10¹¹ is the approximate number of atoms in a bacterium.
What do you understand by mass of element?
The atomic mass of an element is the average mass of the atoms of an element measured in atomic mass unit (amu).
Given,
Mass of a bacterium atom = 10⁻¹⁵ kg.
Mass of a hydrogen atom = 10⁻²⁷ kg.
From the above observation ,
The average mass of an atom of the bacterium is ten times the mass of a hydrogen atom.
Atomic mass 1 bacterium atom = 10 x mass of hydrogen atom
= 10 x 10⁻²⁷ kg.
= 10⁻²⁶ kg.
Thus,
The number of atoms in a bacterium
= [tex]\frac{Total mass}{Atomicmass of 1 bacterium}[/tex]
= [tex]\frac{10^{-15} }{10^{-26} }[/tex]
= [tex]10^{11}[/tex]
The approximate number of atoms in a bacterium is [tex]10^{11}[/tex].
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After watching the video below and based on your personal experiences, is there a difference
between areas in precision of control? Could there be differences between left and right cortex
based on experience such as handedness or specific skills such as playing a guitar?
Based on my personal experiences, I believe that there is a difference between areas in precision of control because there are different part of the brain that controls the functions of the body.
What is the difference between left-handed and right-handed people?From the standpoint of brain lateralization, differences do exist such as based on experience such as handedness or specific skills such as playing a guitar.
Note that Left-handers are said to have reduced or little lateralized brains, which tells us that the two halves of the brain are little different than as seen in the right-handers.
Therefore, I can say that based on my personal experiences, I believe that there is a difference between areas in precision of control because there are different part of the brain that controls the functions of the body.
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When an object is placed 30.0 cm in front of a concave mirror, a real image is formed 60.0 cm
from the mirror's surface. Find the focal length.
Answer=20.0cm
I need the steps I have a final exam and I’m confused can someone help and type it please ?
Answer:
1 / f = 1 / o + 1 / i = (i + o) / o * 1
f = o * i / (o + i) = 60 * 30 / (60 + 30) = 1800 / 90 = 20 cm
Both the object and image are in positive space for a mirror
Answer pleaseeeee
Question 2 of 10
Which is directly proportional to your weight on a planet's surface?
OA. The distance from the surface to the center of the planet
B. Only the mass of the planet
OC. Only your mass
OD. Your mass and the mass of the planet
Answer:
from my side the will be b only the mass of the planet thank you
Answer:
D
Explanation:
W=mg
g=GM/r²(G=Gravitational constant, M=mass of earth, r=Radius of earth)
>>W=m(GM/r²)
W is directly proportional to mass of earth and mass of your body
There are space stations Alpha and Beta located on the line between the planets. Both space stations are at rest with respect to the planets. Alpha is at distance d 4 from planet 1 and Beta is at distance d 3 from planet 2. A projectile of mass m is fired from station Alpha, with its velocity v pointing directly at planet 2. What is the minimum speed v which will permit the projectile to reach station Beta?
The minimum speed v which will permit the projectile to reach station Beta is √ [Gm/3d]
What is gravitational potential energy?If an object is lifted, work is done against gravitational force. The object gains energy.
Given are space stations Alpha and Beta located on the line between the planets. Both space stations are at rest with respect to the planets. Alpha is at distance d 4 from planet 1 and Beta is at distance d 3 from planet 2. A projectile of mass m is fired from station Alpha, with its velocity v pointing directly at planet 2.
The range of the projectile is given by R = v²sin2θ / g
g = gravitational acceleration of Earth
If g = g(p) for planet , range R = v²sin2θ / g(p)..................(1)
The gravitational force of attraction = weight force
Gm² /d² = m g(p)
g(p) = Gm/d².........................(2)
For R = d/3, from equation (1), we have
d/3 = v²sin2θ / g(p)
Plug the expression for g(p) , we get
v = √ [Gm/3dsin2θ ]
For velocity to be minimum, sin2θ =1
So, the minimum velocity will be
v = √ [Gm/3d]
Thus, the minimum speed v which will permit the projectile to reach station Beta is √ [Gm/3d]
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how many times bigger is the radius of a helium atom then the radius of an alpha particle
The radius of a helium atom 4.8 times bigger then the radius of an alpha particle.
