Answer:
= 0.086J
Explanation:
we are given three equal charges q₁ q₂ q₃ = 1.60 μC
sides of triangle = 0.800m
U = 1/4πε₀(3q²/r)
1/4πε₀ = 9.0 × 10⁹ N.m²/C²
U = 9.0 × 10⁹ ((3 * 1.60 × 10⁻⁶)² / (0.800))
= 0.086J
Well, first of all, a car moving around a circular curve is not moving
with uniform velocity. The direction of motion is part of velocity, and
the direction is constantly changing on a curve.
The centripetal force that keeps an object moving in a circle is
Force = (mass of the object) · (speed)² / (radius of the circle)
F = m s² / r
We want to know the radius, to rearrange the formula to give us
the radius as a function of everything else.
F = m s² / r
Multiply each side by 'r': F· r = m · s²
Divide each side by 'F': r = m · s² / F
We know all the numbers on the right side,
so we can pluggum in:
r = m · s² / F
r = (1200 kg) · (20 m/s)² / (6000 N) .
I'm pretty sure you can finish it up from here.
Answer:
bro just go to an easier class
Explanation:
#nerd
Answer:
34 m/s
Explanation:
Potential energy at top = kinetic energy at bottom + work done by friction
PE = KE + W
mgh = ½ mv² + Fd
mg (d sin θ) = ½ mv² + Fd
Solving for v:
½ mv² = mg (d sin θ) − Fd
mv² = 2mg (d sin θ) − 2Fd
v² = 2g (d sin θ) − 2Fd/m
v = √(2g (d sin θ) − 2Fd/m)
Given g = 9.8 m/s², d = 150 m, θ = 28°, F = 50 N, and m = 65 kg:
v = √(2 (9.8 m/s²) (150 m sin 28°) − 2 (50 N) (150 m) / (65 kg))
v = 33.9 m/s
Rounded to two significant figures, her velocity at the bottom of the hill is 34 m/s.
