Answer:
The wagon will move to the right.
Explanation:
From the question given above, the following data were obtained:
Force applied to the left (Fₗ) = 10 N
Force applied to the right (Fᵣ) = 30 N
Direction of the wagon =.?
To determine the direction in which the wagon will move, we shall determine the net force acting on the wagon. This can be obtained as follow:
Force applied to the left (Fₗ) = 10 N
Force applied to the right (Fᵣ) = 30 N
Net force (Fₙ) =?
Fₙ = Fᵣ – Fₗ
Fₙ = 30 – 10
Fₙ = 20 N to the right
From the calculations made above, the net force acting on the wagon is 20 N to the right. Hence the wagon will move to the right.
Answer:
Capacitance is 0.572×10⁻¹⁰ Farad
Explanation:
Radius = R₁ = 6.25 cm = 6.25×10⁻² m
Radius = R₂ = 15 cm = 15×10⁻² m
Dielectric constant = k = 4.8
Electric constant = ε₀ = 8.854×10⁻¹² F/m
ε/ε₀=k
ε=kε₀
![Capacitance\ (C)=\frac{4\pi k\epsilon_0 R_1\times R_2}{R_2-R_1}\\\Rightarrow C=\frac{4\pi 4.8\times 8.854\times 10^{-12}\times 15\times 10^{-2}\times 6.25\times 10^{-2}}{15\times 10^{-2}-6.25\times 10^{-2}}\\\Rightarrow C=0.572\times 10^{-10}\ Farad](https://tex.z-dn.net/?f=Capacitance%5C%20%28C%29%3D%5Cfrac%7B4%5Cpi%20k%5Cepsilon_0%20R_1%5Ctimes%20R_2%7D%7BR_2-R_1%7D%5C%5C%5CRightarrow%20C%3D%5Cfrac%7B4%5Cpi%204.8%5Ctimes%208.854%5Ctimes%2010%5E%7B-12%7D%5Ctimes%2015%5Ctimes%2010%5E%7B-2%7D%5Ctimes%206.25%5Ctimes%2010%5E%7B-2%7D%7D%7B15%5Ctimes%2010%5E%7B-2%7D-6.25%5Ctimes%2010%5E%7B-2%7D%7D%5C%5C%5CRightarrow%20C%3D0.572%5Ctimes%2010%5E%7B-10%7D%5C%20Farad)
∴ Capacitance is 0.572×10⁻¹⁰ Farad
Answer:b
Explanation:
Given
mass of heavy object is 4m
mass of lighter object is m
A person pushes each block with same force F
According to Work Energy theorem Change in kinetic energy of object is equal to Work done by all the object
As launching velocity is same for both the object so heavier mass must possess greater kinetic energy . For same force heavier mass must be pushed 4 times farther than the light block .
![\Delta (K.E.)_H=\frac{1}{2}(4m)v^2](https://tex.z-dn.net/?f=%5CDelta%20%28K.E.%29_H%3D%5Cfrac%7B1%7D%7B2%7D%284m%29v%5E2)
![\Delta (K.E.)_L=\frac{1}{2}(m)v^2](https://tex.z-dn.net/?f=%5CDelta%20%28K.E.%29_L%3D%5Cfrac%7B1%7D%7B2%7D%28m%29v%5E2)
So the correct option is b
Answer:
1 / 2 m v^2 = L m g (1 - cos θ)
This is the KE due to the pendulum falling from a 25 deg displacement
v^2 = 2 L g (1 - cos 25) = 2 * 2 * 9.8 (1 - .906) = 3.67 m^2/s^2
v = 1.92 m/s this is the speed due to an initial displacement of 25 deg
Its speed at the bottom would then be
1.92 + 1.2 = 3.12 m/s since it gains 1.92 m/s from its initial displacement
transverse and longitudinal