Answer:
98N and 147N
Explanation:
We have the following information:
We can find the static fricton force as follow,
Where N is the normal force (mg)
Static friction force at 147N is greater than the force applied hence body does not move.
Friction _____.
1 answer:
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Answer:
= 17º C
Explanation:
This is a calorimetry problem, where heat is yielded by liquid water, this heat is used first to melt all ice, let's look for the necessary heat (Q1)
Let's reduce the magnitudes to the SI system
Ice m = 80.0 g (1 kg / 1000 g) = 0.080 kg
L = 3.33 105 J / kg
Water M = 860 g = 0.860 kg
= 4186 J / kg ºC
Q₁ = m L
Q₁ = 0.080 3.33 10⁵
Q₁ = 2,664 10⁴ J
Now let's see what this liquid water temperature is when this heat is released
Q = M ΔT = M
(T₀₁ -
)
Q₁ = Q
= T₀₁ - Q / M ce
= 26.0 - 2,664 10⁴ / (0.860 4186)
= 26.0 - 7.40
= 18.6 ° C
The initial temperature of water that has just melted is T₀₂ = 0ª
The initial temperature of the liquid water is T₀₁= 18.6
m
+ M
= M
T₀₁ - m
T₀₂o2
= (M To1 - m To2) / (m + M)
= (0.860 18.6 - 0.080 0) / (0.080 + 0.860)
= 17º C
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