Answer:
F = - K x
a) K = 1.3 kg * 9.8 m/s^2 / .096 m = 133 kg/sec^2
b) ω = (K/m)^1/2 angular frequency of SHM
ω = (133 / 1.3)^1/2 = 10.1 / sec
f = 2 π ω = 6.28 * 10.1 / sec = 63.5 / sec
P = 1/f = .0157 sec
Answer:
Acceleration = 5.77 m/s²
Distance cover in 13 seconds = 487.56 meter
Explanation:
Given:
Final velocity of mobile device = 75 m/s
initial velocity of mobile device = 0 m/s
Time taken = 13 seconds
Find:
Acceleration
Distance cover in 13 seconds
Computation:
v = u + at
75 = 0 + (a)(13)
13a = 75
a = 5.77
Acceleration = 5.77 m/s²
s = ut + (1/2)(a)(t²)
s = (0)(t) + (1/2)(5.77)(13²)
Distance cover in 13 seconds = 487.56 meter
The object's speed will remain constant after the it leaves his hand.
So will HIS speed in the opposite direction.
Answer:
Option A is correct.
(The faster object encounters more resistance)
Explanation:
Option A is correct. (The faster object encounters more resistance)
Air resistance depends on various factors:
- Speed of the object
- Cross-sectional area of the object
- Shape of the object
Formula:
As the speed of the object increases the amount of Air resistance/drag increases on the object, as the above formula shows direct relation between Air resistance/drag and velocity i.e F ∝ v^2.