Answer:
Explanation:
Given that
g=9.8m/s²
The spring constant is
k=50N/m
The length of the bungee cord is
Lo=32m
Height of bridge which one end of the bungee is tied is 91m
A steel ball of mass 92kg is attached to the other end of the bungee.
The potential energy(Us) of the steel ball before dropped from the bridge is given as
P.E= mgh
P.E= 92×9.8×91
P.E= 82045.6 J
Us= 82045.6 J
Potential energy)(Uc) of the cord is given as
Uc= ½ke²
Where 'e' is the extension
Then the extension is final height extended by cord minus height of cord
e=hf - hi
e=hf - 32
Uc= ½×50×(hf-32)²
Uc=25(hf-32)²
Using conservation of energy,
Then,
The potential energy of free fall equals the potential energy in string
Uc=Us
25(hf-32)²=82045.6
(hf-32)² = 82045.6/25
(hf-32)²=3281.825
Take square root of both sides
√(hf-32)²=√(3281.825)
hf-32=57.29
hf=57.29+32
hf=89.29m
We neglect the negative sign of the root because the string cannot compressed
Answer:
I think is 2.
Explanation:
(The entire range of wavelengths or frequencies of electromagnetic radiation extending from gamma rays to the longest radio waves and including visible light)
Answer:
(a) 7.72×10⁵ J
(b) 4000 J
(c) 1.82×10⁻¹⁶ J
Explanation:
Kinetic Energy: This can be defined energy of a body due to its motion. The expression for kinetic energy is given as,
Ek = 1/2mv²................... Equation 1
Where Ek = Kinetic energy, m = mass, v = velocity
(a)
For a moving automobile,
Ek = 1/2mv².
Given: m = 2.0×10³ kg, v = 100 km/h = 100(1000/3600) m/s = 27.78 m/s
Substitute into equation 1
Ek = 1/2(2.0×10³)(27.78²)
Ek = 7.72×10⁵ J
(b)
For a sprinting runner,
Given: m = 80 kg, v = 10 m/s
Substitute into equation 1 above,
Ek = 1/2(80)(10²)
Ek = 40(100)
Ek = 4000 J
(c)
For a moving electron,
Given: m = 9.10×10⁻³¹ kg, v = 2.0×10⁷ m/s
Substitute into equation 1 above,
Ek = 1/2(9.10×10⁻³¹)(2.0×10⁷)²
Ek = 1.82×10⁻¹⁶ J
Answer:
Frequency of oscillation, f = 4 Hz
time period, T = 0.25 s
Angular frequency, 
Given:
Time taken to make one oscillation, T = 0.25 s
Solution:
Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:
f = 
f = 
Frequency of oscillation, f = 4 Hz
The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.
Therefore, time period, T = 0.25 s
Angular frequency of oscillation is given by:
