<span>at maximum height the final velocity will be 0
using v=u+at and resolving vertically we get
v=0.6+(-9.81)t
v=0.6-9.81t
0=0.6-9.81t
9.81t=0.6
t=0.6/9.81
t=0.061 to 3sf
Now we need to resolve horizontally to find the horizontal distance
using s=ut+1/2at^2
However we now need the total time taken for the projectile travel and return to the ground. We can assume the time taken for the projectile to reach its maximum height and return to the ground is the same therefore
the total time is 2 x 0.061=0.122seconds. They'll be now horizontal acceleration in this case scenario therefore
Hence s=ut+1/2at^2
since a=0
s=ut
s=0.6 x 0.122
s=0.073m
</span>
To solve this problem it is necessary to apply the concepts related to the centripetal Force, the Force of the weight produced by the gravity of the star and the definition of the velocity as a function of time (period)
In other words, for balance to exist, the force acting on the body (weight) must be equal to the centripetal force that attracts it in motion. That is to say
The value of the given period is 110min or 6600s.
At the same time we know that the speed of a body depending on its period (simple harmonic movement) is subject to
Finally, replacing this value in the event of gravity we have to
"<span>The pointer will be above the zero mark" is the one statement among the following choices given in the question that is true. The correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer that has actually come to your great help.</span>
Answer:
ΔV= -5.833×10⁻³
Negative sign indicates that volume decreases
Explanation:
Given data
System heat gains Q=3220 J
Pressure P=1.32×10⁵Pa
Internal energy increases ΔU=3990 J
To find
Change in volume ΔV
Solution
First we need to find the work done
So
W=Q-ΔU
W=3220J-3990J
W= -770J
Now for the change in volume at constant pressure
ΔV=(W/P)
ΔV= -5.833×10⁻³
Negative sign indicates that volume decreases