Answer:
As an object accelerates i.e., change it's velocity(either direction or speed), the position of the object depends on two factor; If the acceleration was direction based then it might have a zero displacement for eg: if it travels in circle. or it might have a net displacement if it travels in a straight line, quantitatively
where,
s = displacement
u = initial velocity
v = final velocity
a = acceleration
t = time
Now, for the hypothesis;
There is no direct relationship between fan speed and acceleration but anyways generally speaking if we do have a relationship that with more fan speed we have a larger displacement of air i.e., a more force i.e., greater acceleration
Thus, it can be said, well not exactly scientific, that with a greater fan speed there will be greater acceleration. if fan speed is increased then acceleration will be more.
:)
Explanation:
Answer:
The velocity of the shell when the cannon is unbolted is 500.14 m/s
Explanation:
Given;
mass of cannon, m₁ = 6430 kg
mass of shell, m₂ = 73.8-kg
initial velocity of the shell, u₂ = 503 m/s
Initial kinetic energy of the shell; when the cannon is rigidly bolted to the earth.
K.E = ¹/₂mv²
K.E = ¹/₂ (73.8)(503)²
K.E = 9336032.1 J
When the cannon is unbolted from the earth, we apply the principle of conservation of linear momentum and kinetic energy
change in initial momentum = change in momentum after
0 = m₁u₁ - m₂u₂
m₁v₁ = m₂v₂
where;
v₁ is the final velocity of cannon
v₂ is the final velocity of shell
Apply the principle of conservation kinetic energy
Therefore, the velocity of the shell when the cannon is unbolted is 500.14 m/s
Answer:3W
If it takes an amount of work W to move two q point charges from infinity to a distance d apart from each other, then how much work should it take to move three q point charges from infinity to a distance d apart from each other?
A) 2W
B) 3W
C) 4W
D) 6W
Explanation: calculating work done,W, in moving two positive q point charges from infinity to a valued distance d from each other is
W = k(+q)(+q)/ d
k is couloumb's constant
work done in moving 3 equal positive charges from infinity to a finite distance is given by
W₂=W₄=W₆=k(+q)(+q)/ d
Total work done, W' =k(+q)(+q)/ d + k(+q)(+q)/ d + k(+q)(+q)/ d
= W + W + W = 3W
Answer:
0.41
Explanation:
A 49 N block has a mass of 5 kg:
49/9.8
A horizontal force of 50 N giving it an acceleration of 6 m/s^2 means that 30 N is taken up by the acceleration
This leaves 20 N to take care for friction. so the coefficient of friction is:
20/49 = 0.41.
<span>A light-year measures the distance that light travels in 1 year.
Answer : B ) Distance
-Hope this helps.</span>
