
The branch of mechanics which deals with the motion of an object without considering the cause of motion, is known as Kinematics ~
I hope it helps ~
The accurate answer is:
You could expect a white paper to have a red tone when viewed in a red light. This is the case because of the neutrality of the white paper.
Answer:
A. The volume of the object
Explanation:
First of all, it can only PROBABLY be A or D
But why is it only A?
- Mass = The amount of matter in an object (nothing related here)
- Volume = How much space is the object taking (super related here)
It's VOLUME here and not MASS here because:
- It it would be Mass, then the object size could be anything, but the liquid wouldn't move as much higher.
- But because it is Volume, it is taking space, which makes the liquid move
- If the Mass would be the answer, then it would be totally not related, because mass isn't related over here, it matters about size.
<h2>
Hence, A. The volume of the object </h2>
is your answer!!!!
Answer:
Velocity is a change in displacement over change in time and uses the units m/s.
Both are rates of change and can be positive or negative.
Acceleration is a change in velocity over change in time and uses the units m/s².
Explanation:
Velocity is the change in displacement over change in time, this makes it a rate of change. It can be positive or negative because it is a vector quantity. It uses the units m/s because that is a displacement unit over a time unit.
Acceleration is the change in velocity over change in time, this makes it a rate of change. It can be positive or negative because it is also a vector quantity. It uses the units m/s² (m/s/s) because that is a velocity unit over a time unit.
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ =
/ vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m