F = G Mm/r²
mg = G Mm/r²
g = GM/r²
At centre of earth, r=0
g = GM/0
g =0
Answer:
2274 J/kg ∙ K
Explanation:
The complete statement of the question is :
A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.
= mass of metal = 400 g
= specific heat of metal = ?
= initial temperature of metal = 100 °C
= mass of aluminum cup = 100 g
= specific heat of aluminum cup = 900.0 J/kg ∙ K
= initial temperature of aluminum cup = 15 °C
= mass of water = 500 g
= specific heat of water = 4186 J/kg ∙ K
= initial temperature of water = 15 °C
= Final equilibrium temperature = 40 °C
Using conservation of energy
heat lost by metal = heat gained by aluminum cup + heat gained by water
Answer:
0
Explanation:
It’s before the projectile was fired, so nothing has happened yet.
An arrow which shows the direction that the probe should be moving in order for it to enter the orbit is X.
<h3>What is an orbit?</h3>
An orbit can be defined as the curved path through which a astronomical (celestial) object such as planet Earth, in space move around a Moon, Sun, planet or star.
In this scenario, if the scientists want the probe to enter the orbit they should ensure that probe moves in direction X. This ultimately implies that, the probe must move in the same direction as the orbit, in order to enter it.
Read more on orbit here: brainly.com/question/18496962
#SPJ1
Hi there!
Use the equation:
Where m2 and v2 deal with the larger object, and m1 and v1 with the smaller object. Plug in the given values:
v2 = ?
m1 = 0.048 kg (converted)
m2 = 2.95
v1 = 391