What is alpha particle?The structure of alpha particles, also known as alpha rays and alpha radiation, is similar to that of the helium-4 nucleus and has been made up of two protons as well as two neutrons bonded together.
What is helium?For welding metals like aluminum, helium was utilized as an inert gas environment. It is also employed in rocket propulsion.
The radius of a helium atom 4.8 times bigger then the radius of an alpha particle. More precisely than it's ever been, the diameter of the helium atom's nucleus, the alpha particle, had also been measured. Outcomes just point to a size of 1.6782 femtometers, 4.8 times more accurate than earlier readings.
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Injuries occur from muscle imbalances which result from the body's inability to ....................... these types of stress.
Answer:
Tolerate
Explanation:
10 precautions for obtaining the refractive index of a triangular glass prism.
Answer:
Following precautions for obtaining the refractive index of a triangular glass prism:-
Explanation:
(1) All the faces of the prism should be neat and clean.
(2) Pins for holding the paper on the drawing board must be pinned perpendicular to the paper for better handling.
(3) While fixing the pins for checking the refractive index of the prism, make sure that the reflective images of the pins should be aligned to your eye to avoid any type of parallax error.
(4) Pins should not be removed or pinned again during the experiment.
(5) Avoid mishandling of the prism.
(6) Same edge of the prism should be taken as vertex for observations.
(7) For drawing the boundary of the prism, a sharp pencil should be used.
(8) Soft board and pointed pins should be used.
(9) The distance between the pins should be 5 cm or more.
(10) Make sure the glass slab or prism taken are polished and not broken.
Hope this helps!
Please mark me as brainlist.
If the Moon had twice as much mass and still orbits Earth at the same distance, ocean bulges on Earth would be
smaller.
larger.
unequal in size.
not significantly different.
none of the available options
Ocean bulges on Earth would be bigger if the Moon had twice as much mass and yet orbited the planet at the same distance. Option B is correct.
What is ocean bludge?The fluid and moveable ocean water are drawn towards the moon by the gravitational attraction between the moon and the Earth.
The ocean nearest to the moon experiences a bulge as a result, and as the Earth rotates, the affected seas' locations shift.
The Moon's bulges in the oceans would be larger if it had twice the mass and orbited Earth at the same distance.
Hence option B is corect.
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A mass of (200 g) of hot water at (75.0°C) is mixed with cold water of mass M at (5.0°C). The final temperature of the mixture is (25.0°C). What is the mass of the cold water (M)?
Answer:
500 g
Explanation:
Specific heat of water = 1 j/g-c
Heat given up by hot water = 200(75-25)(1) = 10 000 j
this is the heat GAINED by the cold water
10000 j = x ( 25 -5)(1)
x =500g
If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits___, ____, ____, and ____, and all planets will orbit the sun successfully.
Pleas help this is Flvs comprehensive science class
6.01 please please help
So, the complete sentence is If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully for the best conditions.
When the mass of the sun is larger, Earth moves around the sun at a faster pace and When the mass of the sun is smaller, Earth moves around the sun at a slower pace.
When Earth is closer to the sun, its orbit becomes faster and When Earth is farther from the sun, its orbit becomes slower.
When Earth is closer to the sun, there will be a hotter climate. A little movement that takes one closer to the sun could lead to a huge impact, as the sun is very hot.
So, it can be concluded that If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully for the best conditions.
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A house has a roof (colored gray) with the dimensions shown in the drawing. Determine the magnitude of the net force that the
atmosphere applies to the roof when the outside pressure drops suddenly by 13.2 mm of mercury, before the pressure in the attic can
adjust.
The magnitude of the net force that the atmosphere applies to the roof when the outside pressure drops suddenly is 1.9 x 10⁵ N.
What is pressure?The pressure is the amount of force applied per unit area.
Pressure p = Force/Area
Given is a house has a roof (colored gray) with the dimensions shown in the drawing. The outside pressure drops suddenly by 13.2 mm of mercury, before the pressure in the attic can adjust.
The pressure difference ΔP = 13.2 mm of Hg
The length of the roof l = 14.5m
the breadth of the roof h = 4.21m
The force exerted by pressure is
Force, F = P x A
= (13.2 mm of Hg) [(133 N/m²) /1 mm of Hg ](14.5 x 4.21)
= 107,170.6 N
Then the net vertical force
Fnet = 2F cos30
Fnet = 2 ( 107,170.6) cos30
Fnet = 185625 N
Fnet =1.9 x 10⁵ N
The direction of the force is downwards, since the horizontal components of the forces cancel each other.
Hence, magnitude of the net force is 1.9 x 10⁵ N
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If a current uses 31 amperes and has a voltage of 110 volts, how many watts does it dissipate?
Considering the Watt's law, if a current uses 31 amperes and has a voltage of 110 volts, it dissipates 3,410 Watts.
CurrentCurrent is the physical magnitude that expresses the amount of electricity that flows through a conductor in unit time and is measured in amps.
VoltageThe voltage is the difference between the electrical charge that leaves the source and the one that finally reaches the end of the circuit. It is expressed in volts.
Watt's LawWatt's Law refers to the electrical power of an electronic component or device and is defined as the power consumed by the load is directly proportional to the voltage supplied and the current flowing through it. The unit of power is the Watt.
Knowing the voltage and current, this law is expressed as:
P = V×I
Power in this caseIn this case, you know:
V= 110 voltsI= 31 amperesReplacing in Watt's Law:
P = 110 volts× 31 amperes
Solving:
P= 3,410 Watt
Finally, if a current uses 31 amperes and has a voltage of 110 volts, it dissipates 3,410 Watts.
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an ac voltage of 12.0 v is applied to the primary transformer that has 3 times as many turns in the secondary than in the primary. what is the voltage of the secondary?
The voltage of the secondary will be 36 V.From the given conditions primary transformer has 3 times as many turns in the secondary coil.
What is induced voltage?Electromagnetic induction is what causes the induced voltage. Electromagnetic induction is the process of generating emf (induced voltage) by subjecting a conductor to a magnetic field.
In this case, a magnet is pushed in and out of a wire coil attached to a high-resistance voltmeter.
Typically, a transformer's primary winding is attached to the input voltage source and changes electrical power into a magnetic field.
The secondary winding's role is to turn this alternating magnetic field into electricity, generating the necessary output voltage.
Given data;
Primary coil voltage,[tex]\rm V_P = 12 \ v[/tex]
Secondary coil voltage,[tex]\rm V_s = ?[/tex]
No turns in the primary coil,[tex]\rm N_p[/tex]
No turns in the secondary coil,[tex]\rm N_s[/tex]
From the given condition;
[tex]\rm N_s = 3N_p[/tex]
For a transformer,
[tex]\rm \frac{V_p}{V_s}= \frac{N_p}{N_s} \\\\ \rm \frac{12}{V_s}= \frac{N_p.}{3N_p} \\\\ V_s = 36 \ V[/tex]
Hence the voltage of the secondary will be 36 V.
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How to do projects on newton’s law
An electric field is described by the vector e=yj -xi. what is the electric flux through a rectangular box in the x-z plane that is bounded by x between 0 and 1, and z between 0 and 1? (You need to identify dA to do this problem.)
The flux of [tex]\vec E = -x\,\vec\imath + y\,\vec\jmath[/tex] is given by the surface integral
[tex]\displaystyle \iint_S \vec E \cdot d\vec\sigma[/tex]
where [tex]S[/tex] is the given square region, which we can parameterize by
[tex]\vec s(x, z) = x\,\vec\imath + z\,\vec k[/tex]
with [tex]0\le x\le 1[/tex] and [tex]0\le z\le 1[/tex]. The area element is
[tex]d\vec\sigma = \vec n \, dx\,dz[/tex]
where [tex]\vec n[/tex] is the normal vector to [tex]S[/tex]. Depending on the orientation of [tex]S[/tex], this vector could be
[tex]\vec n = \dfrac{\partial\vec s}{\partial x} \times \dfrac{\partial\vec s}{\partial z} = -\vec\jmath[/tex]
or [tex]-\vec n = \vec \jmath[/tex]; either way, the integral reduces to
[tex]\displaystyle \iint_S \vec E \cdot d\,\vec\sigma = \int_0^1 \int_0^1 (-x\,\vec\imath + z\,\vec k) \cdot (\pm\vec\jmath) \, dx\,dz = \boxed{0}[/tex]
A room has a 60 Watt, a 100 Watt and a 150 Watt light bulb. How much does it
cost to use all the lamps for 2.5 hours at $0.08/ kWh?
