The force of gravity and the force of air
Answer: Did you ever find the answer I’m on the same question
Explanation:
Answer:
θ = 29.38°
Explanation:
The centripetal force is given by the formula;
F_c = F_n(sin θ) = mv²/r
Now, the vertical component of the normal force is; F_n(cos θ)
Now, this vertical component is also expressed as; F_n(cos θ) = mg
Thus, the slope is;
F_n(sin θ)/F_n(cos θ) = (mv²/r)/mg
tan θ = v²/rg
v² = rg(tan θ)
The initial speed will be gotten from the relation;
(v_o)² = μ_s(gr)
Plugging rg(tan θ) for (v_o)², we have;
μ_s(gr) = rg(tan θ)
rg will cancel out to give;
μ_s = (tan θ)
Thus, θ = tan^(-1) μ_s
μ_s is coefficient of static friction given as 0.563
θ = tan^(-1) 0.563
θ = 29.38°
Let m = mass of earth
<span>let r = radius of earth </span>
<span>let a = acceleration due to gravity on earth </span>
<span>G = constant (it doesn't change) </span>
<span>If you plug in the mass and radius of the new planet, you get </span>
<span>a = G (.5m)/(2r)^2 </span>
<span>a = 1/8 (Gm/r^2) </span>
<span>so the acceleration on the new planet is 1/8 what it is here.</span>
To solve this exercise we will use the concept related to heat loss which is mathematically given as
Where,
m = mass
= Specific Heat
Change in temperature
Replacing with our values we have that
Specific heat of mercury
Replacing
Therefore the heat lost by mercury is 0.09J
