1 ) The coefficient of static friction μs by using the equation (2) = 0.26
3 ) The coefficient of kinetic friction μk by using the equation (1) = 0.19
1 ) The coefficient of static friction
μ[tex]_{s}[/tex] = [tex]f_{s} _{max}[/tex] / F[tex]_{n}[/tex]
μ[tex]_{s}[/tex] = 125 / 50 * 9.8
μ[tex]_{s}[/tex] = 0.26
3 ) The coefficient of kinetic friction
μ[tex]_{k}[/tex] = [tex]f_{k}[/tex] / F[tex]_{n}[/tex]
μ[tex]_{k}[/tex] = 94 / 50 * 9.8
μ[tex]_{k}[/tex] = 0.19
4 ) Yes, from ( 1 ) and ( 3 ),
μ[tex]_{s}[/tex] > μ[tex]_{k}[/tex]
5 ) μ[tex]_{k}[/tex] = [tex]f_{k}[/tex] / F[tex]_{n}[/tex]
μ[tex]_{k}[/tex] = 188 / 100 * 9.8
μ[tex]_{k}[/tex] = 0.19
6 ) No the μ[tex]_{k}[/tex] from ( 3 ) and ( 5 ) are exactly the same.
7 ) Since μ[tex]_{k}[/tex] from ( 3 ) and ( 5 ) are exactly the same, coefficient of friction is constant.
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A light ray travels through water and reflects off a glass surface back into the water. How do the phases of the incident light and the reflected light compare?.
The light travels through water and gets reflected off on the glass surface into the water. There had been a 180° phase change between the incident and the reflected wave. This is called Total internal reflection (TIR).
In total internal reflection, in physics, a ray of light in a medium such as water or glass is completely reflected back into the medium from the surrounding surfaces. This phenomenon occurs when the angle of incidence is greater than a certain critical angle called the critical angle.
TIR only occurs when both of the following two conditions are met
Light is in a denser medium and is approaching a less dense medium. The angle of incidence should be greater than the so-called critical angle.Thus, the phases which include the TIR are the incident and the reflected phase and the incident light hits the surface while the reflected light reflects back.
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A 50 kg child is on a carnival ride with a radius of 5 meters. If the ride spins at a velocity of 5
m/s, what is the centripetal force on the child?
Explanation:
Find a force is F=mv^2/r, so F=50*5^2/5 = 250N
A 4.3-kg bat is traveling 21 m/s just before it strikes a 2.5-kg baseball that’s moving 37 m/s. After the collision, the bat travels in the same direction at 19 m/s. What is the baseball’s speed just after impact?
The speed of the baseball is 30.38m/s .
We are given that,
The mass of bat = m₁ = 4.3 kg
The mass of baseball = m₂ = 2.5 kg
The initial velocity of bat = v₁ = 37 m/s
The final velocity of bat = v₁' = 19 m/s
So that the speed in baseball can be calculated from the law of conservation of momentum which can be given as,
mₙvₙ = m₁v₁ +m₂v₂
(m₁ + m₂)vₙ = m₁v₁ + m₂v₂
Where, vₙ is the speed of the baseball after impact, mₙ is the total mass of the system ,
(4.3kg + 2.5kg)vₙ = (4.3kg × 37m/s) + (2.5kg × 19m/s)
(6.8kg) vₙ = 159.1kg-m/s + 47.5kg-m/s
vₙ = (206.6kg-m/s)/6.8kg
vₙ = 30.38 m/s
Therefore, the speed of the baseball would be 30.38m/s after impact.
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a child is sledding down a hill. the child has 200 j of potential energy and 500 j of kinetic energy at one point on the hill. how much more kinetic energy does the child gain for the remainder of the motion to the bottom of the hill?
The additional kinetic energy that the child gains after reaching the bottom of the hill is 200 Joule. The result is obtained by using the principle of Conservation of Mechanical Energy.
What is Conservation of Mechanical Energy?The principle of conservation of energy states that "The total energy is neither increased nor decreased in any process. Energy can be transformed from one form to another, and transferred from one object to another, but the total amount remains constant."