The cost to use all lamps is $62.
room has 60 watts, 100 watts, and 150 watt light bulbs.Duration of using bulbs=2.5 hours power =$0.08kWh
power=energy/time
(60+ 100+150)=energy/2.5
energy=775 joule
cost for 1 hour=$0.08
= 775 * 0.08
= $62
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Answer:
6.2 cents
Explanation:
(60+100+150/1000)*2.5*8=6.2 cents
Joe runs 10 m north, 20 m south, 9m south, and then 15 m north. What is Joe's
displacement?
Answer:
Joes displacement is 54m
What is an advantage of using renewable energy resources?
A. They can be converted directly into electric energy.
B. They do not waste energy in heat.
C. They can be replenished in a human lifetime.
D. They do not produce greenhouse gases.
143°C = _____
416 K
-130 K
0 K
143 K
For which length of wire are the reading of resistance most precise
Answer:
Explanation:
There are different ways to investigate the factors that affect resistance. In this practical activity, it is important to:record the length of the wire accuratelymeasure and observe the potential difference and currentuse appropriate apparatus and methods to measure current and potential difference to work out the resistance
5, the following equation shows the position of a particle in time t, x=at2i + btj where t is in second and x is in meter. A=2m/s2, b=1m/s. Find
A, the average velocity of the particle in the time interval t1=2sec and t2=3sec
B, the velocity and acceleration at any time t.
C, the average acceleration in the time interval given in part (a)
Explanation:
hi
please help me. ineed answers
Gerund/Infinitive
1. Put the verbs in brackets into the correct infinitive form or the -ing form.
1. Jane went on____(sleep) for another two hours.
2. He told us his name and went on____ (introduce) us to his wife.
3. She tried (finish) her homework, but it was too difficult.
4. You should try (eat) more fruit. It's good for your health.
5. He regrets (argue) with his best friend.
6.We regret (inform) you that tonight's performance will be cancelled.
7. Oh, no! I forgot____(lock) the front door.
8. I'll never forget_(meet) my favourite film star.
9.Claire likes (ski). She says it's very exciting.
10. I like__(go) to the dentist every six months.
11.I must remember (post) these letters today.
12. I remember__(read) the book, but I don't know who wrote it.
13. I'm sorry for___(forget) your birthday. It was awful of me.
14. I'm sorry.(say) that you have failed the exam.
15. She is afraid___(climb) the tree in case she falls.
16.Mary never wears her diamond ring. She is afraid of (lose) it.
17. I have stopped . (buy) some food before continuing our journey.
18. We stopped (watch) horror films because they give me nightmares
19. Tom stopped__(pick up) his washing on the way home.
20 If you don't stop. (smoke), you'll make yourself ill.
21.Will you quit___(complain)! It doesn't help.___(solve) the problem.
22.If you ever decide___(sell) your car, let me____(know).
23.I would like you.__(water) the plants for me at the weekend.
24.I clearly remember__(set) my alarm clock before__(go) to bed last night.
25. These plants require(water) every day.
26. I resent you (speak) to me like that! Have some respect.
27. It would be good for the children__ (play) outdoors more often.
28. I promised __(take) Jill to the party, but I don't feel like__(go) now.
29. Don't waste your time (look for) the document.Ask Mr Gale.
30. Please, excuse his__(leave) so early. He wants __(catch up) with his (study).
31. Try__(phone) John at the office if he's not at home.
32. I tried my best__(finish), but there just wasn't enough time.
33. He was promoted in 1990 and went on__(become) a company director.
34. The band went on__(play) even after the lights had gone out.
35. "Why is the baby crying?" "I think he wants (feed)."
36. Sharon wants__(talk) to you.
37. Jane was afraid__(show) her school report to her parents.
38. I'm afraid of__(lose) my way in the forest.
39. I regret__ (inform) you that your husband has been arrested.
A block of mass 4 kg is attached to a spring of force constant (k) 1000 N/m, and undergoes simple harmonic motion. What is the period of the motion?
a. 0.1 s
b. 0.4 s
с. 0.2 s
d. 0.3 s
Answer:
0.4
Explanation:
ω = km
ω = 1000 x 4
ω = 4000 divide this by 10000 and you get 0.4s
4. A ball is thrown with an initial speed vi at an angle θi with the horizontal. The
horizontal range of the ball is R, and the ball reaches a maximum height R
6
. In
terms of R and g, find (a) the time interval during which the ball is in motion,
(b) the ball’s speed at the peak of its path, (c) the initial vertical component of
its velocity, (d) its initial speed, and (e) the angle θi
Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).
At time [tex]t[/tex], the horizontal position [tex]x[/tex] and vertical position [tex]y[/tex] of the ball are given respectively by
[tex]x = v_i \cos(\theta_i) t[/tex]
[tex]y = v_i \sin(\theta_i) t - \dfrac g2 t^2[/tex]
and the horizontal velocity [tex]v_x[/tex] and vertical velocity [tex]v_y[/tex] are
[tex]v_x = v_i \cos(\theta_i)[/tex]
[tex]v_y = v_i \sin(\theta_i) - gt[/tex]
The ball reaches its maximum height with [tex]v_y=0[/tex]. At this point, the ball has zero vertical velocity. This happens when
[tex]v_i \sin(\theta_i) - gt = 0 \implies t = \dfrac{v_i \sin(\theta_i)}g[/tex]
which means
[tex]y = \dfrac R6 = v_i \sin(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g - \dfrac g2 \left(\dfrac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \dfrac{6{v_i}^2 \sin^2(\theta_i)}g - \dfrac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \dfrac{3{v_i}^2 \sin^2(\theta_i)}g[/tex]
At the same time, the ball will have traveled half its horizontal range, so
[tex]x = \dfrac R2 = v_i \cos(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g \\\\ \implies R = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g[/tex]
Solve for [tex]v_i[/tex] and [tex]\theta_i[/tex] :
[tex]\dfrac{3{v_i}^2 \sin^2(\theta_i)}g = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0[/tex]
Since [tex]0^\circ<\theta_i<90^\circ[/tex], we cannot have [tex]\sin(\theta_i)=0[/tex], so we're left with (e)
[tex]3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \dfrac23 \\\\ \implies \boxed{\theta_i = \tan^{-1}\left(\dfrac23\right) \approx 33.7^\circ}[/tex]
Now,
[tex]\cos\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac3{\sqrt{13}}[/tex]
[tex]\sin\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac2{\sqrt{13}}[/tex]
so it follows that (d)
[tex]R = \dfrac{2{v_i}^2 \times\frac3{\sqrt{13}} \times \frac2{\sqrt{13}}}g \\\\ \implies {v_i}^2 = \dfrac{13Rg}{12} \\\\ \implies \boxed{v_i = \sqrt{\dfrac{13Rg}{12}}}[/tex]
Knowing the initial speed and angle, the initial vertical component of velocity is (c)
[tex]v_y = \sqrt{\dfrac{13Rg}{12}} \sin\left(\tan^{-1}\left(\dfrac23\right)\right) \\\\ \implies v_y = \sqrt{\dfrac{13Rg}{12}} \times \dfrac2{\sqrt{13}} \\\\ \implies \boxed{v_y = \sqrt{\dfrac{Rg}3}}[/tex]
We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)
[tex]v_x = \sqrt{\dfrac{13Rg}{12}} \times \dfrac3{\sqrt{13}} \\\\ \implies v_x = \dfrac{\sqrt{3Rg}}{2}[/tex]
Then with [tex]v_y=0[/tex], the ball's speed [tex]v[/tex] is
[tex]v = \sqrt{{v_x}^2 + {v_y}^2} \\\\ \implies v = v_x \\\\ \implies \boxed{v = \dfrac{\sqrt{3Rg}}2}[/tex]
Finally, in the work leading up to part (e), we showed the time to maximum height is
[tex]t = \dfrac{v_i \sin(\theta_i)}g[/tex]
but this is just half the total time the ball spends in the air. The total airtime is then
[tex]2t = \dfrac{2 \times \sqrt{\frac{13Rg}{12}} \times \frac2{\sqrt{13}}}g \\\\ \implies 2t = 2\sqrt{\dfrac R{3g}}[/tex]
and the ball is in the air over the interval (a)
[tex]\boxed{0 < t < 2\sqrt{\frac R{3g}}}[/tex]
Which scientific method could I use to test this hypothesis? "If the mass and the volume of an object are known, then its density can be calculated dividing the object's mass by its volume."
Answer:
Multiple so you can test multiple hypothesis at once
Explanation:
because
Answer:
To test the hypothesis, we need to make an observation or perform an experiment associated with the prediction. For instance, in this case, we would plug the toaster into a different outlet and see if it toasts.
inspired by answer up there
sorry i couldn't add details because think its bad
Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories.
A plane leaves with an acceleration of 6.34 m/s squared and takes 1.5 hours to stop. What is the speed of the plane? What was the distance it traveled?
The answer to this question is Initial velocity of plane will be 34236 m/s and 92437.2 Km is the distance travelled by it.
Three equation of motion are:-
v = u + ats = ut + (1/2)at²v² - u² = 2asWhere v is final velocity, u in initial velocity, s is the displacement by the object, a is the acceleration and t denotes the time.
In question we have given deceleration as 6.34 m/s² and time as 1.5 hour which is equal to 5400 seconds.
Applying equation 1 to find the initial speed of plane
v = u + at
0 = u + (-6.34 × 5400) {v=0 as plane will stop after 5400 sec}
u = 6.34 × 5400
u = 34236 m/s
Initial velocity of plane is 34236 m/s
Applying equation 2 to find the displacement of plane in that time period
s = ut + (1/2)at²
s = ( 34236 × 5400 ) - ( (1/2) × 6.34 × 5400² )
s = 5400 × ( 34236 - ((1/2) × 6.34 × 5400) )
s = 5400 × ( 34236 - 17118 )
s = 5400 × 17118 metres
s = 5.4 × 17118 Km
s = 92437.2 Km
Distance travelled by plane is 92437.2 Km
So, the initial velocity of plane will be 34236 m/s and the displacement of plane in that time period will be 92437.2 Km.
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what will happen to the pressure in a gas if you compressed it into a volume that was one third the size? why?
Answer:
Pressure will triple (Boyle's Law)
Explanation:
Assuming a constant temperature: (compressing gases usually raises the temp significantly):
P1V1 = P2V2
P1V1 / V2 = P2
Now change V2 to 1/3 V2:
P1V1 / (1/3 V2 ) = P2 / (1/3 )
P1V1/ (1/3 v2 ) = 3 P2 <======= THE PRESSURE WILL TRIPLE
This is Boyle's Law
A diver shines a light up to the surface of a flat glass-bottomed boat at an angle of 37◦ relative to the normal. If the indices of refraction of air, water, and glass are 1.0, 1.33, and 1.4 respectively, at what angle does the light leave the glass (relative to its normal)? Answer in units of ◦ .
Answer:
Approximately [tex]53^{\circ}[/tex], assuming that the upper and lower surfaces of the glass on this boat are parallel.
Explanation:
Assume that the upper and lower surfaces of the glass at the bottom of this ship are parallel. Refer to the diagram attached. The two normals would also be parallel to each other.
The following angles would be alternate interior angles between the two normals:
The angle at which the light enters the glass, andThe angle at which the light leaves the glass.Since the two normals are parallel to each other, these two angles would have the same value. Let [tex]\theta_{\text{glass}}[/tex] denote the value of both of these angles.
Let [tex]\theta_{\text{src}}[/tex] denotes the angle at which a beam of light leaves the original medium (angle of incidence.) Let [tex]\theta_{\text{dst}}[/tex] denote the angle at which this beam of light enters the new medium.
Let [tex]n_\text{src}[/tex] and [tex]n_\text{dst}[/tex] denote the refractive indices of the original and the new medium, respectively. By Snell's Law:
[tex]\begin{aligned}\frac{\sin(\theta_{\text{dst}})}{\sin(\theta_{\text{src}})} = \frac{n_{\text{src}}}{n_{\text{dst}}}\end{aligned}[/tex].
