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marta [7]
3 years ago
13

The driver of a car traveling 110 km/h slams on the brakes so that the car undergoes a constant acceleration, skidding to a comp

lete stop in 4.5 sec. What is the average acceleration of the car during braking?
Physics
1 answer:
Ber [7]3 years ago
5 0

Answer: -6.80\ ms^{-2}

Explanation:

We know that the formula for acceleration is given by:

a=\dfrac{v-u}{t} , where v = Final velocity

u= Initial velocity

Given :  The driver of a car traveling 110 km/h slams on the brakes so that the car undergoes a constant acceleration.

i.e. u=   110 km/h =\dfrac{110\times1000}{3600}\approx35.6\ m/s  [∵  1 km= 100 meters and 1 hour = 3600 seconds]

v=  0  m/s ( At brake , final velocity becomes 0)

t=4.5 seconds  

Substitute all the values in the formula , we get

a=\dfrac{0-30.56}{4.5}\approx-6.80\ ms^{-2}

Hence, the average acceleration of the car during braking is -6.80\ ms^{-2}.

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Answer: The value of the celsius temperature of the cube is 472.2°c.

Explanation:        

The expression for the power radiated is as follows;

P=A\epsilon\sigma T^{4}

Here, A is the area, \sigma is the stefan's constant,\epsilon is the emissivity and T is the temperature.

It is given in the problem that A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the same emissivity as the sphere.

Then the expression for the radiated power for the cube and the sphere can be expressed as;

A_{1}\epsilon \e\sigma T_{1}^{4}=A_{2}\epsilon \e\sigma T_{2}^{4}

Here, A_{1} is the area of the sphere, A_{2} is the area of the cube,T_{1}  is the temperature of the sphere and T_{2}  is the temperature of the cube.

The radiated powers and emissivity of the cube and the sphere are same.

A_{1}T_{1}^{4}=A_{2}T_{2}^{4}

The area of the sphere is A_{1}=4\pi \times r^{2}.

Here, r is the radius of the sphere.

The area of the cube is A_{2}=6\times a^{2}.

Here, a is the edge of the cube.

Put A_{1}=4\pi \times r^{2} and A_{2}=6\times a^{2}.

T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}  ....(1)

The masses and the densities of the sphere and the cube are same. Then the volumes are also same.

V_{1}=V_{2}

Here,V_{1},V_{1} are the volumes of the sphere and the cube.

\frac{4}{3}\pi r^{3}=a^{3}

\frac{r}{a}=(\frac{3}{4\pi })^{\frac{1}{3}}  

Put this value in the equation (1).

T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}T_{2}=T_{1}(\frac{2\pi }{3}\times ((\frac{3}{4\pi })^{\frac{1}{3}})^{2})^{\frac{1}{4}}

Put T_{1}=500°c.

T_{2}=(500)(\frac{2\pi }{3}\times (\frac{3}{4\pi })^{\frac{2}{3}})^{\frac{1}{4}}

T_{2}=472.2^{\circ}c

Therefore, the value of the celsius temperature of the cube is 472.7°c.    

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