(a) At a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).
(b) If the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.
<h3>
Conservation of mechanical energy</h3>
The effect of height and gravity on speed on the given planet Epislon is determined by applying the principle of conservation of mechanical energy as shown below;
ΔK.E = ΔP.E
¹/₂m(v²- u²) = mg(hi - hf)
¹/₂(v²- u²) = g(0 - hf)
v² - u² = -2ghf
v² = u² - 2ghf
where;
- v is the final velocity at upper level
- u is the initial velocity
- hf is final height
- g is acceleration due to gravity
when u² = 2gh, then v² = 0,
when gravity reduces, u² > 2gh, and v² > 0
Thus, at a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).
<h3>Final speed</h3>
v² = u² - 2ghf
where;
- u is the initial speed = 5 m/s
- g is acceleration due to gravity and its less than 9.8 m/s²
- v is final speed
- hf is equal height
Since g on Epislon is less than 9.8 m/s² of Earth;
5² - 2ghf > 3 m/s
Thus, if the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.
Learn more about conservation of mechanical energy here: brainly.com/question/6852965
The net force on particle particle q1 is 13.06 N towards the left.
<h3>
Force on q1 due to q2</h3>
F(12) = kq₁q₂/r₂
F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)
F(12) = -14.41 N (towards left)
<h3>Force
on q1 due to q3</h3>
F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)
F(13) = 1.352 N (towards right)
<h3>Net force on q1</h3>
F(net) = 1.352 N - 14.41 N
F(net) = -13.06 N
Thus, the net force on particle particle q1 is 13.06 N towards the left.
Learn more about force here: brainly.com/question/12970081
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Answer : The heat change of the cold water in Joules is, 
Explanation :
First we have to calculate the mass of cold water.
As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.


Now we have to calculate the heat change of cold water.
Formula used :

where,
Q = heat change of cold water = ?
m = mass of cold water = 45 g
c = specific heat of water = 
= initial temperature of cold water = 
= final temperature = 
Now put all the given value in the above formula, we get:


Therefore, the heat change of cold water is 
For the answer to the question above
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I hope my answer helped you.