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ratelena [41]
3 years ago
12

A driver starts from rest on a straight test track that has markers every 0.14 km. The driver presses on the accelerator and for

the entire period of the test holds the car at constant acceleration. The car passes the 0.14 km post at 8.0 s after starting the test.
(a) What was the car's acceleration?
(b) What was the car's speed as it passed the 0.14 km post?
Physics
1 answer:
kari74 [83]3 years ago
5 0

Answer:

(A) Constant acceleration will be a=1.479m/sec^2

(B) Velocity of the car when it passes 0.14 km is 20.349 m/sec

Explanation:

It is given that driver starts from rest so initial velocity of the driver u = 0 m/sec

Distance traveled s = 0.14 km = 140 m

Time taken to cross 0.14 km is 8 sec

So time taken t = 8 sec

(A) From second equation of motion s=ut+\frac{1}{2}at^2

So 140=0\times 8+\frac{1}{2}\times a\times 8^2

a=1.479m/sec^2

(B) From third equation of motion

v^2=u^2+2as

v^2=0^2+2\times 1.479\times 140

v^2=414.12

v = 20.349 m/sec

So speed of the car when it passed 0.14 km is 20.349 m/sec

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A car changes velocity at a constant acceleration of 2.5m/s to reach 43.7m/s in 2.7 s how fast was the car moving when it began
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A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w
frosja888 [35]

Answer:

(a) n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) n_{2} = 1.349

(c) v_{1} = 2.04\times 10^{8}\ m/s

(d) v_{2} = 2.22\times 10^{8}\ m/s

Solution:

As per the question:

Refractive index of medium 1, n_{1} = 1.47

Angle of refraction for medium 1, \theta_{1} = 59^{\circ}

Angle of refraction for medium 2, \theta_{1} = 69^{\circ}

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

n_{1}sin\theta_{1} = n_{2}sin\theta_{2}

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Now,

n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}

n_{2} = 1.349

(c) To calculate the velocity of light in medium 1:

We know that:

Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}

Thus for medium 1

n_{1} = \frac{c}{v_{1}

v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s

(d) To calculate the velocity of light in medium 2:

For medium 2:

n_{2} = \frac{c}{v_{2}

v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s

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