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ladessa [460]
2 years ago
9

What is the net force acting on a car cruising at a constant velocity of 70 km/h (a) on a level road, (b) on an uphill road?

Physics
1 answer:
xeze [42]2 years ago
6 0

The net force on the car for either case will be zero, since the car is moving at a constant velocity.

<h3>Net force on the car</h3>

The net force on the car is the sum of all the forces acting on the car.

∑F = ma

<h3>When the car is on a level road</h3>

When the car is on a level road, the only two forces acting on the car is the applied force and frictional force of the road.

F - Ff = 0

The net force will be zero, since the car is moving at a constant velocity.

<h3>When the car is on uphill</h3>

The forces acting on the car at the uphill is the applied force, weight of the car acting downwards and the frictional force acting opposite direction.

F - Wsinθ - μFₙcosθ = 0

The net force will be zero, since the car is moving at a constant velocity.

Learn more about net force at constant velocity here: brainly.com/question/14392124

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Answer:

The normal strain along an axis oriented 45° from the positive x axis in the clockwise direction is -ε₀/2

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\epsilon_{x}=\epsilon_{o}\\\\\epsilon_{y}=-2\epsilon_{o}\\\\\gamma_{xy}=0\\\\\theta=-45^{o}\\\\\epsilon_{x_{1}}=?

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Substituting the values:

\epsilon_{x_{1}}=\epsilon_{o}cos^{2}(-45)-2\epsilon_{o}sin^{2}(-45)+0\\\\\epsilon_{x_{1}}=\frac{\epsilon_{o}}{2}-2\frac{\epsilon_{o}}{2}\\\\\epsilon_{x_{1}}=-\frac{\epsilon_{o}}{2}

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3) A driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefticient of friction between
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Answer:

A) 37 m

Explanation:

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F=ma=-\mu mg

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m = 1000 kg is the mass of the car

\mu = 0.80 is the coefficient of friction

a is the deceleration of the car

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The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

a=-\mu g

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