Answer:
0.07756 m
Explanation:
Given mass of object =0.20 kg
spring constant = 120 n/m
maximum speed = 1.9 m/sec
We have to find the amplitude of the motion
We know that maximum speed of the object when it is in harmonic motion is given by
where A is amplitude and
is angular velocity
Angular velocity is given by
where k is spring constant and m is mass
So 

Answer:
Explanation:
Given:
Force, f = 5 N
Velocity, v = 5 m/s
Power, p = energy/time
Energy = mass × acceleration × distance
Poer, p = force × velocity
= 5 × 5
= 25 W.
Note 1 watt = 0.00134 horsepower
But 25 watt,
0.00134 hp/1 watt × 25 watt
= 0.0335 hp.
5) 204 meters
6)
A) 150 miles
B)241 km
Answer: 2.49×10^-3 N/m
Explanation: The force per unit length that two wires exerts on each other is defined by the formula below
F/L = (u×i1×i2) / (2πr)
Where F/L = force per meter
u = permeability of free space = 1.256×10^-6 mkg/s^2A^2
i1 = current on first wire = 57A
i2 = current on second wire = 57 A
r = distance between both wires = 26cm = 0.26m
By substituting the parameters, we have that
Force per meter = (1.256×10^-6×57×57)/ 2×3.142 ×0.26
= 4080.744×10^-6/ 1.634
= 4.080×10^-3 / 1.634
= 2.49×10^-3 N/m