Answer:
11.8 m/s
Explanation:
At the top of the hill, there are two forces on the car: weight force pulling down (towards the center of the circle), and normal force pushing up (away from the center of the circle).
Sum of forces in the centripetal direction:
∑F = ma
mg − N = m v²/r
At the maximum speed, the normal force is 0.
mg = m v²/r
g = v²/r
v = √(gr)
v = √(9.8 m/s² × 14.2 m)
v = 11.8 m/s
If the rod is in rotational equilibrium, then the net torques acting on it is zero:
∑ τ = 0
Let's give the system a counterclockwise orientation, so that forces that would cause the rod to rotate counterclockwise act in the positive direction. Compute the magnitudes of each torque:
• at the left end,
τ = + (50 N) (2.0 m) = 100 N•m
• at the right end,
τ = - (200 N) (5.0 m) = - 1000 N•m
• at a point a distance d to the right of the pivot point,
τ = + (300 N) d
Then
∑ τ = 100 N•m - 1000 N•m + (300 N) d = 0
⇒ (300 N) d = 1100 N•m
⇒ d ≈ 3.7 m
Well formation of metallic bond depends on free electrons.smaal sized atoms hold their electrons more firmly as compared to large size atoms ,this z due to distance of outer shell electrons by nucleus..in this way no of free electrons affect strength of metallic bond..smaal sized atoms release less free electrons..
The height of the oil column above the water in the vessel is determined as 2 cm.
<h3>
Pressure of the vessel</h3>
The pressure of the vessel due to water, oil and silver poured into the vessel is determined from mercury column.
let level of mercury = 20 cm + 0.5 cm = 20.5 cm
20.5 cmHg = 205 mmHg
1 mmHg = 133.32 Pa
205 mmHg = 27,330.6 Pa
<h3>Height of the liquids in the vessel</h3>
P = ρgh
where;
ρ is the density of water, oil and silver respectively
ρ = 1000 kg/m³ + 881 kg/m³ + 10,800 kg/m³ = 12,681 kg/m³
h = P/(ρg)
h = (27,330.6) / (12,681 x 9.8)
h = 0.22 m
h = 22 cm
<h3>Height of oil column</h3>
Oil is less dense than water and will float on water.
Height of oil column = 22 cm - 20 cm = 2 cm
Learn more about density here: brainly.com/question/6838128
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