Let [tex]\theta_{\text{water}}[/tex] denote the angle at which the beam of light in this question leaves the water. Let [tex]\theta_{\text{air}}[/tex] denote the angle at which this beam of light enters the air. It is given that [tex]\theta_{\text{water}} = 37^{\circ}[/tex], while [tex]\theta_{\text{air}}[/tex] is the value that needs to be found.
Let [tex]n_{\text{air}}[/tex], [tex]n_{\text{water}}[/tex], and [tex]n_{\text{glass}}[/tex] denote the refractive index of air, water, and glass, respectively. By Snell's Law:
[tex]\begin{aligned}\frac{\sin(\theta_{\text{glass}})}{\sin(\theta_{\text{water}})} = \frac{n_{\text{water}}}{n_{\text{glass}}}\end{aligned}[/tex].
[tex]\begin{aligned}\frac{\sin(\theta_{\text{air}})}{\sin(\theta_{\text{glass}})} = \frac{n_{\text{glass}}}{n_{\text{air}}}\end{aligned}[/tex].
Thus:
[tex]\begin{aligned} & \frac{\sin(\theta_{\text{air}})}{\sin(\theta_{\text{water}})} \\ =\; & \frac{\sin(\theta_{\text{glass}})}{\sin(\theta_{\text{water}})}\times \frac{\sin(\theta_{\text{air}})}{\sin(\theta_{\text{glass}})} \\ =\; & \frac{n_{\text{water}}}{n_{\text{glass}}}\times \frac{n_{\text{glass}}}{n_{\text{air}}} \\ =\; & \frac{n_{\text{water}}}{n_{\text{air}}}\end{aligned}[/tex].
Since [tex]\theta_{\text{water}} = 37^{\circ}[/tex]:
[tex]\begin{aligned} & \sin(\theta_{\text{air}})\\ =\; & \sin(\theta_{\text{water}}) \times \frac{\sin(\theta_{\text{air}})}{\sin(\theta_{\text{water}})} \\ =\; & \sin(\theta_{\text{water}})\times \frac{n_{\text{water}}}{n_{\text{air}}} \\ =\; & \sin(37^{\circ}) \times \frac{1.33}{1.0} \\ \approx \; & 0.800 \end{aligned}[/tex].
Therefore:
[tex]\begin{aligned}\theta_{\text{air}} &= \arcsin(\sin(\theta_{\text{air}})) \\ & \approx \arcsin(0.800) \\ &\approx 53^{\circ} \end{aligned}[/tex].
In other words, this beam of light would leave the glass at approximately [tex]53^{\cic}[/tex] from the normal.
How to put a new command in the command prompt
Typing in the commands and instructions and pressing enter will put a new command in the command prompt.
What is a Command?This refers to a words or phrase which causes the computer to execute certain tasks or functions.
Typing the exact instruction and entering it will ensure that the chosen tasks are found in the command prompt.
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The high point of the ripple seen in a pond after throwing in a pebble is the...
Select one:
a. crest.
b. trough.
c. wavelength.
d. A & B
The crest is the highest point of a pebble-induced ripple in a body of water. Option a is correct.
What is wavelength?The distance between two successive troughs or crests is known as the wavelength. The highest point of the wave is the crest, while the trough is the lowest.
The high point of the ripple seen in a pond after throwing in a pebble is the crest.
Hence option a is correct.
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the road from city A to city B is described by a car with Vm 40 km / h. When the car turns (from B to A) the average speed is 60 km / h. Find the average round trip speed.
Answer:
Let distance between two cities is x km . So total distance travelled bus x+x=2x km and time taken was=(x/40)+(x/60)=(3x+2x)/120=5x/120=x/24 hrs . So avg speed is 2x/(x/24)=48 km per hr.
or
dt1=60km1h
dt2=30km1h
2dt1+t2=r¯
d=60t1
d=30t2
60t1=30t2
2t1=t2
2dt1+(2t1)=r¯
2(60t1)t1+2t1=r¯
2(60t1)=r¯(t1+2t1)
120=r¯+2r¯
r¯=1203
r¯=40km/h
