Answer:
μ = 0.109
Explanation:
Draw a free body diagram of the crate. There are four forces:
Weight force mg pulling down.
Normal force N pushing up.
Applied force P pulling at θ above the horizontal.
Friction force Nμ pushing to the left.
Sum of the forces in the y direction:
∑F = ma
N + P sin θ − mg = 0
N = mg − P sin θ
Sum of the forces in the x direction:
∑F = ma
P cos θ − Nμ = ma
P cos θ − ma = Nμ
μ = (P cos θ − ma) / N
μ = (P cos θ − ma) / (mg − P sin θ)
Given:
P = 585 N
θ = 28.0°
m = 125 kg
a = 3.30 m/s²
μ = (585 cos 28.0° − 125 kg × 3.30 m/s²) / (125 kg × 9.8 m/s² − 585 sin 28.0°)
μ = 0.109
Lets say sphere 1 has a charge of 12 + and sphere 2 has a charge of 0 +. After they are touched Sphere 1 becomes 6 + and sphere 2 6 +. So 6 - 12 = a change of 6 -, while 6 - 0 = a change of 6 + Therfore,
Answer: The sign of the charge change / transfered are opposites.
Answer:
a) v = 88.54 m/s
b) vf = 26.4 m/s
Explanation:
Given that;
m = 1400.0 kg
a)
by using the energy conservation
loss in potential energy is equal to gain in kinetic energy
mg × ( 3200-2800) = 1/2 ×m×v²
so
1400 × 9.8 × 400 = 0.5 × 1400 × v²
5488000 = 700v²
v² = 5488000 / 700
v² = 7840
v = √7840
v = 88.54 m/s
b)
Work done by all forces is equal to change in KE
W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)
we substitute
1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf² -0 )
488000 = 700 vf²
vf² = 488000 / 700
vf² = 697.1428
vf = √697.1428
vf = 26.4 m/s
The correct answer is “C” ultrasound. Hope this helps!
