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torisob [31]
3 years ago
6

A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes

and the car comes to a rest uniformly in a distance of 240 m. What are the magnitude and direction of the net force applied to the car to bring it to rest?
Physics
1 answer:
Nitella [24]3 years ago
3 0

Answer:

The value of F= - 830 N

Since the force is negative, it implies direction of the force applied was due south.

Explanation:

Given data:

Mass = 1000-kg

Distance, d = 240 m

Initial velocity, v1 = 20.0 m/s

Final velocity, v2 = 0 (since the car came to rest after brake was applied)

v2²= v1² + 2ad (using one of the equation of motion)

0=  20² + (2 x a x  240)

0= 400 + 480 a

a = - 400/480

a = - 0.83 m/s²

Then, imputing the value of a into

F = ma

F = 1000 kg x ( - 0.83 m/s²)

F= - 830 N

The car was driving toward the north, and since the force is negative, it implies direction of the force applied was due south.

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Answer:

The wires are connected to both terminals of the battery, so they form a closed loop. Most circuits have devices such as light bulbs that convert electrical energy to other forms of energy. ... When the switch is turned on, the circuit is closed and current can flow through it.

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Answer:

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By Considering the vertical distance and both vertical and horizontal final velocity, the time t = 0.45 s and Velocity V = 6.7 m/s

Given that a Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table.

Height h = 1.0 m

As the ball rolls off the table, it will be fallen under gravity. Where

g = 9.8 m/s^{2}

Initial vertical velocity u_{y} = 0

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Considering the vertical distance, the formula to use to calculate the time will be;

h = ut + 1/2gt^{2}

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It takes 0.45 seconds to hit the floor if no one sneezes.

To calculate its velocity when it hits the floor, we will need to calculate for both vertical and horizontal final velocity and find the resultant velocity of the two.

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V_{y} = 4.41 m/s

Horizontal component

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but a = 0

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Final velocity V = \sqrt{5^{2} + 4.41^{2}  }

V = 6.67 m/s

Therefore, it will hit the floor at a velocity of 6.7 m/s

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