In the same way, the Conservation of Mechanical Energy states that the total mechanical energy of a system is conserved. It can be expressed as
[tex]E_{1} = E_{2}[/tex]
[tex]K_{1} + U_{1} = K_{2} + U_{2}[/tex]
[tex]\frac{1}{2} mv_{1} ^{2} + mgh_{1} = \frac{1}{2} mv_{2} ^{2} + mgh_{2}[/tex]
Where
E₁ = mechanical energy at point 1E₂ = mechanical energy at point 2K₁ = kinetic energy at point 1K₂ = kinetic energy at point 2U₁ = potential energy at point 1U₂ = potential energy at point 2m = mass of an objectv₁ = velocity of object at point 1v₂ = velocity of object at point 2h₁ = height of object at point 1h₂ = height of object at point 2At one point, a child has:
U₁ = 200 JK₁ = 500 JHow much more kinetic energy does the child gain at the bottom of the hill?
At the bottom of the hill, the height of the child is zero, h₂ = 0. So, the potential energy is
[tex]U_{2} = mgh_{2}[/tex]
[tex]U_{2} = mg(0)[/tex]
[tex]U_{2} = 0[/tex]
The kinetic energy at the bottom of the hill is
[tex]K_{1} + U_{1} = K_{2} + U_{2}[/tex]
[tex]500 + 200 = K_{2} + 0[/tex]
[tex]K_{2} = 700 J[/tex]
The kinetic energy added for the remainder of the motion is
[tex]K = K_{2} - K_{1}[/tex]
[tex]K = 700 - 500[/tex]
[tex]K = 200 J[/tex]
Hence, the additional kinetic energy that the child gains after reaching the bottom of the hill is 200 Joule.
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What diagrams can help support my answer and explain why is does it support?
Sunburn is caused by UV radiation by damaging DNA, as the following diagram shows:
Visible light and UV radiation differ in wavelength:
Since only Sunscreen A blocks UVA and UVB radiation, then it effectively prevents sunburn.
On which object will Earth's gravity act with the greatest magnitude? *
An apple
A cereal bowl
A watermelon
A TV remote
Answer: The watermelon
Explanation: The watermelon has a larger mass than the rest of the three.
When tightening a bolt, you push perpendicularly on a wrench with a force of 165 n exerting a torque 23. 1 n. M relative to the center of the bolt. At what distance from the center of the bolt was the force applied?.
The torque was applied at distance of 14cm from the center of the bolt.
The perpendicular force on the wrench applied is 165N.
The total torque experienced by the bolt is 23.1 N-m.
The torque on any body is given by,
T = Fr
Where,
T is the torque on the body,
F is the force applied in the circular manner for the body,
r is the perpendicular distance of the force applied on the body.
So, now putting the values for the bolt,
23.1 = 165r
r = 0.14m
So, we can conclude here that the force was applied at a distance of 14 cm from the center of the bolt.
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what is the direction of the magnetic force on a positive charge moving into the page in a b field that is facing to the right?
According to the right hand rule, the thumb of the right hand should point in the direction of the magnetic force acting on a positive moving charge.
Which magnetic field direction is positive?It is demonstrated that the magnetic line's direction is from north to south and the electric line's direction is from positive to negative.
What direction does the coil's left-hand side force come from?A downward force is produced when current flows through the left hand of the coil, and an upward force is produced when current flows through the right hand of the coil.
What direction does this magnetic force go in?The direction of the magnetic field's force is opposite the direction that it acts.
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An object is launched at a velocity of 20 m/s in a direction making an angle of 22° upward with the horizontal.What is the maximum height achieved by the projectile?
Given data:
Initial velocity,
[tex]u=20\text{ m/s}[/tex]Angle of projection,
[tex]\theta=22\degree[/tex]The maximum height achieved by the projectile is given as,
[tex]H=\frac{u^2\sin ^2\theta}{2g}[/tex]Here, g is the acceleration due to gravity.
Substituting all known values,
[tex]\begin{gathered} H=\frac{(20\text{ m/s})^2\times\sin ^2(22\degree)}{2\times(9.8\text{ m/s}^2)} \\ \approx2.86\text{ m} \end{gathered}[/tex]Therefore, the maximum height achieved by the projectile is 2.86 m.
