Answer:
a) The velocity of rock at 1 second, v = 9.8 m/s
b) The velocity of rock at 3 second, v = 29.4 m/s
c) The velocity of rock at 5.5 second, v = 53.9 m/s
Explanation:
Given data,
The rock is dropped from a bridge.
The initial velocity of the rock, u = 0
a) The velocity of rock at 1 second,
Using the first equation of motion
v = u + gt
v = 0 + 9.8 x 1
v = 9.8 m/s
b) The velocity of rock at 3 second,
v = u + gt
v = 0 + 9.8 x 3
v = 29.4 m/s
c) The velocity of rock at 5.5 second,
v = u + gt
v = 0 + 9.8 x 5.5
v = 53.9 m/s
I assume the 100 N force is a pulling force directed up the incline.
The net forces on the block acting parallel and perpendicular to the incline are
∑ F[para] = 100 N - F[friction] = 0
∑ F[perp] = F[normal] - mg cos(30°) = 0
The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.
Then
F[friction] = 100 N
F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N
If µ is the coefficient of static friction, then
F[friction] = µ F[normal]
⇒ µ = (100 N) / (84.9 N) ≈ 1.2
For a flower to be pollinated, pollen from an anther (which is located at the top of the stamen) needs to reach a stigma (at the top of the pistle.) Some plants are genetically capable of pollinating themselves if their own pollen reaches their own stigma; some plants are not capable of self pollination under any circumstances.
For plants that can genetically self pollinate, but would prefer not to, they can avoid this by having their pistil and pollen/stamens mature at different times. If the stamens mature first, the pollen will be dispersed by animals or wind or whatever dispersal mechanism it relies on. Then by the time the pistil is ready to be pollinated, there is no pollen left in that flower to land on the stigma.
In order to compute the torque required, we may apply Newton's second law for circular motion:
Torque = moment of inertia * angular acceleration
For this, we require the angular acceleration, α. We may calculate this using:
α = Δω/Δt
The time taken to achieve rotational speed may be calculated using:
time = 1 revolution * 2π radians per revolution / 3.5 radians per second
time = 1.80 seconds
α = (3.5 - 0) / 1.8
α = 1.94 rad/s²
The moment of inertia of a thin disc is given by:
I = MR²/2
I = (0.21*0.1525²)/2
I = 0.002
τ = 1.94 * 0.002
τ = 0.004
The torque is 0.004
Answer: 37.5 kg in 3 s.f.
Explanation:
