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3 years ago

8

For an object to be in static equilibrium, the net force must be equal to zero and and The center of gravity must be above the s

upport point. The moment of inertia must be zero. The normal force must equal the weight. The net torque must be equal to zero.

1 answer:

iren [92.7K]3 years ago

8 0

Answer:

The normal force must equal the weight.

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An car is brought to rest in a distance of 652 m using a constant acceleration of -2.7 m/s2. What was the

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if somebody swim to the right for 30 meters then swims to left for 15 meters and then swims back to the right for 50 meters what

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50m

Explanation:

Displacement is the length of path traveled which is measured from start to the finishing of the path.

Analysis of the journey;

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2 years ago

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E. The refracted ray is vertically polarized whereas the reflected ray is horizontally polarized.

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2 years ago

Does crystallization break down rocks at earths surface to form sediment

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No

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2 years ago

As a city planner, you receive complaints from local residents about the safety of nearby roads and streets. One complaint conce

WINSTONCH [101]

Answer:

a) x₁ = 290.50 feet , x₂ = 169.74 feet , b) v_max= 41 mph

Explanation:

For this exercise we will work in two parts, the first with Newton's second law to find the acceleration of vehicles

X Axis fr = m a

Y Axis N-W = 0

N = W = mg

The force of friction has the expression

fr = μ N

We replace

μ mg = ma

a = μ g

g = 32 feet / s²

Let's calculate the acceleration for each coefficient and friction

μ a (feet / s2)

0.599 19.168

0.536 17,152

0.480 15.360

0.350 11.200

These are the acceleration values, for the maximum distance we use the minimum acceleration (a₁ = 11,200 feet / s²) and for the minimum braking distance we use the maximum acceleration (x₂ = 19,168 feet / s²)

v² = v₀² - 2 a x

When the speed stops it is zero

x₁ = v₀² / 2 a₁

Let's reduce speed

v₀ = 55mph (5280 foot / 1 mile) (1h / 3600s) = 80,667 feet / s²

Let's calculate the maximum braking distance

x₁ = 80.667² / (2 11.2)

x₁ = 290.50 feet

The minimum braking distance

x₂ = 80.667² / (2 19.168)

x₂ = 169.74 feet

b) maximum speed to stop at distance x = 155 feet

0 = v₀² - 2 a x

v₀ = √2 a x

We calculate the speed for the two accelerations

v₀₁ = √ (2 11.2 155)

v₀₁ = 58.92 feet / s

v₀₂ = √ (2 19.168 155)

v₀₂ = 77.08 feet / s

To stop at the distance limit in the worst case the maximum speed must be 58.92 feet / s = 40.85 mph = 41 mph

5 0

3 years ago