Answer:
Explanation:
a)
Ff = μmgcosθ
Ff = 0.28(1600)(9.8)cos(-84)
Ff = 458.9217...
Ff = 460 N
b) ignoring the curves required at top and bottom which change the friction force significantly, especially at the bottom where centripetal acceleration will greatly increase normal forces and thus friction force.
W = Ffd
W = 458.9217(-49.4/sin(-84)
W = 22,795.6119...
W = 23 kJ
c) same assumptions as part b
The change in potential energy minus the work of friction will be kinetic energy.
KE = PE - W
½mv² = mgh - (μmgcosθ)d
v² = 2(gh - (μgcosθ)(h/sinθ))
v = √(2gh(1 - μcotθ))
v = √(2(9.8)(49.4)(1 - 0.28cot84))
v = 30.6552...
v = 31 m/s
Answer:
(for small oscillations)
Explanation:
The total energy of the pendulum is equal to:
For small oscillations, the equation can be re-arranged into the following form:
Where:
, measured in radians.
If the amplitude of pendulum oscillations is increase by a factor of 4, the angle of oscillation is 
and the total energy of the pendulum is:
The factor of change is:
Answer:
0.8 seconds
Explanation:
F=ma
Let x be the seconds the force is applied.
m = 20kg
F = 50 Newtons (kg*m/sec^2)
acceleration, a, is provided for x seconds to increase the speed from 1 m/s to 3 m/s, an increase of 2m/s
Let's calculate the acceleration of the cart:
F=ma
(50 kg*m/s^2) = (20kg)*a
a = 2.5 m/s^2
---
The acceleration is 2.5 m/s^2. The cart increases speed by 2.5 m/s every second.
We want the number of seconds it takes to add 2.0 m/sec to the speed:
(2.5 m/s^2)*x = 2.0 m/s
x = (2.0/2.5) sec
x = 0.8 seconds
Answer:
0.00091
Explanation:
(9x10^9) (2.6x10^-6) (1.4x10^-6) / 36
(9,000,000,000) (0.0000026) (0.0000014) /36
|
23,400(0.0000014) /36
|
0.03276 /36
|
0.00091
The answer is (A) hope it helps
