<span>f2 = f0/4
The gravity from the planet can be modeled as a point source at the center of the planet with all of the planet's mass concentrated at that point. So the initial condition for f0 has the satellite at a distance of 2r, where r equals the planet's radius.
The expression for the force of gravity is
F = G*m1*m2/r^2
where
F = Force
G = Gravitational constant
m1,m2 = masses involved
r = distance between center of masses.
Now for f2, the satellite has an altitude of 3r and when you add in the planet's radius, the distance from the center of the planet is now 4r. When you compare that to the original distance of 2r, that will show you that the satellite is now twice as far from the center of the planet as it was when it started. So let's compare the gravitational attraction, before and after.
f0 = G*m1*m2/r^2
f2 = G*m1*m2/(2r)^2
f2/f0 = (G*m1*m2/(2r)^2) / (G*m1*m2/r^2)
The Gm m1, and m2 terms cancel, so
f2/f0 = (1/(2r)^2) / (1/r^2)
f2/f0 = (1/4r^2) / (1/r^2)
And the r^2 terms cancel, so
f2/f0 = (1/4) / (1/1)
f2/f0 = (1/4) / 1
f2/f0 = 1/4
f2 = f0*1/4
f2 = f0/4
So the gravitational force on the satellite after tripling it's altitude is one fourth the original force.</span>
Answer:
53.63 μA
Explanation:
radius of solenoid, r = 6 cm
Area of solenoid = 3.14 x 6 x 6 = 113.04 cm^2 = 0.0113 m^2
n = 17 turns / cm = 1700 /m
di / dt = 5 A/s
The magnetic field due to the solenoid is given by
B = μ0 n i
dB / dt = μ0 n di / dt
The rate of change in magnetic flux linked with the solenoid =
Area of coil x dB/dt
= 3.14 x 8 x 8 x 10^-4 x μ0 n di / dt
= 3.14 x 64 x 10^-4 x 4 x 3.14 x 10^-7 x 1700 x 5 = 2.145 x 10^-4
The induced emf is given by the rate of change in magnetic flux linked with the coil.
e = 2.145 x 10^-4 V
i = e / R = 2.145 x 10^-4 / 4 = 5.36 x 10^-5 A = 53.63 μA
Answer:
44 1/3 cm
Explanation:
The cube has an edge length of ∛0.027 m = 0.3 m, so a center of mass (CoM) 15 cm above the floor.
The sphere's center of mass is 40 cm above the top of the cube, so is 70 cm above the floor. The weighted average of the CoM locations is ...
((15 cm)(0.700 kg) +(70 cm)(0.800 kg))/(0.700 kg +0.800 kg)
= (10.5 kg·cm +56 kg·cm)/(1.500 kg) = 44.333... cm
The center of mass of the two-object system is 44 1/3 cm above the floor.
_____
<em>Comment on the units</em>
We're not familiar with "hcm" as a unit. We presume that you can convert the given answer to the units you desire.
<h3>Question:</h3>
A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.
<h3>
Answer:</h3>
1.6nT [in the negative z direction]
<h2>
Explanation:</h2>
The magnetic field, B, due to a distance of finite value b, is given by;
B = (μ₀IL) / (4πb
) -----------(i)
Where;
I = current on the wire
L = length of the wire
μ₀ = magnetic constant = 4π × 10⁻⁷ H/m
From the question,
I = 20A
L = 2.0cm = 0.02m
b = 5.0m
Substitute the necessary values into equation (i)
B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0 
)
B = (10⁻⁷ x 20 x 0.02) / (5.0 )
B = (10⁻⁷ x 20 x 0.02) / (5.0 
)
B = (10⁻⁷ x 20 x 0.02) / (25.0)
B = 1.6 x 10⁻⁹T
B = 1.6nT
Therefore, the magnetic field at the point x = 5.0m on the x-axis is 1.6nT.
PS: Since the current is directed in the positive y direction, from the right hand rule, the magnetic field is directed in the negative z-direction.
Answer:
The plate's surface charge density is 
Explanation:
Given that,
Speed = 9800 km/s
Distance d= 75 cm
Distance d' =15 cm
Suppose we determine the plate's surface charge density?
We need to calculate the surface charge density
Using work energy theorem
Here, final velocity is zero
...(I)
We know that,
...(II)
From equation (I) and (II)
Charge is negative for electron
Put the value into the formula
Hence, The plate's surface charge density is
