This problem involves Newton's universal law of gravitation and the equation to follow would be.
F = GM₁M₂/r²
Given: M₁ = 0.890 Kg; M₂ = 0.890 Kg; F = 8.06 x 10⁻¹¹ N; G = 6.673 X 10⁻¹¹ N m²/Kg²
Solving for distance r = ?
r = √GM₁M₂/F
r = √(6.673 x 10⁻¹¹ N m₂/Kg²)(0.890 Kg)(0.890 Kg)/ 8.06 x 10⁻¹¹ N
r = 0.81 m
Answer:
14 m/s
Explanation:
u = 0, h = 10 m, g = 9.8 m/s^2
Use third equation of motion
v^2 = u^2 + 2 g h
Here, v be the velocity of ball as it just strikes with the ground
v^2 = 0 + 2 x 9.8 x 10
v^2 = 196
v = 14 m/s
Answer:
Explanation:
First of all, I used the specific heat of water as 4182 J/(kgC) and the specific heat of ethyl alcohol (EtOH) as 2440 J/(kgC); that means that we need the masses in kg, not g.
120.g = .1200 kg of ethyl alcohol. Now for the formula:
where spheat is specific heat.
Filling that horrifying-looking formula in with some values:
and
and
16(4182x + 292.8) = 83640x + 2928 and
66912x + 4684.8 = 83640x + 2928 and
1756.8 = 16728x so
x = .105 kg and the amount of water added is 105 g