Answer: 7.78m/s
Explanation: As the the skier slide down the height, we assume the motion of a body, slidind down an incline plane.
Force down the plane= [email protected]
Frictional force= umg
u= coefficient of friction
Net force on skier = [email protected] umg
ma = [email protected]
a = g([email protected] - u) = 9.8 (sin 25- 0.2)
a = 9.8 × (0.4226-0.2) = 9.8 × 0.2226
a = 2.18m/s²
Using the formula V² = U² + 2aH
Where H = (10.4+ 3.5)=total height of descent before landing, U= 0.
V = √ 2 × 2.18× 13.9 = √60.604
V = 7.78m/s
Answer:
2. displacement is maximum
Explanation:
This makes sense as the spring is stretched the farthest at this point
F = kx
The largest force makes the largest acceleration.
Here I =0.4 amp and t=2hr or 2×60×60 sec
We known I=nQ/t where Q is charge and n is no of charge.
n=It/Q
0.4x(2x60x60)/1.6x10^-19
Answer:
Explanation:
Weight will be highest at the pole where there is no tangential velocity requiring centripetal acceleration. Also, due to the slight bulge of the earth at the equator, the distance from the surface to the center of mass of earth is slightly shorter there meaning gravity is slightly stronger. Fg = GmM/R²
Weight will be lowest at the equator where the object is moving about the earth axis at an angular velocity of 2π/(24(3600)) rad/s ( about 7.27e-5 rad/s)
This means that some of the (already weaker, see above) gravity there is required to supply the needed centripetal acceleration to keep the object on the ground.
The height function would be s(t) = -16t^2 + 64t + 32.
Maximum height ---> set s'(t) = 0.
-32t + 64 = 0. ---> t = 2. s(2) = -16(2)^2 + 64(2) + 32 = the maximum height is 96 feet.
To find the velocity when it hits the ground, set s(t) = 0.
-16t^2 + 64t + 32 = 0. Divide by -16.
t^2 - 4t - 2 = 0. The only positive solution is x = 4.45.
s'(4.45) = -32(4.45) + 64 = It's going -78.4 feet/second when it hits the ground.
