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gladu [14]
3 years ago
9

Resistance to the motion of an automobile consists of road friction, which is almost independent of speed, and air drag, which i

s proportional to speed-squared. For a certain car with a weight of 12000 N, the total resistant force F is given by F = 300 + 1.8v2, with F in newtons and v in meters per second. Calculate the power (in horsepower) required to accelerate the car at 0.96 m/s^2 when the speed is 82 km/h.
Physics
1 answer:
Troyanec [42]3 years ago
6 0

Answer:

73.52983 Hp

Explanation:

m = Mass of car = \dfrac{W}{g}

W = Weight of car = 12000 N

a = Acceleration = 0.96 m/s²

Velocity of the car

v=82\times \dfrac{1000}{3600}\\\Rightarrow v=\dfrac{82}{3.6}\\\Rightarrow v=\dfrac{41}{1.8}

From the question

F=300+1.8v^2\\\Rightarrow F=300+1.8\times \left(\dfrac{41}{1.8}\right)^2\\\Rightarrow F=1233.88\ N

Balancing the forces

\dfrac{P}{v}-1233.88=ma\\\Rightarrow P=(ma+12000)v\\\Rightarrow P=\left(\dfrac{12000}{9.81}\times 0.96+1233.88\right)\dfrac{41}{1.8}\\\Rightarrow P=54853.26055\ W=\dfrac{54853.26055}{746}\\\Rightarrow P=73.52983\ Hp

The power is 73.52983 Hp

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To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:

I=I_0 cos^2\theta

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I_ {0} indicates the intensity of the light before passing through the polarizer,

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Since we have two objects the law would be,

I=I_0cos^2\theta_1*cos^2(\theta_2-\theta_1)

Replacing the values,

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3 years ago
38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari
kolbaska11 [484]

Answer:

Final volumen first process V_{2} = 98,44 cm^{3}

Final Pressure second process P_{3} = 1,317 * 10^{10} Pa

Explanation:

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P_{1} = \frac{n_{1} R T_{1} }{V_{1} }

where:

P is the pressure, in Pa

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T is the temperature in Kelvin

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P_{1} = P_{2}

\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }

\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }

V_{2} = \frac{T_{2} V_{1} }{T_{1} }

Replacing : T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}

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Replacing on the ideal gases formula the pressure at this piont is:

P_{2} = 3,92 * 10^{9} Pa

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T = \frac{P V }{n R }

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\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }

P_{2} V_{2}  = P_{3} V_{3}

P_{3} = \frac{P_{2} V_{2}}{V_{3}}

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Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant
irakobra [83]

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Here

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M =V \rho

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V = Volume

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Now considering the volume of the star as a Sphere we have

V = \frac{4}{3} \pi R^3

Replacing at the previous equation we have,

M = (\frac{4}{3}\pi R^3)\rho

Now replacing the mass at the gravitational acceleration formula we have that

a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho

a_g = \frac{4}{3} G\pi R\rho

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration.  So centripetal acceleration of the star is

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\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R

Considering the constant values we have that

\Theta = \text{Constant} \times R

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As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

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A hockey ball accelerates from 0.m/s to 25m/s in 0.05 seconds what is the acceleration of the ball ?
S_A_V [24]

Answer:

500

Explanation:

25/0.05 .

I think this ?

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