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Yakvenalex [24]

2 years ago

8

A 7780 ‑kg car is travelling at 30.7 m/s when the driver decides to exit the freeway by going up a ramp. After coasting 404 m al

ong the exit ramp, the car's speed is 13.3 m/s, and it is h = 13.1 m above the freeway. What is the magnitude of the average drag force F drag exerted on the car?

1 answer:

vredina [299]2 years ago

6 0

Answer:

4899.47 N

Explanation:

mass of the car = 7780 kg, u, initial speed = 30.7 m/s, v, final speed = 13.3 m/s

d = 404 m and h = 13.1 m

initial kinetic energy of the car = potential energy attained by the car climbing the ramp + the work done by friction against the motion + final kinetic energy

1/2 mu² = mgh + (F × d) + 1/2mv²

3666286.1 J = 998796.4 J + ( 404 F) + 688102.1 J

404 F = 3666286.1 J - 998796.4 J - 688102.1 J

F = 1979387.6 / 404 = 4899.47 N

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Law Incorporation [45]

The best explanation for the difference in time is: A. The difference in weight doesn't affect the time, but they are affected differently by air resistance.

<h3>What is weight?</h3>

Weight can be defined as the force acting on an object or a physical body due to the effect of gravity. Also, the weight of an object (body) is typically measured in Newton.

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4 0

1 year ago

A flute player hears four beats per second when she compares her note to a 523 HzHz tuning fork (the note C). She can match the

laiz [17]

Answer:

527 Hz

Solution:

As per the question:

Beat frequency of the player, $\Delta f = 4\ beats/s$

Frequency of the tuning fork, f = 523 Hz

Now,

The initial frequency can be calculated as:

$\Delta f = f - f_{i}$

$f_{i} = f \pm \Delta f$

when

$f_{i} = f + \Delta f = 523 + 4 = 527 Hz$

when

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But we know that as the length of the flute increases the frequency decreases

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7 0

3 years ago

Read 2 more answers

Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m

Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

$m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s$

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$V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}$

Substituting,

$V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}$

$V_{cm} = 1.034m/s$

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where here $v_f$ is the velocity after the collision.

$v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}$

$v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}$

$v_f = 1.034m/s$

8 0

3 years ago

it takes car 1 minute to go from rest to 30 m/s east. What is the acceleration of this car? Please Show Work.

ivann1987 [24]

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7 0

3 years ago

Two long straight parallel wires separated by a distance of 20 cm carry currents of 30 A and 40 A in opposite directions. What i

irakobra [83]

Answer:

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now the magnetic field due to 30 A current is given as

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$B_2 = \frac{(2 \times 10^{-7})(40)}{0.25}$

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7 0

3 years ago