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sdas [7]
2 years ago
8

Determine the maximum theoretical speed that may be achieved over a distance of 66 m by a car starting from rest, knowing that t

he coefficient of static friction is 0.80 between the tires and the pavement and that 60 percent of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels.
Physics
1 answer:
My name is Ann [436]2 years ago
3 0

Answer:

v=32.49 m/s

Explanation:

Given that

Distance ,d= 66 m

Initial speed of the car ,u = 0 m/s

Coefficient of friction ,μ = 0.8

Lets take the total mass of the car = m

The acceleration of the car is given as

a = μ g                               ( g= 10 m/s² )

Now by putting the values in the above equation we get

a= 0.8 x 10 m/s²

a= 8 m/s²

We know that ,final speed is given as

v²= u ²+ 2 a d

Now putting the value

v²=0² + 2 x 8 x 66

v²= 1056

v=32.49 m/s

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On a winter day with a temperature of -10°C, 500g of snow (water ice) is brought inside where the temperature is 18 °C. The snow
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Answer:

Explanation:

Mass of ice m = 500g = .5 kg

Heat required to raise the temperature of ice by 10 degree

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= .5 x 2093 x 10 J

10465 J

Heat required to melt the ice

= mass of ice x latent heat

0.5 x 334 x 10³ J

167000 J

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Using newtons second law of motion, how fast for 100 KG object accelerates 350 N of force is applied to
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3.5m/s^2

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From Newton's second Law of Motion

F = ma

Where F is the applied force, m is the mass of the object and a is the acceleration.

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350N = 100×a

a = 350/100

a = 3.5m/s^2

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Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
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Answer:

\frac{1}{10}M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒I_{1} \times W_{1} =I_{2} \times W_{2}

  • I_{1} is the moment of interia of earth before impact
  • W_{1} is the angular velocity of earth about an axis passing through the center of earth before impact
  • I_{2} is moment of interia of earth and asteroid system
  • W_{2} is the angular velocity of earth and asteroid system about the same axis

let  W_{1}=W

since \text{Time period of rotation}∝\frac{1}{\text{Angular velocity}}

⇒ if time period is to increase by 25%, which is \frac{5}{4} times, the angular velocity decreases 25% which is \frac{4}{5}  times

therefore W_{1} = \frac{4}{5} \times W_{1}

I_{1}=\frac{2}{5} \times M\times R^{2}(moment of inertia of solid sphere)

where M is mass of earth

           R is radius of earth

I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where M_{1} is mass of asteroid

⇒ \frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})

\frac{1}{2} \times M\times R^{2}= (\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})

M_{1}\times R^{2}= \frac{1}{10} \times M\times R^{2}

⇒M_{1}=}\frac{1}{10} \times M

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It is C because less than one percent of water is fresh water
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