An electroscope is a device that detects charge on
What methods of charging does John use?
objects that touch it or are brought near it. John rubs a
piece of cotton fabric on a plastic rod and then louches
John charges the rod through induction, then charges
the rod to the electroscope.
the electroscope through friction.
• John charges the rod through conduction, then
charges the electroscope through induction.
• John charges the rod through friction, then charges
the electroscope through induction.
John charges the rod through friction, then charges
the electroscope through conduction.
John charges the rod through friction, then charges the electroscope through induction.
What is induction charging?A charging technique called induction charging involves charging a thing without actually contacting it to another charged object.
The charged particle is held close to an uncharged conductive substance that is grounded on a neutrally charged material during the charging by induction process. When a charge is transferred between two objects, an uncharged conductive substance produces an oppositely polarized charge.
The rod is firstly charged by friction with the cotton fabric then John charges the rod through induction, then charges the rod to the electroscope.
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a projectile lands at the same height from which it was launched. which initial velocity will result in the greatest horizintal displacment of the projectile
The initial velocity that will result in the greatest horizontal displacement of the projectile is V cosθ
The required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ. Since we are provided with a projectile that lands at the same height from which it was launched, so the component of the velocity in the vertical direction is usin θ, and the component along the horizontal direction is ucosθ.Neglecting the air resistance, the velocity will be constant for the object throughout the flight, so the initial velocity will be equal to the final velocity.Therefore the initial velocity will be, V₀ = V cosθ
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An object rolls off a 8 m tall table with a velocity of 3 and lands 6 away from the base of the table. What is the vertical acceleration of the object? (Answer must be negative!!!)
Answer:
-9.81 m/s/s or -4 m/s/s
Explanation:
Technically, the only vertical component of force and therefore acceleration on the object after it leaves the table will be acceleration due to gravity. So we can say the value of acceleration is g, -9.81m/s/s.
Though after checking your parameters, maybe you are on another planet with a different gravitation constant.
in this case, assuming your other values are in the standard m/s and m, we can determine how long the ball should be in the air to land 6 meters away and thus use that to find downward acceleration.
6m = 3m/s • t
2s = t
8 = 1/2 g • (2)^2
2m/s/s = 1/2 g
4 m/s/s = g
this would approximate to doing this experiment on mercury
A monochromatic laser is exciting hydrogen atoms from the n=2 state to the n=5 state.
A monochromatic laser is responsible for exciting hydrogen atoms from their n=2 state to their n=5 state in which :
wavelength λ of the laser =435 nm
longest wavelength = λmax = 4052 nm
shortest wavelength = λmin= 95 nm
To do excitation, the energy of the laser should be greater than or equal to the energy of transition of the hydrogen atom.
In this case, the states of energy are explained by the Bohr model.
En = - 13,606 /n² [eV]
therefore the energy of the transition is
ΔE = E 5 - E2
ΔE = 13.606 (1 n2^2 - 1 n5^5 )
ΔE = 13.606(1 /2^2 - 1 /5^5).
ΔE = 2,85726 eV
Using the Planck's equation
E = h f
Light’s relation to wavelength and frequencies
c = λ f
f = c /λ
λ = h c / E
reducing energy to the SI system
E = 2,85726 eV (1.6x10⁻¹⁹ J /1 eV) = 4.5716 x 10⁻¹⁹ J
hence, λ = 6,626 x 10⁻³⁴ 3 x 10⁸ /4 ,5716 x 10⁻¹⁹
λ = 4.348x 10⁺⁷ m (10⁹ nm / 1 m)
λ = 435 nm
photon emission [e] , to the base state n = 1, from n = 5.
Initial state, n = 5
final state, n = 4
ΔE = 13.606 (1/4² - 1/5²)
ΔE = 0.306 eV
λ =h c / E
λ = 4052 nm
n = 5
final / level ΔE (eV) λ (nm)
4 0.306 4052
3 0.9675 1281
2 2,857 435
1 13.06 95
n = 4
3 0.661 1876
2 2,551 486
1 11,905 104
n = 3
2 1.89 656
1 12.09 102.5
n = 2
1 10.20 121.6.
