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:
Explanation:
a ) Between r = 0 and r = r₁
Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .
b ) From r = r₁ to r = r₂
At distance r , charge contained in the sphere of radius r
volume charge density x 4/3 π r³
q = Q x r³ / R³
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q x r³ / ε₀R³
E= Q x r / (4πε₀R³)
E ∝ r .
c )
Outside of r = r₂
charge contained in the sphere of radius r = Q
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q / ε₀
E = Q / 4πε₀r²
E ∝ 1 / r² .
Explanation:
Let us first calculate long does it take to go 12m at 30m/s( assumed speed)
12/30 = 0.4 seconds
horizontal distance the ball drop in that time
H= (0)(0.4)+1/2(-9.8)(0.4)2
H= -0.78m
negative sign shows that the height of the ball at the net from the top.
Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m
As 1.62m>0.9m so the ball will clear the net.
H_1= V0y t’ + ½ g t’^2
-2.4= (0)t’ + ½ (-9.8) t’^2
t’= 0.69s
X’=V0x t’
X’=(30)(0.96)
X’= 20.7m
Answer:
-0.79 J
Explanation:
We are given that
We have to find the work done by the electric force on the moving point charge.
Work done,
Where 
Using the formula
Kinetic energy is the energy of mass in motion, the kinetic energy that an object has is because of it's motion.Heavier objects that are moving have more kinetic energy than lighter ones.
