THE BALL'S ACCELERATION IS 130.90![\frac{m}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bs%7D)
Explanation:
According to Newton's second law "the force (F) acting on an object is equal to the mass (m) of an object times its acceleration (a)", meaning that when the soccer player kicks a ball, a force is acting on the ball, therefore increasing it's acceleration
SO FOR CALCULATING THIS WE WILL USE NEWTON LAW,
FORCE = MASS × ACCELERATION
WE ARE GIVEN
FORCE = 57.6N
MASS= .44KG
SO HERE WE APPLYING FORMULA ,WE WILL GET
ACCELERATION = ![\frac{FORCE}{MASS}](https://tex.z-dn.net/?f=%5Cfrac%7BFORCE%7D%7BMASS%7D)
ACCELERATION = ![\frac{57.6N}{.44KG}](https://tex.z-dn.net/?f=%5Cfrac%7B57.6N%7D%7B.44KG%7D)
ACCELERATION =130.90![\frac{m}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bs%7D)
Answer:
![1.475\times 10^{-13}\ C/m^3](https://tex.z-dn.net/?f=1.475%5Ctimes%2010%5E%7B-13%7D%5C%20C%2Fm%5E3)
Explanation:
= Permittivity of free space = ![8.85\times 10^{-12}\ F/m](https://tex.z-dn.net/?f=8.85%5Ctimes%2010%5E%7B-12%7D%5C%20F%2Fm)
A = Area
h = Altitude = 600 m
Electric flux through the top would be
(negative as the electric field is going into the volume)
At the bottom
![120A](https://tex.z-dn.net/?f=120A)
Total flux through the volume
![\phi=120-110\\\Rightarrow \phi=10A](https://tex.z-dn.net/?f=%5Cphi%3D120-110%5C%5C%5CRightarrow%20%5Cphi%3D10A)
Electric flux is given by
![\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=10A\epsilon_0](https://tex.z-dn.net/?f=%5Cphi%3D%5Cdfrac%7Bq%7D%7B%5Cepsilon_0%7D%5C%5C%5CRightarrow%20q%3D%5Cphi%5Cepsilon_0%5C%5C%5CRightarrow%20q%3D10A%5Cepsilon_0)
Charge per volume is given by
![\rho=\dfrac{q}{v}\\\Rightarrow \rho=\dfrac{10A\epsilon_0}{Ah}\\\Rightarrow \rho=dfrac{10\epsilon_0}{h}\\\Rightarrow \rho=\dfrac{10\times 8.85\times 10^{-12}}{600}\\\Rightarrow \rho=1.475\times 10^{-13}\ C/m^3](https://tex.z-dn.net/?f=%5Crho%3D%5Cdfrac%7Bq%7D%7Bv%7D%5C%5C%5CRightarrow%20%5Crho%3D%5Cdfrac%7B10A%5Cepsilon_0%7D%7BAh%7D%5C%5C%5CRightarrow%20%5Crho%3Ddfrac%7B10%5Cepsilon_0%7D%7Bh%7D%5C%5C%5CRightarrow%20%5Crho%3D%5Cdfrac%7B10%5Ctimes%208.85%5Ctimes%2010%5E%7B-12%7D%7D%7B600%7D%5C%5C%5CRightarrow%20%5Crho%3D1.475%5Ctimes%2010%5E%7B-13%7D%5C%20C%2Fm%5E3)
The volume charge density is ![1.475\times 10^{-13}\ C/m^3](https://tex.z-dn.net/?f=1.475%5Ctimes%2010%5E%7B-13%7D%5C%20C%2Fm%5E3)
Answer:
Yes
Explanation:
Heat affects the magnets because it confuses and misaligns the magnetic domains, causing magnetism to decrease
Answer:
Explanation:
Given that,
Mass of the thin hoop
M = 2kg
Radius of the hoop
R = 0.6m
Moment of inertial of a hoop is
I = MR²
I = 2 × 0.6²
I = 0.72 kgm²
Period of a physical pendulum of small amplitude is given by
T = 2π √(I / Mgd)
Where,
T is the period in seconds
I is the moment of inertia in kgm²
I = 0.72 kgm²
M is the mass of the hoop
M = 2kg
g is the acceleration due to gravity
g = 9.8m/s²
d is the distance from rotational axis to center of of gravity
Therefore, d = r = 0.6m
Then, applying the formula
T = 2π √ (I / MgR)
T = 2π √ (0.72 / (2 × 9.8× 0.6)
T = 2π √ ( 0.72 / 11.76)
T = 2π √0.06122
T = 2π × 0.2474
T = 1.5547 seconds
T ≈ 1.55 seconds to 2d•p
Then, the period of oscillation is 1.55seconds