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Aleks [24]
2 years ago
7

The drag force, fd, imposed by the surrounding air on a vehicle moving with velocity v is given by fd = cdaρv 2/2 where cd is a

constant called the drag coefficient, a is the projected frontal area of the vehicle, and ρ is the air density. an automobile is moving at v = 80 kilometers per hour with cd = 0.28, a = 2.3 m2, and ρ = 1.2 kg/m3.
Physics
1 answer:
kiruha [24]2 years ago
7 0

Answer: 191 N

Explanation:

Given that,

Velocity v = 80 km/hr

Area a = 2.3\ m^{2}

Air density \rho = 1.2\ kg/m^{3}

Constant cd = 0.28

We know the drag force,

F_{d} = cda\dfrac{1}{2}\rho v^{2}

F_{d} = 0.28\times 2.3\ m^{2}\times\dfrac{1}{2}\times1.2\ kg/m^{3} \times (\dfrac{800}{36}\ m/s)^{2}

F_{d} = 190.8 N

F_{d} = 191 N

Hence, the drag force is 191 N.








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creativ13 [48]

Answer: a) 456.66 s ; b) 564.3 m

Explanation: The time spend to cover any distance a constant velocity is given by:

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The slower student time is: t=780m/0.9 m/s= 866.66 s

For the faster students t=780 m/1,9 m/s= 410.52 s

Therefore the time difference is 866.66-410.52= 456.14 s

In order to calculate the distance that faster student should  walk

to arrive 5,5 m before that slower student, we consider the follow expressions:

distance =vslower*time1

distance= vfaster*time 2

The time difference is 5.5 m that is equal to 330 s

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2 years ago
The image shows one complete cycle of a mass on a spring in simple harmonic motion. An illustration of a mass on a vertical spri
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D. "The net force is zero, so the acceleration is zero"

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6 0
2 years ago
A 500 kg motorcycle accelerates at a rate of 2 m/s .how much force was applied to the motorcycle?
Aleksandr [31]

Answer:

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A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the
yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE          

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0     (initial KE = 0 because the ball starts from rest)  (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0
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A car is traveling at 118 km/h. What is its speed in mm/h?
Afina-wow [57]
1km=1000m=1000000mm
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