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3 years ago

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Assume light is traveling through an optical medium of n1 and is incident on a boundary with a different optical medium with n2.

If the speed of light in the second medium is half the value compared to that of the first one, then one properly concludes that

1 answer:

Elza [17]3 years ago

5 0

That n2 = 2*n1. That is, the index of refraction is twice as big in medium 2 since v=c/n

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A car traveling at 50 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 61 cm (with respect to the

Karolina [17]

Answer:

6957.04N

Explanation:

Using

vf2=vi2+2ad

But vf = 0 .

So convert 50km/hr to m/s, and you need to convert 61 cmto m

(50km/hr)*(1hr/3600s)*(1000m/km) = 13.9m/s

61cm * (1m/100cm) = .61m

So n

0 = (13.9m/s)^2 + 2a(.61m)

a = 158.11m/s^2

So

using F = ma

F = 44kg(158.11m/s^2) = 6957.04N

3 0

3 years ago

Read 2 more answers

I pooped in the bathtub? I tryed to poop in my hand so i can carry it to the fish tank so they can feed thereselfs but it fell t

cupoosta [38]

Answer:

Sorry don't know

Explanation:

Ask google

8 0

3 years ago

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

<em>Current flowing in the wire, I = 4.00 mA</em>
<em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
<em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
<em>Length of wire, L = 2.00 m</em>
<em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>

<em />

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0

3 years ago

A body with the inertial

Andrews [41]

Answer:

Explanation:

Hi there,

To get started, recall the kinematic equations from either a textbook, equation sheet, etc. Kinematic equations are used when acceleration is <em>constant,</em> as stated in the prompt.

Best way to use kinematic equations is to see which variable you are looking for, then which variable is unknown to you and is not needed for that equation.

a) average velocity

Takes the form of:

$v_a_v_g=\frac{d_t_o_t_a_l}{t}=\frac{v+v_0}{2}$ this is the literal definition of average velocity; initial plus final divided by 2.

We know total displacement and total time elapsed, so we will use the middle form of the equation:

$v_a_v_g=\frac{1640m}{40s}=41 \ m/s$

b) the final velocity

We can still use the average velocity formula, as the other two equations that include final velocity have acceleration variable which is unknown as of now.

Solve for final velocity:

$v=(2v_a_v_g)-v_o = 2(41 \ m/s) - (8 m/s) = 74 m/s\\$ this makes sense, since a velocity later in time is higher than a velocity earlier in time. It is increasing with increasing time because of acceleration.

c) the acceleration

There are two equations that can be used to solve this, but we will use the less time-consuming one, but both produce same answer:

$a = \frac{v-v_0}{t_t_o_t_a_l} = \frac{(74-8)m/s}{40s} =1.65 m/s^{2}$

Notice, change in velocity over change in time, and acceleration is constant. When acceleration is constant, it models a linear function, and acc. is just slope!

Study well and persevere. If you liked this solution, hit Thanks or give a rating!

thanks,

3 0

3 years ago

On a calm day with no wind, you can run a 1500-m race at a velocity of 4.0 m/s. If you run the same race on a day when you have

lesantik [10]

Answer:

The time taken to finish the race is 750 s.

Explanation:

The velocity of the person on the day of wind is slowed down by 2.0 m/s. So the person's velocity on the day of wind is 4-2=2 m/s.

The relation between time, speed and distance is t=v/d

Given d=1500 m and calculated v= 2 m/s.

t=1500/2

t=750 s.

Learn more about distance formula.

brainly.com/question/11954435

#SPJ10

3 0

2 years ago

Read 2 more answers