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Lostsunrise [7]

3 years ago

7

Light going from glass (index of refraction = 1.4) to ethanol has a critical angle of incidence of 65°. What is the index of ref

raction of ethanol?

1 answer:

sukhopar [10]3 years ago

6 0

Answer:

1.27

Explanation:

$n_{g}$ = Index of refraction of glass = 1.4

$n_{e}$ = Index of refraction of ethanol = ?

i = Angle of incidence = critical angle = 65° deg

r = Angle of refraction = 90° deg

Using Snell's law

$n_{g}$ Sini = $n_{e}$ Sinr

(1.4) Sin65 = $n_{e}$ Sin90

$n_{e}$ = 1.27

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2 years ago

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates

melisa1 [442]

Answer:

$v_{2.6b}=11.18\ m.s^{-1}$

$v_{7.2b}=14.19\ m.s^{-1}$

$s_{bb}=226.3305\ m$

$a_{db}=-3.7386\ m.s^{-2}$ negative sign denotes deceleration.

$t_b=21.3956\ s$

$a_y=1.1065\ m.s^{-2}$

Explanation:

Given:

initial speed of blue car, $u_b=0\ m.s^{-1}$

initial speed of yellow car, $u_y=0\ m.s^{-1}$

acceleration rate of blue car, $a_b=4.3\ m.s^{-2}$

time for which the blue car accelerates, $t_{ab}=3.3\ s$

time for which the blue car moves with uniform speed before decelerating, $t_{ub}=14.3\ s$

total distance covered by the blue car before coming to rest, $s_b=253.26 \ m$

distance at which the the yellow car intercepts the blue car just as the blue car come to rest, $s_y=253.26 \ m$

1)

<u>Speed of blue car after 2.6 seconds of starting the motion:</u>

Applying the equation of motion:

$v_{2.6b}=u_b+a_b.t$

$v_{2.6b}=0+4.3\times 2.6$

$v_{2.6b}=11.18\ m.s^{-1}$

<u>Speed of blue car after 7.2 seconds of starting the motion:</u>

∵The car accelerates uniformly for 3.3 seconds after which its speed becomes uniform for the next 14.3 second before it applies the brake.

so,

$v_{7.2b}=u+a_b\times t_{ab}$

$v_{7.2b}=0+4.3\times 3.3$

$v_{7.2b}=14.19\ m.s^{-1}$

<u>Distance travelled by the blue car before application of brakes:</u>

This distance will be $s_{bb}=$ (distance travelled during the accelerated motion) + (distance travelled at uniform motion)

<em>Now the distance travelled during the accelerated motion:</em>

$s_{ab}=u_b.t_{ab}+\frac{1}{2} a_{b}.t_{ab}^2$

$s_{ab}=0\times 3.3+0.5\times 4.3\times 3.3^2$

$s_{ab}=23.4135\ m$

<em>Now the distance travelled at uniform motion:</em>

$s_{ub}=14.19\times 14.3$

$s_{ub}=202.917\ m$

Finally:

$s_{bb}=s_{ab}+s_{ub}$

$s_{bb}=23.4135+202.917$

$s_{bb}=226.3305\ m$

<u>Acceleration of the blue car once the brakes are applied</u>

Here we have:

initial velocity, $u=14.19\ m.s^{-1}$

final velocity, $v=0\ m.s^{-1}$

distance covered while deceleration, $s_{db}=s_b-s_{bb}$

$\Rightarrow s_{db}=253.26 -226.3305=26.9295\ m$

Using the equation of motion:

$v^2=u^2+2a_{db}.s_{db}$

$0^2=14.19^2+2\times a_{db}\times 26.9295$

$a_{db}=-3.7386\ m.s^{-2}$ negative sign denotes deceleration.

<u>The total time for which the blue car moves:</u>

$t_b=t_a+t_{ub}+t_{db}$ ........................(1)

<em>Now the time taken to stop the blue car after application of brakes:</em>

Using the eq. of motion:

$v=u+a_{db}.t_{db}$

$0=14.19-3.7386\times t_{db}$

$t_{db}=3.7956\ s$

Putting respective values in eq. (1)

$t_b=3.3+14.3+3.7956$

$t_b=21.3956\ s$

<u>For the acceleration of the yellow car:</u>

We apply the law of motion:

$s_y=u_y.t_y+\frac{1}{2} a_y.t_y^2$

<em>Here the time taken by the yellow car is same for the same distance as it intercepts just before the stopping of blue car.</em>

Now,

$253.26=0\times 21.3956+0.5\times a_y\times 21.3956^2$

$a_y=1.1065\ m.s^{-2}$

7 0

3 years ago