Answer:
a) 2.5 m/s²
b) 6.12 m/s
Explanation:
Tension of rope = T = 356N
Weight of material = W = 478 N
Distance from the ground = s = 7.5 m
Acceleration due to gravity = g = 9.81 m/s²
Mass of material = m = 478/9.81 = 48.72
Final velocity before the bundle hits the ground = v
Initial velocity = u = 0
Acceleration experienced by the material when being lowered = a
a) W-T = ma
⇒478-356 = 48.72×a
⇒a = 2.5 m/s²
∴ Acceleration achieved by the material is 2.5 m/s²
b) v²-u² = 2as
⇒v²-0 = 2×2.5×7.5
⇒v² = 37.5
⇒v = 6.12 m/s
∴ Velocity of the material before hitting the ground is 6.12 m/s
The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation.
is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
Our final equation is:
Colatitude is:
The answer is:
Answer:
A) 
B) 
Explanation:
Given:
mass of car, 
A)
frequency of spring oscillation, 
We knkow the formula for spring oscillation frequency:
Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.
<u>So, the stiffness of each spring is (as they are identical):</u>
B)
given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:
A) 300cm/h
B)1 hr=60 min
300/60=5
5cm/min
C)1m=100cm
300/100=3
3m/h
Answer: A. Her speed is 4.4 m/s, and her velocity is 0 m/s.
Explanation: i took the test on edgenuity
