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2 years ago

15

A beam of light shining from water to air strikes the surface at an angle greater than 48° and produces total internal reflectio

n. If the incident beam is at a 50° angle, at what angle is the reflected beam?

1 answer:

shepuryov [24]2 years ago

3 0

<span>To find the solution for this problem, let us remember that the angle of incidence and the angle of reflection are always the same. Thus, the reflected beam would be at an angle of 50 degrees as wel</span>l.

<span> Total internal reflection means that the total amount of incident light at the boundary between two media gets reflected, and therefore no beam of light is transmitted to </span>the air in this case.

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6. An object accelerates from rest to 70 m/s in 3.5 s. What is the acceleration of the object?

sweet-ann [11.9K]

Answer:

The acceleration of the object is 20 meters per second square = 20 m/s^2

Explanation:

Recall that acceleration is defined as the change in velocity divided the time it takes for the change. Therefore , if the object accelerates from rest (zero velocity) to 70 m/s , the change in velocity is (70 m/s - 0 m/s = 70 m/s)

which divided by the 3.5 seconds it took for the change, gives:

acceleration = (70 m/s / 3.5 s ) = 20 m/s^2

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3 years ago

After the driver first notices the obstacle, the car moves uniformly for a time interval t1−t0=t before the brakes are applied.

loris [4]

Answer:

V(t1-t0)

Explanation:

Moving 'uniformly' means constant velocity (speed). the formula for constant speed motion is $V = \frac{distance}{time}$ =( change in position/ change in time)

where,

V is speed

given in the statement :

change in time = t = t1-t0

let the constant speed be ' V '

disance = X = X1-X0

applying the above mentioned formula: V = $\frac{X}{t}$

V = X/t

X = Vt

the distance X1-X0 = Vt =V(t1-t0)

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3 years ago

A runner has an original velocity of 6 m/s and slows to a final velocity of 0 m/s. If the runner covers a

myrzilka [38]

Answer:

4 s

Explanation:

Given:

Δx = 12 m

v₀ = 6 m/s

v = 0 m/s

Find: t

Δx = ½ (v + v₀) t

12 m = ½ (0 m/s + 6 m/s) t

t = 4 s

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2 years ago

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

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3 years ago

The noise floor, also known as additive white Gaussian noise (AWGN), is a continuous noise level that appears over a wide spectr

harkovskaia [24]

Answer:

correct option is a. True

Explanation:

solution

the noise floor is AWGN ( additive white Gaussian noise )

and when viewed in the frequency domain, it is the continuous noise level

because as they have a uniform power over all the frequency.

so that it is additive white Gaussian noise

as we can say given statement is True

correct option a true

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3 years ago