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2 years ago

Index of refraction

Physics

1 answer:

rodikova [14]2 years ago

5 0

Answer:

The relation is n = c/v where c is speed of light in a vacuum....v is the speed of light in the new medium ......n is the refractive index of the material

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In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th

qaws [65]

Answer:

The distance between both cars is 990 m

Explanation:

The equations for the position and the velocity of an object moving in a straight line are as follows:

x = x0 + v0 * t + 1/2 * a * t²

v = v0 + a * t

where:

x = position of the car at time "t"

x0 = initial position

v0 = initial speed

t = time

a = acceleration

v = velocity

First let´s find how much time it takes the driver to come to stop (v = 0). We will consider the origin of the reference system as the point at which the driver realizes she must stop. Then x0 = 0

With the equation of velocity, we can obtain the acceleration and replace it in the equation of position, knowing that the position will be 250 m at that time.

v = v0 + a*t

v-v0 / t = a

0 m/s - 71.0 m/s / t =a

-71.0 m/s / t = a

Replacing in the equation for position:

x = v0* t +1/2 * a * t²

250 m = 71.0 m/s * t + 1/2 *(-71.0 m/s / t) * t²

250 m = 71.0 m/s * t - 1/2 * 71.0 m/s * t

250m = 1/2 * 71.0m/s *t

t = 2 * 250 m / 71.0 m/s = 7.04 s

It takes the driver 7.04 s to stop.

Then, we can calculate how much time it took the driver to reach her previous speed. The procedure is the same as before:

v = v0 + a*t

v-v0 / t = a now v0 = 0 and v = 71.0 m/s

(71.0 m/s - 0 m/s) / t = a

71.0 m/s / t =a

Replacing in the position equation:

x = v0* t +1/2 * a * t²

390 m = 0 m/s * t + 1/2 * 71.0 m/s / t * t² (In this case, the initial position is in the pit, then x0 = 0 because it took 390 m from the pit to reach the initial speed).

390m * 2 / 71.0 m/s = t

t = 11.0 s

In total, it took the driver 11.0s + 5.00 s + 7.04 s = 23.0 s to stop and to reach the initial speed again.

In that time, the Mercedes traveled the following distance:

x = v * t = 71.0 m/s * 23.0 s = 1.63 x 10³ m

The Thunderbird traveled in that time 390 m + 250 m = 640 m.

The distance between the two will be then:

distance between both cars = 1.63 x 10³ m - 640 m = 990 m.

3 0

2 years ago

A 1.0 kg object is attached to a string 0.50 m. It is twirled in a horizontal circle above the ground at a speed of 5.0 m/s. A b

aivan3 [116]

50 N The centripetal force upon an object is expressed as F = mv^2/r So let's substitute the known values and calculate F = mv^2/r F = 1.0 kg * (5.0 m/s)^2 / 0.5 m F = 1.0 kg * 25 m^2/s^2 / 0.5 m F = 25 kg*m^2/s^2 / 0.5 m F = 50 kg*m/s^2 F = 50 N So the answer is 50 N which matches one of the available choices.

6 0

2 years ago

Calculate the energy in the form of heat (in kJ) required to change 75.0 g of liquid water at 27.0 °C to ice at –20.0 °C. Assume

REY [17]

Explanation:

The given data is as follows.

mass, m = 75 g

$T_{1} = 0^{o}C$

$T_{2} = 27^{o}C$

Specific heat of water = 4.18

First, we will calculate the heat required for water is as follows.

q = $m C \times (T_{1} - T_{2})$

= $75 g \times 4.18 J/g^{o}C \times (0 - 27)^{o}C$

= 8464.5 J/mol

= 8.46 kJ ......... (1)

Also, it is given that $T_{3} = -20^{o}C$ = (20 + 273) K = 293 K and specific heat of ice is 2.108 kJ/kg K.

Now, we will calculate the heat of fusion as follows.

q = $mC \times (T_{3} - T_{1})$

= $0.075 kg \times 2.108 kJ/kg K \times (-293 - 0) K$

= -46.32 kJ ......... (2)

Now, adding both equations (1) and (2) as follows.

8.46 kJ - 46.32 kJ

= -37.86 kJ

Therefore, we can conclude that energy in the form of heat (in kJ) required to change 75.0 g of liquid water at $27.0^{o}C$ to ice at $-20.0^{o}C$ is -37.86 kJ.