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3 years ago

7

You have an incident ray from a medium with a n1=1, through a medium with a n2 =3.325. If the incident angle is equal to 0.478 r

ad, what is the refraction angle? (Answer is in radians)

1 answer:

sesenic [268]3 years ago

6 0

Answer:

0.139 rad

Explanation:

We use Snell's law $n_1sin\theta_1=n_2sin\theta_2$ , where if $n_1$ is the <em>refractive index</em> of the medium containing the <em>incident ray</em>, $\theta_1$ would be the <em>incident angle</em>, and if $n_2$ is the <em>refractive index</em> of the medium containing the <em>refracted ray</em>, $\theta_2$ would be the <em>refraction angle</em>, which we want, so we do:

$sin\theta_2=\frac{n_1}{n_2}sin\theta_1$

And finally:

$\theta_2=arcsin(\frac{n_1}{n_2}sin\theta_1)$

We then insert our values:

$\theta_2=arcsin(\frac{n_1}{n_2}sin\theta_1)=Arcsin(\frac{1}{3.325}sin(0.478rad))=arcsin(0.13834714686
)=0.139 rad$

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show mathematically why an 80,000 pound ( 36,000 kg ) big rig taveling 2 mph (0.89 m/s) has the SAME MOMENTUM as a 4,000 pound (

bekas [8.4K]

The momentum value for both Big rig and sport utility vehicle are the same as 16000 pound mph

<u>Explanation:</u>

Given data are as follows

For Big rig, Mass = 80,000 pound (36,000 kg)

velocity = 2 mph (0.89 m/s)

where mph is meter per hour

For sport utility vehicle, Mass = 40,000 pound (1800 kg)

velocity = 40 mph (18 m/s)

The formula to find the Momentum of an object is

Momentum = Mass × Velocity (Kilogram meter per second)

Momentum for Big rig = 80,000 × 2 (pound mph)

= 1,60,000 pound mph

Momentum for sport utility vehicle = 4000 × 40 (pound mph)

= 1,60,000 pound mph

Hence it is mathematically proved that

The momentum of big rig = The momentum of sport utility vehicle

1,60,000 pound mph = 1,60,000 pound mph

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4 years ago

Bill was really bored at work. During his break, he decided to check his carotid artery. He placed both fingers on his neck and

Sergeeva-Olga [200]

Answer:

72 beats per minute

Explanation:

Heart beat causes the flow of blood round the body. This heart beat can be felt as pulse in the wrist or neck carotid artery. The heart rate which is measured in beats per minute (BPM) is used to determine the number of heart beats per minute.

You can calculate your BPM using the carotid artery found in the neck close to the windpipe.

Given that for 20 seconds, Bill had a total of 24 beats.

60 seconds = 1 minute.

Hence, Bill's BPM = (24 beats per 20 seconds) * (60 seconds per minute) = 72 beats per minute

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3 years ago

A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters [cm]. The rover picks up the rock and lifts it

Makovka662 [10]

Answer: 5166.347

Explanation:

The specific gravity of a solid $SG$ (also called relative density) is the ratio of the density of that solid $\rho_{rock}$ to the density of water $\rho_{water}=1 kg/m^{3}$ (normally at $4\°C$ ):

$SG=\frac{\rho_{rock}}{\rho_{water}}$ (1)

On the other hand, the density of the rock is calculated by:

$\rho_{rock}=\frac{m_{rock}}{V_{rock}}$ (2)

Where:

$m_{rock}$ is the mass of the rock

$V_{rock}=\frac{4}{3} \pi r^{3}$ is the volume of the rock, since is spherical

Well, we already know the value of $\rho_{water}$ , but we need to find $\rho_{rock}$ in order to find the rock's specific gravity; and in order to do this, we firsly have to find $m_{rock}$ and then calculate $V_{rock}$ :

In the case of the mass of the rock, we can calclate it by the following equation:

$W_{rock}=m_{rock}g_{mars}$ (3)

Where:

$W_{rock}$ is the weight if the rock in mars

$g_{mars}=3.7 m/s^{2}$ is the acceleration due gravity in Mars

Isolating $m_{rock}$ :

$m_{rock}=\frac{W_{rock}}{g_{mars}}$ (4)

$m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}}$ (5)

To find $W_{rock}$ we can use the following equation of the potential gravitational energy $U$ :

$U=W_{rock}H$ (6)

Where:

$U=2 J=2 Nm$ is the potential energy

$H=20 cm \frac{1m}{100 cm}=0.2 m$ is the height at which the rock has the mentioned potential energy

Isolating $W_{rock}$ :

$W_{rock}=\frac{U}{H}$ (7)

$W_{rock}=\frac{2 Nm}{0.2 m}$ (8)

$W_{rock}=10 N$ (9)

Substituting (9) in (5):

$m_{rock}=\frac{10 N}{3.7 m/s^{2}}$ (10)

$m_{rock}=2.702 kg$ (11)

Substituting (11) in (2):

$\rho_{rock}=\frac{2.702 kg}{V_{rock}}$ (12) At this point we only need to find the volume of the rock, knowing its diameter is $d=10 cm$ , hence its radius is $r=\frac{d}{2}=5 cm$

$V_{rock}=\frac{4}{3} \pi (5 cm)^{3}$ (13)

$V_{rock}=523.59 cm^{3} \frac{1 m^{3}}{(100 cm)^{3}}=0.000523 m^{3}$ (14)

Substituting (14) in (12):

$\rho_{rock}=\frac{2.702 kg}{0.000523 m^{3}}$ (15)

$\rho_{rock}=5166.34 kg/m^{3}$ (16)

Substituting (16) in (1):

$SG=\frac{5166.34 kg/m^{3}}{1 kg/m^{3}}$ (17)

Finally we obtain the specific gravity of the rock:

$SG=5166.347$

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3 years ago

Which of the following particles is NOT a candidate for dark matter?

murzikaleks [220]

its C) Axion

hope it helps

7 0

3 years ago

Sprinting near the end of a race, a runner with a mass 63 kg accelerates from a speed of 25 m/s to a speed of 26 m/s in 54 s. To

gavmur [86]

Answer:

1.1655 N

Explanation:

Given that,

Initial speed, u = 25 m/s

Final speed, v = 26 m/s

Time taken, t = 54 s

So, Applying equation of motion as:

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{26-25}{54}\ m/s^2\\\Rightarrow a=0.0185\ m/s^2$

According to the Newton's second law of motion:-

$Force=Mass\times Acceleration$

Mass = 63 kg

So,

$Force=63\times 0.0185\ kgm/s^2$

<u>Force = 1.1655 N</u>

5 0

4 years ago