Exactly what the first person said. Hope this helps!
Answer:
7. Net constant force down the ramp
Explanation:
After the car is released and starts moving up the ramp, the only force that is applied on the car is weight because of the gravity, we were told that the friction force is neglected. the force because of the weight is given by:
where θ is the angle of the ramp.
as you can see those values won't change, so the force remains constant down the ramp.
Answer:
No, there wasn't any variation in the light intensity at 360 degrees.
During the rotation, rotating through an angle of 90° gradually brought the intensity to a maximum. Rotating by another 90° degrees brought the intensity to a minimum at some point. Rotating by another 90° brought it back to its maximum and then another 90° brought it to its initial intensity.
Answer: h = 0.52m
Explanation:
Using the equation of out flow;
A1 × V1 = A2 ×V2
Where A1 = area of the first nozzle
A2 = area of the second nozzle
V1= velocity of flow out from the first nozzle
V2 = velocity of flow out from 2nd nozzle
But AV= area of nozzle × velocity of water = volume of water per second(m³/s).
Now we can set A×V = Area of nozzle × height of rise.
Henceb A1× h1 = A2 × h2 ( note the time cancel on both sides)
D1 = 20mm= 0.02m; h1 = 0.13m
D2 = 10mm = 0.01m; h2= ?
h2 = π(D1/2)²× h1 /π(D2/2)²
h2 = (0.02/2)² × 0.13/(0.01/2)²
= (0.01)² ×0.13 /(0.005)²
= 1.3 × 10^-5/(5 × 10^-3)²
= 1.3 × 10^-5/25 × 10^-6
= (1.3/25) 10^-5 × 10^6
= 0.052× 10
= 0.52m
