Using the formula:
a = (Vf - Vi) / t
Our initial velocity is 0 m/s, and our final velocity is 8.15 m/s, with a time period of 5 seconds:
a = (8.15 - 0.0) / 5
a = 1.63 m/s^2
If you know the acceleration due to gravity on the Moon, you can confirm this answer. The recorded gravitational acceleration on the Moon is 1.62 m/s^2.
Horizontal distance covered by a projectile is X = Vix *T
where Vix is the initial horizontal component of velocity and T is time taken by the projectile
Vix = ViCos theta
In question they said that initial velocity and angle is same on earth and moon
so Vix would remains same
now let's see about time taken T
time taken to reach the highest point
Vfy = Viy +gt
at highest point vertical velocity become zero so Vfy =0
0 = Vi Sin theta + gt
t = Vi Sintheta /g
Total time taken to land will be twice of that
On earth
Te= 2t
Te = 2Sinθ/g
on moon g is one-sixth of g(earth)
Tm = 2Sinθ/(g/6)
Tm = 6(2Sinθ/g)
Tm = 6Te
so total time taken by the projectile on moon will be six times the time taken on earth
From first equation X = Vix*T
we can see that X will also be 6 times on moon than earth
so projectile will cover 6 times distance on moon than on earth
Explanation:
I want to say option B - Both forces can act without objects touching.
A. True. You can use displacement to determine the volume of solids and liquids.
Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
