Potential energy is measured using formula Ep=mgh
m=mass (kg)
g= acceleration due to gravity (which is 9.8 on earth)
h= height in metres above ground
For this question
m=0.1
g=9.8
h=1
So Ep=0.1(9.8)(1)
Ep=0.98 Joules
When it is dropped all of this potential energy is converted into kinetic energy which can be measured using formula
Ek=1/2m(v^2) (v=final velocity)
Since all potential energy in this q is converted to kinetic we know Ek=0.98Joules and our mass is the same (0.1kg)
So when we sub everything in we get
0.98=1/2(0.1)(v^2)
0.98=0.05(v^2)||divide both side by 0.05
19.6=v^2 ||square root both sides
v=4.4 m/s
Answer:
q = 3.87 x 10⁵ C
Explanation:
given,
Electric field, E = 8.60 x 10¹ = 86 N/C
radius of earth, R = 6371 Km = 6.371 x 10⁶ m
Coulomb constant, K = 9 x 10⁹ N · m²/C²
Charge on the earth = ?
the electric field at the point
inserting all the values
q = 3.87 x 10⁵ C
The electric charge on the earth is equal to 3.87 x 10⁵ C
<span>A wave with a large amplitude has a lot of a.vibration b.speed <u> c.energy</u></span>
Answer:
0.94 m³/s
Explanation:
From the question given above, the following data were obtained:
Air flow (in ft³/min) = 2×10³ ft³/min
Air flow (in m³/s) =.?
Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:
35.315 ft³/min = 1 m³/min
Therefore,
2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min
2×10³ ft³/min = 56.63 m³/min
Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:
1 m³/min = 1/60 m³/s
Therefore,
56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min
56.63 m³/min = 0.94 m³/s
Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.
False the North Star never changes it position
