Answer: 9938.8 km
Explanation:
1 pound-force = 4.48 N
30.0 pounds-force = 134.4 N
The force of gravitation between Earth and object on the surface of is given by:
Where M is the mass of the Earth, m is the mass of the object, R (6371 km) is the radius of the Earth.
At height, h above the surface of the Earth, the weight of the object:
we need to find "h"
taking the ratio of two:
Hence, Pete would weigh 30 pounds at 9938.8 km above the surface of the Earth.
Answer:
5 mg, 
Explanation:
First of all, let's rewrite the mass in grams using scientific notation.
we have:
m = 0.005 g
To rewrite it in scientific notation, we must count by how many digits we have to move the dot on the right - in this case three. So in scientific notation is
If we want to convert into milligrams, we must remind that
1 g = 1000 mg
So we can use the proportion
and we find
Answer:
a) I = 13.04 A
b) R = 8.82 ohms
c) 1291.87 kilocalories are generated an hour.
Explanation:
let P be the power of the heater, V be the voltage of the heater, I be the current of the heater, R be the resistance.
a) we know that:
P = I×V
I = P/V
= (1500)/(115)
= 13.04 A
Therefore, the current of the heater is 13.04 A
b) we now have voltage and current, according to Ohm's law:
R = V/I
= (115)/(13.04)
= 8.82 ohms
Therefore, the resistance of the heating coil is 8.82 ohms.
c) the number of kilocalories generated in one hour by the heater is just the energy the heater produces in one hour which is given by:
E = P×t
= (1500)(1×60×60)
= 5400000 J
since 1 calorie = 4.81 J
1 kilocalorie = 0.001 calories
E = 5400000/4.18 ≈ 1291866.029 calories ≈1291.87 kilocalories
Therefore, 1291.87 kilocalories are produced/generated in one hour.
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
