in the same direction as the displacement vector and the motion
Answer:
Step 1: We make the assumption that 125 is 100% since it is our output value.
Step 2: We next represent the value we seek with $x$.
Step 3: From step 1, it follows that $100\%=125$.
Step 4: In the same vein, $x\%=125$.
Step 5: This gives us a pair of simple equations:
$100\%=125(1)$.
Explanation:
M = mass of aluminium = 1.11 kg
= specific heat of aluminium = 900
= initial temperature of aluminium = 78.3 c
m = mass of water = 0.210 kg
= specific heat of water = 4186
= initial temperature of water = 15 c
T = final equilibrium temperature = ?
using conservation of heat
Heat lost by aluminium = heat gained by water
M
(
- T) = m
(T -
)
(1.11) (900) (78.3 - T) = (0.210) (4186) (T - 15)
T = 48.7 c
<span>more lines = a lot of electrons returning back to ground state from same level</span>
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;
