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antiseptic1488 [7]
3 years ago
7

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

r (mph). At full power, the car can accelerate from zero to 32.0mph in time 1.10s .A)At full power, how long would it take for the car to accelerate from 0 to 64.0mph ? Neglect friction and air resistance. =4.40sPart BA more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 32.0mph in time 1.10s , how long would it take to go from zero to 64.0mph ?am not sure how to do part B
Physics
1 answer:
Inessa05 [86]3 years ago
7 0

Answer:

a) 4.40 s

b) 2.20 s

Explanation:

Given parameters are:

At constant power  ,

initial speed of the car, v_0=0

final speed of the car, v=32 mph

At full power,

initial speed of the car, v_0=0

final speed of the car, v=64 mph

a)

At constant power, KE = \frac{1}{2} mv^2

At full power, KE = \frac{1}{2} m(2v)^2

So KE_f = 4KE_i

So, time to reach 64 mph speed is 4 times more than the initial time

t = 4*1.10 =4.40 s

b)

v=v_0+at\\a=\frac{v-v_0}{t}=\frac{32-0}{1.1/3600}=104727.27 miles/hours^2

For final 64 mph speed,

v=v_0+at\\t=\frac{v-v_0}{a}=\frac{64-0}{104727.27} = 6.111*10^{-4} hours = 6.111*10^{-4}*3600=2.20 s

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FrozenT [24]

Answer:

<em>Speed of the electron is 2.46 x 10^8 m/s</em>

<em></em>

Explanation:

momentum of the electron before relativistic effect = M_{0} V

where M_{0} is the rest mass of the electron

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under relativistic effect, the mass increases.

under relativistic effect, the new mass M will be

M = M_{0}/ \sqrt{1 - \beta ^{2}  }

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V is the speed with which the electron travels.

The new momentum will therefore be

==> M_{0}V/ \sqrt{1 - \beta ^{2}  }

It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have

1.75M_{0} V = M_{0}V/ \sqrt{1 - \beta ^{2}  }

the equation reduces to

1.75 = 1/ \sqrt{1 - \beta ^{2}  }

square both sides of the equation, we have

3.0625 = 1/(1 - \beta ^{2} )

3.0625 - 3.0625\beta ^{2} = 1

2.0625 = 3.0625\beta ^{2}

\beta ^{2} = 0.67

β = 0.819

substitute for  \beta = V/c

V/c = 0.819

V = c x 0.819

V = 3 x 10^8 x 0.819 = <em>2.46 x 10^8 m/s</em>

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A block on a horizontal frictionless plane is attached to a spring, as shown below. The block oscillates along the x-axis with s
AleksandrR [38]

The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.

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D. At x=0, it's acceleration is at a maximum

Explanation:

As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.

Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.

From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.

The same can be said as the box travels backward from point A to -A

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A cat jumps from 2.5-meter-tall bookshelf to a 1.3-meter-tall countertop. If the cat
UkoKoshka [18]

The change in Potential energy of the cat is 176.4 J.

<h3 /><h3>Potential Energy:</h3>

This is the energy due to the position of a body. The S.I unit is Joules (J)

The formula for change in potential energy.

<h3 /><h3>Formula:</h3>
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<h3>Where:</h3>
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From the question,

<h3>Given:</h3>
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Substitute these values into equation 1

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Learn more about Potential energy here: brainly.com/question/1242059

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P = 4.96 × 10⁵ Pa

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3 years ago
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