Answer:
Explanation:
Before it hits the ground:
The initial potential energy = the final potential energy + the kinetic energy
mgH = mgh + 1/2 mv²
gH = gh + 1/2 v²
v = √(2g (H - h))
v = √(2 * 9.81 m/s² * (0.42 m - 0.21 m))
v ≈ 2.0 m/s
When it hits the ground:
Initial potential energy = final kinetic energy
mgH = 1/2 mv²
v = √(2gH)
v = √(2 * 9.81 m/s² * 0.42 m)
v ≈ 2.9 m/s
Using a kinematic equation to check our answer:
v² = v₀² + 2a(x - x₀)
v² = (0 m/s)² + 2(9.8 m/s²)(0.42 m)
v ≈ 2.9 m/s
Answer:
If the force applied is larger than 185.2 N, yes.
Explanation:
In order to move the table, the pushing force must be larger than the frictional force. The frictional force is given by:
where
is the coefficient of static friction
is the mass of the table
is the gravitational acceleration
Substituting,
So, we are able to move the table if we push with a force larger than 185.2 N.
Answer:
Scenario 1 is the correct answer.
Explanation:
The sound of the drumstick hitting the metal bar will get to me in a shorter amount of time in Scenario 1 . The sound wave will travel faster in the metal bar than through the air because the speed of sound waves in solid is faster than it is in gases.
Answer:
W = 34.64 ft-lbs
Explanation:
given,
Horizontal force = 4 lb
distance of push, d = 10 ft
angle of ramp, θ = 30°
Work done on the box = ?
We know,
W = F.d cos θ
W = 4 x 10 x cos 30°
W = 40 x 0.8660
W = 34.64 ft-lbs
Hence, work done on the box is equal to W = 34.64 ft-lbs
Answer:
Driving force increases, friction forces increase, the driving force is bigger than friction 12.
Explanation:
