The first one is A and the second one would be C
Answer:
266.67Watts
Explanation:
Time = 2.5hr to seconds
3600s = 1hr
2.5hrs = 3600×2.5= 9000s
Force = 32N
Distance = 75km to m
1000m = 1km
75km = 1000×75 = 75000m
Power = workdone / time
Work = force × distance
Therefore work = 32N × 75000m
Work = 2400000Nm
Power = work ➗ time
Power = 2400000Nm ➗ 9000s
Power = 266.67Watts
Watts is the S. i unit of power
I hope this was helpful, please mark as brainliest
Answer:
96%
Explanation
Let A the total area of the galaxy, is modeled as a disc:
A = πR^2 = π (25 kpc)^2
And let a be the area that astronomers are able to see:
a = πr^2 = π(5 kpc)^2
The percentage that can be seen is equal to 100 times the ratio of the areas, of the galaxy and the "visible" part:
P = 100 a/A = (5/25)^2 = 100/25 = 4%
Therefore, the percentage of the galaxy not included, i.e. not seen is:
(100-4)% = 96%
Answer:
Approximately 
(assuming that the projectile was launched at angle of 
above the horizon.)
Explanation:
Initial vertical component of velocity:
.
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing 
is the same as the altitude 
at which this projectile was launched: 
.
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 
(upwards,) the vertical velocity right before landing would be 
(downwards.) The change in vertical velocity is:
.
Since there is no drag on this projectile, the vertical acceleration of this projectile would be 
. In other words, 
.
Hence, the time it takes to achieve a (vertical) velocity change of 
would be:
.
Hence, this projectile would be in the air for approximately .