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a child pulls a friend in a little red wagon with constant speed. if the child pulls with a force of 16 n for 10.0 m, and the handle of
The work done by the child to pull the wagon is found to be 67.61N.
The child pulls the wagon with a force 16N and the handle of the wagon is inclined at the angle of 25°. The wagon is puled till a distance of 10m.
So, the force will have to component, one vertical and one is horizontal,
The horizontal component will move the wagon,
So, the work done by the child is given by,
W = F.S.sin(25°)
F is the force and S is the distance, Putting values,
W = 16 x 10 x 0.42
W = 67.61 N.
The work done by the child in pulling the wagon is 67.61 N.
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Complete Question - A child pulls a friend in a little red wagon with constant speed. If the child pulls with a force of 16 N for 10.0 m, and the handle of the wagon is inclined at an angle of 25° above the horizontal, how much work does the child do on the wagon?
Which one of these objects has kinetic energy?
an Iphone laying on a table
a frisbee sitting on a roof
a ball rolling down the driveway
Answer: A ball rolling down the driveway.
Explanation:
Kinetic is when an object is in motion. Potential is when an object rests.
why do we spread our fingers wide open while sitting before the fire or heater?
Answer: This is an evolutionary adaption
Explanation:
As a result, practically all the heat you receive from a campfire when you are sitting next to one is transmitted through thermal radiation. Because of this, the side of your body that is facing the fire warms up, while the side that is facing away from it remains cool.
What our fingers wide open while sitting in a heater?Since all the hot air is rising, you can actually fit your hand through the flame's base, which is also where the cold air from the surrounding space is drawn in. The flame's base is extremely chilly. You may reach through it and touch it with some soot.
Therefore, Heat is transferred through convection, which is the large-scale movement of molecules in fluids like liquid and gas. Consequently, one becomes warm while seated close to a fire.
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A lemming take off from the roof of a building 56m tall and lands 45m from the base. What was the lemming’s initial speed ?
Given data
*The given height of the building h = 56 m
*The distance from the base is d = 45 m
*The value of the acceleration due to gravity is g = 9.8 m/s^2
The formula for the time taken by the lemming is given as
[tex]t=\sqrt[]{\frac{2h}{g}}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} t=\sqrt[]{\frac{2\times56}{9.8}} \\ =3.38\text{ s} \end{gathered}[/tex]The formula for the lemming's initial speed is given as
[tex]v=\frac{d}{t}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v=\frac{45}{3.38} \\ =13.3\text{ m/s} \end{gathered}[/tex]Hence, the lemming's initial speed is v = 13.3 m/s
a container holds a gas consisting of 1.50 moles of oxygen molecules. One in a million of these molecules has lost a single electron. What is the net charge of the gas?
A container holds a gas consisting of 1.50 moles of oxygen molecules. One in a million of these molecules has lost a single electron. the net charge of the gas will be + 9.033 * [tex]10^{17}[/tex]
If a negative charge is lost by an atom , then due to conservation of charge , that atom will acquire a positive charge . Similarly , if positive charge is lost then that atom will acquire a negative charge .Now in above question
Total number of moles = 1.5
number of molecules in 1.5 moles of oxygen = 1.5 * 6.022 *[tex]10^{23}[/tex] = 9.033 *[tex]10^{23}[/tex] molecules
9.033 *[tex]10^{23}[/tex] / [tex]10^{6}[/tex] million molecules = 9.033 * [tex]10^{17}[/tex] million molecule
1 in a million has lost a single electron
9.033 * [tex]10^{17}[/tex] million molecule will loose 9.033 * [tex]10^{17}[/tex] electron
the net charge of the gas will be + 9.033 * [tex]10^{17}[/tex]
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4. A 1.5 m tall man is standing 2m away from the concave lens (remember f is negative) of a peephole with a focal length of 3.0 cm. a) What is the distance to the image in centimeters? b) What is the magnification of the image in meters?
Here,
Size of object = 1.5 m;
Object distance (u) = - 2m = -200cm
focal length (f) = - 3.0 cm
Image distance (v) = ?
