Answer is 12 milesBecause yeah
Answer:
The average current that this cell phone draws when turned on is 0.451 A.
Explanation:
Given;
voltage of the phone, V = 3.7 V
electrical energy of the phone battery, E = 3.15 x 10⁴ J
duration of battery energy, t = 5.25 h
The power the cell phone draws when turned on, is the rate of energy consumption, and this is calculated as follows;

where;
P is power in watts
E is energy in Joules
t is time in seconds

The average current that this cell phone draws when turned on:
P = IV

Therefore, the average current that this cell phone draws when turned on is 0.451 A.
Answer:
(3) The period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.
(4) he gravitational force between the Sun and Neptune is 6.75 x 10²⁰ N
Explanation:
(3) The period of a satellite is given as;

where;
T is the period of the satellite
M is mass of Earth
r is the radius of the orbit
Thus, the period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.
(4)
Given;
mass of the ball, m₁ = 1.99 x 10⁴⁰ kg
mass of Neptune, m₂ = 1.03 x 10²⁶ kg
mass of Sun, m₃ = 1.99 x 10³⁰ kg
distance between the Sun and Neptune, r = 4.5 x 10¹² m
The gravitational force between the Sun and Neptune is calculated as;

There is a lot of glare off of the ice, due to the sun and it also is always good to have eye protection in case you fall face first :P
Answer:
the can's kinetic energy is 0.42 J
Explanation:
given information:
Mass, m = 460 g = 0.46 kg
diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m
velocity, v = 1.1 m/s
the kinetic energy of the can is the total of kinetic energy of the translation and rotational.
KE =
I ω^2 + 
where
I =
and ω = 
thus,
KE =
(
)^2 + 
=
+ 
=
+ 
= 
=
= 0.42 J