Potential Energy= 24m * 14kg * 9.8N/kg = 3292.8J
Answer:
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Answer: d= 0.57* l
Explanation:
We need to check that before ladder slips the length of ladder the painter can climb.
So we need to satisfy the equilibrium conditions.
So for ∑Fx=0, ∑Fy=0 and ∑M=0
We have,
At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal
At the top of ladder, N₂ acting horizontal
And Between somewhere we have the weight of painter acting downward equal to= mg
So, we have N₁=mg
and also mg*d*cosФ= N₂*l*sin∅
So,
d=
* tan∅
Also, we have f₁=N₂
As f₁= чN₁
So f₁= 0.357 * 69.1 * 9.8
f₁= 241.75
Putting in d equation, we have
d=
* tan 58
d= 0.57* l
So painter can be along the 57% of length before the ladder begins to slip
Answer:
47.8 °C
Explanation:
Use the heat equation:
q = mCΔT
where q is the heat absorbed/lost,
m is the mass of water,
C is the specific heat capacity,
and ΔT is the change in temperature.
Here, q = 100 kJ, m = 0.5 kg, and C = 4.184 kJ/kg/°C.
100 kJ = (0.5 kg) (4.184 kJ/kg/°C) ΔT
ΔT = 47.8 °C
Answer:

Explanation:
Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:

Where
and
are the orbital periods of Mercury and Earth respectively. We have
and
. Replacing this and solving for
