Answer:
fr = ½ m v₀²/x
Explanation:
This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.
The best way to solve this exercise is to use the energy work theorem
W = ΔK
Where work is defined as the product of force by distance
W = fr x cos 180
The angle is because the friction force opposes the movement
Δk =
–K₀
ΔK = 0 - ½ m v₀²
We substitute
- fr x = - ½ m v₀²
fr = ½ m v₀²/x
Would be A 1012 N/C because The magnitude of the electric field at distance r from a point charge q is E=k
e
q/r
2
, so
E=
(5.11×10
−11
m)
2
(8.99×10
9
N.m
2
/C
2
)(1.60×10
−19
C)
=5.51×10
11
N/C∼10
1
2N/C
making (e) the best choice for this question.
The normal force is always perpendicular to the surface. So it would be straight to the left of the wall
Air for a diver comes out of a high pressure tank at - Same- pressure compared to the water around the diver (metered by the regulator).
This means the lungs are inflated with - Highly pressurized- gas.
This does not adversely affect the diver when deep underwater, because the entire environment around the diver is at -Same - pressure.
If the diver suddenly surface, the air in the alveoli in the lungs will still be at - a higher - pressure compared to the air around the diver, which will be at - a lower - pressure.
The gas in the diver's lungs will - expand - and can damage the alveoli.
The kinetic energy (KE) of a 0.155 kg arrow that is shot from ground level, upward at 31.4 m/s, when it is 30.0 m above the ground is 30.85 J
Assuming air friction is negligible,
a = - 9.8 m / s²
u = 31.4 m / s
s = 30 m
v² = u² + 2 a s
v² = 31.4² + ( 2 * - 9.8 * 30 )
v² = 985.96 - 588
v² = 397.96 m / s
KE = 1 / 2 m v²
KE = 1 / 2 * 0.155 * 397.96
KE = 0.0775 * 397.96
KE = 30.85 J
Therefore, the kinetic energy ( KE ) when it is 30.0 m above the ground is 30.85 J
To know more about kinetic energy
brainly.com/question/24360064
#SPJ1
