Notice that because the package would be travelling downward, the value of should be negative.
.
Subtract the initial velocity from the new value to find the change in velocity. Divide this change in velocity by acceleration (rate of change in velocity) to find the duration of the motion:
.
the gravitational force of attraction between
them is 2.0 × 10^20 newtons. What would be the
magnitude of this gravitational force of attraction
if Earth and the Moon were separated by a
distance of 1.92 × 10^8 meters?
(1) 5.0 × 10^19 N (3) 4.0 × 10^20 N
(2) 2.0 × 10^20 N (4) 8.0 × 10^20 N
B. 3.8 m/s2
C. 2.4 m/s2
D. 9.8 m/s2
Acceleration due to gravity on this planet will be 3.802 m / s^2
Equation of motion are defined as equations that describe the behavior of a physical system in terms of its motion as a function of time
Using equation of motion
u=0
s= 2.3 m
t = 1.1 sec
to find = g (acceleration due to gravity on this planet)
s = u t + 1/2 (a ) (t ^(2))
s = 1/2 (g) (t^2)
2.3 = 1/2 (g) (1.1^2)
g = 2 * 2.3 /(1.1)^2
g = 4.6 /1.21= 3.802
correct answer is b) g = 3.802 m / s^2
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