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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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This same experiment is performed on the international space station. What is the primary issue with performing this experiment
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Difference in experimental data.

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3 years ago
A car with mass 950 kg and a speed of 16 m/s approaches an intersection. A 1300 kg minivan traveling at 21 m/s is heading for th
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Answer:

V_f = 13.8863 \angle 60.89\°

Explanation:

Our values are,

m_1 = 950Kg\\v_1 = 16m/s \\m_2 =1300Kg\\v_2 = 21m/s

We have all the values to apply the law of linear momentum, however, it is necessary to define the two lines in which the study will be carried out. Being an intersection the vehicle of mass m_1 approaches through the X axis, while the vehicle of mass m_2 approaches by the y axis. In the collision equation on the X axis, we despise the velocity of object 2, since it does not come in this direction.

m_1v_1=(m_1+m_2)v_fcos\theta

For the particular case on the Y axis, we do the same with the speed of object 1.

m_2v_2=(m_1+m_2)v_fsin\theta

By taking a final velocity as a component, we can obtain the angle between the two by relating the equations through the tangent

Tan\theta = \frac{m_2v_2}{m_1v_1}\\Tan\theta = \frac{1300*21}{950*16}\\\theta = tan^{-1}(1.7960)\\\theta = 60.89\°

Replacing in any of the two functions, given above, we will find the final speed after the collision,

(950)(16)=(950+1300)V_fcos(60.89)

V_f= \frac{(950)(16)}{(950+1300)cos(60.89)}

V_f = 13.8863 \angle 60.89\°

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A solid box made from a material whose density is rhoS floats with two thirds of its volume submerged in a liquid whose density
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Answer:

\frac{\rho_S}{\rho_L} = \frac{2}{3}

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density of the solid box material = \rho_s

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V\rho_sg= \frac{2V}{3}\rho_L g

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