7. How long does it take a ball rolling down a hill to change its speed from 3 m/sec to 34.5 m/sec
1 answer:
The time elapsed is 9 seconds
Explanation:
The motion of the ball is a uniformly accelerated motion (a motion with constant acceleration), so we can use the following suvat equation:
where :
v is the final velocity of the ball
u is the initial velocity
a is the acceleration
t is the time elapsed
For the ball in this problem, we have:
u = 3 m/s is the initial velocity
v = 34.5 m/s is the final velocity
is the acceleration
Solving for t, we find the time taken for this change in velocity:
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The gravitation force is 1/6 that of earth, and the ice in the same soft drink would float with 9/10 submerged option (B) is correct.
<h3>What is gravitational force?</h3>All mass-bearing objects are attracted by gravitational force. Because it consistently attempts to bring masses together rather than push them apart, the gravitational force is referred to as attractive.
As we know, the gravitational force is given by:
Where G is the gravitational constant.
m1 and m2 are masses.
r is the distance between the masses.
Weight of ice cube = weight of soft drink displaced
mg = ρVg
m = ρV
g does not affect astronauts on earth in a lunar module.
Thus, the gravitation force is 1/6 that of earth, and the ice in the same soft drink would float with 9/10 submerged option (B) is correct.
An astronaut on Earth notes that in her soft drink an ice cube floats with 9/10 of its volume submerged. If she were instead in a lunar module parked on the Moon where the gravitation force is 1/6 that of Earth, the ice in the same soft drink would float:A) with more than 9/10 submerged. B) with 9/10 submerged.C) with 6/10 submerged.D) totally submerged.
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Answer:
Hence the pressure is
Explanation:
Given data
Q=1500 J system gains heat
ΔV=- 0.010 m^3 there is a decrease in volume
ΔU= 4500 J internal energy decrease
We know work done is
W= Q- ΔU
=1500-4500= -3000 J
The change in the volume at constant pressure is
ΔV= W/P
there fore P = W/ΔV= -3000/-0.01= 3×10^5
Hence the pressure is