Answer:1 trip around the earth is an angular displacement of 2*pi
3.6525*10^2 days
I
Explanation:24 h/1 day * 3.600*10^3 s/1h = 3.156*10^7 s
Angular speed = angular displacement / time
Angular speed = 2*pi rads / 3.156*10^7 s = 1.9910*10^-7 rad/s
Answer:
3.76 m/s
Explanation:
Instantaneous velocity: This can be defined as the velocity of an object in a non uniform motion. The S.I unit is m/s.
v' = dx(t)/dt..................... Equation 1
Where v' = instantaneous velocity, x = distance, t = time.
Given the expression,
x(t) = 28.0 m + (12.4 m/s)t - (0.0450 m/s³)t³
x(t) = 28 + 12.4t - 0.0450t³
Differentiating x(t) with respect to t.
dx(t)/dt = 12.4 - 0.135t²
dx(t)/dt = 12.4 - 0.135t²
When t = 8.00 s.
dx(t)/dt = 12.4 - 0.135(8)²
dx(t)/dt = 12.4 - 8.64
dx(t)/dt = 3.76 m/s.
Therefore,
v' = 3.76 m/s.
Hence, the instantaneous velocity = 3.76 m/s
Answer:
6.5 x 10^32 eV
Explanation:
mass of particle, mo = 1 g = 0.001 kg
velocity of particle, v = half of velocity of light = c / 2
c = 3 x 10^8 m/s
Energy associated to the particle
E = γ mo c^2
Convert Joule into eV
1 eV = 1.6 x 10^-19 J
So, 
Answer:
a. A = 0.1656 m
b. % E = 1.219
Explanation:
Given
mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150 m/s , k = 500 N / m
a.
To find the amplitude of the resulting SHM using conserver energy
ΔKe + ΔUg + ΔUs = 0
¹/₂ * m * v² - ¹/₂ * k * A² = 0
A = √ mB * vₓ² / k
vₓ = mb * u₁ / mb + mB
vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518
A = √ 4.0 kg * (1.852 m/s)² / (500 N / m)
A = 0.1656 m
b.
The percentage of kinetic energy
%E = Es / Ek
Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5
Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N
% E = 13.72 / 1125 = 0.01219 *100
% E = 1.219
Answer:0.0704 kg
Explanation:
Given
initial Absolute pressure
=210+101.325=311.325
as the volume remains constant therefore
therefore Gauge pressure is 337.44-101.325=236.117 KPa
Initial mass 
Final mass 
Therefore 
=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back
