Answer:
U = √Rg/sin2θ
Explanation:
Using the formula for "range" in projectile motion to derive the average speed before the ball hits the ground.
Range is the distance covered by the body in the horizontal direction from the point of launch to the point of landing.
According to the range formula,
R = U²sin2θ/g
Cross multiplying we have;
Rg = U²sin2θ
Dividing both sides by sin2θ, we have;
U² = Rg/sin2θ
Taking the square root of both sides we have;
√U² = √Rg/sin2θ
U = √Rg/sin2θ
Therefore, his average speed if he is to meet the ball just before it hits the ground is √Rg/sin2θ
Both valves are closed during the power stroke.
While the fuel is burning in the cylinder, you want
all the force of the expanding gases to push the
piston down ... you don't want any of the gases
or their pressure escaping.
If either of the valves was open, even just a crack,
then part of the gases would go blooey out the valve,
and some pressure would be lost that's supposed to be
pushing the piston.
Answer: 20 kgm/s
Explanation:
Given that M1 = M2 = 10kg
V1 = 5 m/s , V2 = 3 m/s
Since momentum is a vector quantity, the direction of the two object will be taken into consideration.
The magnitude of their combined
momentum before the crash will be:
M1V1 - M2V2
Substitute all the parameters into the formula
10 × 5 - 10 × 3
50 - 30
20 kgm/s
Therefore, the magnitude of their combined momentum before the crash will be 20 kgm/s
Answer:
A
Explanation:
houses use alternating current source
Finding acceleration= final velocity-initial velocity/ time taken (or A= V-U/T)
Final speed= 2m
Initial speed= 0m
Time taken= 2 seconds
2-0/2 so it’ll be 1m/s
2-0=0
2/2=