Using lens formula
[tex]\begin{gathered} \frac{1}{f}=\text{ }\frac{1}{v}-\frac{1}{u}; \\ \frac{1}{-3}=\frac{1}{v}-\frac{1}{-200}; \\ \frac{1}{-3}=\text{ }\frac{1}{v}\text{ +}\frac{1}{200} \\ \frac{1}{v}=\text{ -}\frac{1}{3}-\frac{1}{200}=-\text{ }\frac{203}{600} \\ v=\text{ - }\frac{600}{203}=\text{ -2.96 cm} \end{gathered}[/tex]Now magnification is given by
[tex]Magnification=\text{ }\frac{v}{u}=\text{ }\frac{2.96}{200}=0.0148[/tex]Final answer is :-
Image distance = - 2.96 cm & magnification = 0.0148
A runner covers the last straight stretch of a race in 8s.During that time he speeds up from 7 m/s to 9m/s. What is the acceleration
Given
Time taken is t=8 s.
The initial speed is u=7 m/s
The final speed is v=9 m/s.
To find
The acceleration
Explanation
We know the acceleration is the ratio of the difference in the speed to the time taken.
Thus,
[tex]\begin{gathered} a=\frac{v-u}{t} \\ \Rightarrow a=\frac{9-7}{8} \\ \Rightarrow a=\frac{2}{8}=\frac{1}{4}=0.25\text{ m/s}^2 \end{gathered}[/tex]Conclusion
The acceleration is
[tex]0.25\text{ m/s}^2[/tex]this is a 3 part question6) (a) Your heart beats with a frequency of 1.45 Hz. How many beats occur in a minute? (b) If the frequency of your heartbeat increases, will the number of beats in a minute increase, decrease, or stay the same? (c) How many beats occur in a minute if the frequency increases to 1.55 Hz?
Given,
The initial frequency of the heartbeat, f₁=1.45 Hz
The increased heartbeat, f₂=1.55 Hz
The frequency of the heartbeat can be described as the number of occurence of the heartbeat per second. That is every second, the heart beats 1.45 times.
(a)
Thus for a minute, the number of the heartbeats is,
[tex]\begin{gathered} N_1=f_1\times60 \\ =1.45\times60=87 \end{gathered}[/tex]Thus 87 beats occur for a minute.
(b)The increase in the frequency of the heartbeat implies the increase in the number of the heartbeat for every second. And hence the beats in a minute increase when the frequency of the heartbeat increases.
(c)
The number of the beats per minute after the increase of the frequency is,
[tex]\begin{gathered} N_2=f_2\times60 \\ =1.55\times60 \\ =93 \end{gathered}[/tex]Thus after the increase in the frequency, 93 beats occur in a minute.
Rudolph's mother is a physicist. She gave him a present for his birthday, but
she told him he can only open it if he can calculate its mass. She told him that
the normal force the table exerts on the gift is 55 N. What is the mass of the
gift?
The mass of the gift if it exert a normal force of 55 N is 5.61 kg.
What is mass?Mass can be defined as the quantity of matter a body contains.
To calculate the mass of the gift, we use the formula below.
Formula:
m = N/g................. Equation 1Where:
m = Mass of the giftN = Normal forceg = Acceleration due to gravityFrom the question,
Given:
N = 55 Ng = 9.8 m/s²Substitute these values into equation 1
m = 55/9.8 m = 5.61 kg.Hence, the mass of the gift is 5.61 kg.
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what happens to the molecules when carbon dioxide is heated
When the molecules of carbon dioxide is heated, it is disintegrated into molecules of carbon monoxide and oxygen atoms.
CO2 → CO + O
This is the basic reaction expected to happen by carbon dioxide molecules are heated. However it may also disintegrate into C2, O2, C, etc.
Since heat is absorbed in this reaction, it is an example of endothermic reaction. In an exothermic reaction heat is released. These are the basic two types of chemical reactions on basis of heat.
Therefore, when the molecules of carbon dioxide is heated, it is disintegrated into molecules of carbon monoxide and oxygen atoms.
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What does it mean to tare a balance and why do you think it is important to complete this before you begin measuring mass?.
When weighing chemicals on a balance, the term "tare" is used. Pressing the tare button enables you to record only the weight of the substance being measured, not the weight of the container it is being measured in.
