First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg.
F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N
Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m
Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356 N
Ratio = 0.356 N/589.18 N
<em>Ratio = 6.04</em>
Magnetic flux can be calculated by the product of the magnetic field and the area that is perpendicular to the field that it penetrates. It has units of Weber or Tesla-m^2. For the first question, when there is no current in the coil, the flux would be:
ΦB = BA
A = πr^2
A = π(.1 m)^2
A = π/100 m^2
ΦB = 2.60x10^-3 T (π/100 m^2 ) ΦB = 8.17x10^-5 T-m^2 or Wb (This is only for one loop of the coil)
The inductance on the coil given the current flows in a certain direction can be calculated by the product of the total number of turns in the coil and the flux of one loop over the current passing through. We do as follows:
L = N (ΦB ) / I
L = 30 (8.17x10^-5 T-m^2) / 3.80 = 6.44x10^-4 mH
PART A)
Electrostatic potential at the position of origin is given by
here we have
now we have
Now work done to move another charge from infinite to origin is given by
here we will have
so there is no work required to move an electron from infinite to origin
PART B)
Initial potential energy of electron
Now we know
now by energy conservation we will have
So here initial total energy is sufficient high to reach the origin
PART C)
It will reach the origin
Answer:
41.27m/s
Explanation:
According to law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the velocity after impact
Given
m1 = 0.2kg
u1 = 43.7m/s
m2 = 45.9g = 0.0459kg
u2 = 30.7m/s
Required
Velocity after impact v
Substitute the given parameters into the formula
0.2(43.7)+0.0459(30.7) = (0.2+0.0459)v
8.74+1.409 = 0.2459v
10.149 = 0.2459v
v = 10.149/0.2459
v = 41.27m/s
Hence the speed of the golf ball immediately after impact is 41.27m/s