Why is tare the balance prior to use important?Weighing by difference is automatically accomplished by tarring a balance. When a balance is tared with an object on the balance pan, the weight of the object is automatically subtracted from each reading up until the balance is re-tared or zeroed.
Testing a balance is crucial because you want an accurate result when you weigh an object. The calculations could all be off if a balance is not set to zero.
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I need to find out number 51.Can you help me ?
v= constant
mass of the cart = 1.5 kg
m2 (added)= 7.5 kg
When using their brakes,
cars disengage the engine so that it no longer
provides any forward force.
A(n) 2006.1 kg car is coasting along a level
road at 31.5 m/s. A constant braking force
is applied, such that the car is stopped in a
distance of 62.1 m.
What is the magnitude of the braking force?
Answer in units of N.
The magnitude of the braking force is 0
What is acceleration?
velocity changes with time at an accelerating rate in both speed and direction. In the case of simple harmonic motion, the spring's elastic force serves as the restoring force. The application of Hooke's law, F = - kx, allowed for its discovery. Here, k is the spring constant and x is the spring's deformation (change of its length from equilibrium position).
We can see that this force is variable and dependent on the mass-spring system's x displacement. The minus sign indicates that the force's direction is the opposite of the displacement, and as a result, the force causes the mass of the back to return to its equilibrium position.
As with any force, restoring force can be calculated using Newton's second law, F = ma, where m is the body's mass and an is its acceleration.
Now we can write:
The body's displacement x is zero when it reaches equilibrium. Acceleration is therefore 0.
By the way, when the amplitude of the this mass-spring system reaches its maximum, the acceleration will be at its maximum value. The mass-spring pendulum swings in the opposite direction at these points. We refer to those as inflection points.
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A bus goes 500km east from town A to town B in the morning and comes back halfway in the evening travelling west. What is the distance and displacement of the bus?
Given:
Distance the bus goes from town A to town B = 500 km east
Given that the bus comes back halfway in the evening travelling west, let's find the distance and the displacement of the bus.
• To find the distance of the bus, we have:
[tex]\begin{gathered} d=500+(\frac{500}{2}) \\ \\ d=500+250 \\ \\ d=750\operatorname{km} \end{gathered}[/tex]Therefore, the distance the bus covered is 750 km
• To find the displacement, we have:
[tex]\begin{gathered} Displacement\text{ = 500 - (}\frac{500}{2}) \\ \\ \text{Displacement = 500 - 250} \\ \\ \text{Displacement = 250 km} \end{gathered}[/tex]Therefore, the displacement of the bus is 250 km east
ANSWER:
Distance = 750 km
Displacement = 250 km east
PLS HURRY
The photo shows a roller coaster. Assume the system is closed. Which roller-coaster car has the least potential energy due to gravity?
A. Car A
B. Car B
C. Car C
D. Car D
Answer: it's rather A or B
Explanation: those are the moments it stops but most likely it is answer A.
the speed of light in a material is what is the critical angle of light ray at the interface between material and vacuum
This problem indicates that the speed of light in a material medium is 0.5 10⁸ m / s, they ask to find the critical angle between the material and the vacuum.
Let's find the refractive index of the material
n = c / v
n = 3 10⁸ / 0.5 10⁸
n = 6
When the material passes from one medium to another, it must comply with the law of refraction.
n₁ sin θ = n₂ sin θ₂
for the angle criticize the angles tea2 = 90
t = sin⁻¹n₂/ n₁
The vacuum replacement index is n₂ = 1
t = sin⁻¹ (1/6)
t = 9.59º
Therefore, the angle of refractive index will be 9.59°.
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[NOTE: THIS IS AN INCOMPLETE QUESTION. THE COMPLETE QUESTION IS: The speed of light in a material is 0.50 c. what is the critical angle of a light ray at the interface between the material and a vacuum?]
Which factor will increase the resistance of a conductor?
Length is directly proportional to resistance. When length increases resistance increases and vice versa.
Cross sectional area is inversely proportional to resistance. When cross sectional area of a conductor increases resistance decreases and vice versa.
Different materials have different resistances.
Temperature is directly proportional to resistance. When temperature increases resistance increases and vice versa.