The final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.
To find the answer, we need to know about the thermodynamic processes.
How to find the final temperature of the gas?Any processes which produce change in the thermodynamic coordinates of a system is called thermodynamic processes.In the question, it is given that, the tank is rigid and non-conducting, thus, dQ=0.The membrane is raptured without applying any external force, thus, dW=0.We have the first law of thermodynamic expression as,[tex]dU=dQ-dW[/tex]
Here it is zero.[tex]dU=0[/tex],
As we know that,[tex]dU=C_pdT=0\\\\thus, dT=0\\\\or , T=constant\\\\i.e, T_1=T_2[/tex]
Thus, the final temperature of the system will be equal to the initial temperature,[tex]T_1=T_2=100^0C=373K[/tex]
How much work is done?We found that the process is isothermal,Thus, the work done will be,[tex]W=RT*ln(\frac{V_2}{V_1} )=373R*ln(\frac{4}{\frac{4}{3} })\\ \\W=409.8R J[/tex]
Where, R is the universal gas constant.
What is a reversible process?Any process which can be made to proceed in the reverse direction is called reversible process.During which, the system passes through exactly the same states as in the direct process.Thus, we can conclude that, the final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.
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The system's final temperature will be 373K, which is the same as its starting temperature. The system exerts 409.8R Joules of work.
We need to understand the thermodynamic processes in order to locate the solution.
How can I determine the gas's final temperature?Thermodynamic processes are any actions that result in modifications to a system's thermodynamic coordinates.Given that the tank is stiff and non-conducting, the answer to the question is that dQ=0.Without using any external force, the membrane is torn; hence, dW=0.The first law of thermodynamics is expressed as follows:[tex]dU=dQ-dW[/tex] , It is 0 here.
As we are aware,[tex]dU=C_pdT=0\\dT=0\\T=constant\\T_1=T_2=373K[/tex]
As a result, the system's final temperature will be equal to its starting temperature.
How much work is expended?The process is isothermal, as we discovered.As a result, the work will be,[tex]W=RT ln(\frac{V_2}{V_1} )=373R*ln(3 )\\W=409.8R Joules[/tex]
R is the gaseous universal constant.
A reversible process is what?Reversible processes are any operations that have the ability to be reversed.The system goes through the exact same states as it did during the direct procedure throughout this time.Thus, we can draw the conclusion that the system's end temperature will be 373K, the same as its starting temperature. The system exerts 409.8R Joules of work.
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Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.240 A/s, the induced emf in the second coil has a magnitude of 1.60×10−3 V.
a) What is the mutual inductance of the pair of coils?
b) If the second coil has 30 turns, what is the flux through each turn when the current in the first coil equals 1.25 A ?
c) If the current in the second coil increases at a rate of 0.365 A/s , what is the magnitude of the induced emf in the first coil?
The mutual inductance of the pair of coils is 6.67 x 10⁻³ H.
The the flux through each turn is 2.78 x 10⁻⁴ Tm².
The magnitude of the induced emf in the first coil is 2.435 x 10⁻³ V.
Mutual inductance of the coilM = -NΦ/I
M = emf/I
M = -(1.6 x 10⁻³)/(-0.24)
M = 6.67 x 10⁻³ H
Flux in the second coilM = NΦ/I
MI = NΦ
Φ = MI/N
Φ = (6.67 x 10⁻³ x 1.25)/(30)
Φ = 2.78 x 10⁻⁴ Tm²
Induced emf in the first coilemf = MI
emf = 6.67 x 10⁻³ x 0.365
emf = 2.435 x 10⁻³ V
Thus, the mutual inductance of the pair of coils is 6.67 x 10⁻³ H.
The the flux through each turn is 2.78 x 10⁻⁴ Tm².
The magnitude of the induced emf in the first coil is 2.435 x 10⁻³ V.
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A fisherman and his young son are in a boat on a small pond. Both are wearing life jackets. The son is holding a large helium filled balloon by a string. Consider each action below independently and indicate whether the level of the water in the pond, Rises, Falls, is Unchanged or Cannot tell.
The son pops the helium balloon.
The fisherman knocks the tackle box overboard and it sinks to the bottom.
The son finds a cup and bails some water out of the bottom of the boat
The fisherman lowers himself in the water and floats on his back.
The fisherman lowers the anchor and it hangs one foot above the bottom of the pond.
The son gets in the water, looses his grip on the string, letting the balloon escape upwards.
Based on the weight of the objects in the water, the water level in the given scenario is as follows;
unchangedrisesfallsunchangedrisesrisesWhat changes will be observed in each of the given scenario?The rise or fall of a fluid when an object is placed in it is determined by the density, mass, and volume of the object.
According to Archimedes' principle, the upthrust or upward force acting on a body fully or partially immersed in a fluid, is equal to the weight of the fluid displaced.
Considering the given situations:
The son pops the helium balloon - water level is unchanged because the weight of the balloon is negligible.Fisherman knocks the tackle box overboard and it sinks to the bottom - water level will rise since the tackle box has significant weightSon finds a cup and bails some water out of the bottom of the boat - water level will fall since some volume of water is being removed and added to the boat.Fisherman lowers himself in the water and floats on his back - water level is unchanged because no additional weight is added to the waterFisherman lowers the anchor and it hangs one foot above the bottom of the pond - water level will rise since the anchor will displace some volume of water equal to its weight in the pondSon gets in the water, looses his grip on the string, letting the balloon escape upwards- water level will rise since the weight of the balloon and string are significant.In conclusion, the rise in water level depends on the weight of the objects.
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100gm o2 gas is pressurized to 20 degree Celsius. done Also, how much heat energy will be converted into mechanical energy?
The heat energy that will be converted into mechanical energy is 1.83 kJ.
Heat capacity of the O2 gas
The heat energy that will be converted into mechanical energy is calculated as follows;
Q = mcΔθ
where;
m is mass = 100 g = 0.1 kgΔθ is change in temperaturec is specific heat capacityat 20 ⁰C = 293 K, C = 0.915 kJ/kg K
Q = (0.1 kg)(0.915)(20 )
Q = 1.83 kJ
Thus, the heat energy that will be converted into mechanical energy is 1.83 kJ.
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Which statement is a postulate of general relativity?
The speed of light is constant for all observers.
Observers will see the same laws of physics whether at rest or in uniform motion.
A gravitational field is the same as an object moving at the speed of light.
Observers will see the same laws of physics in any frame of motion whether accelerated or not
Answer:
The speed of light is constant for all observers
Explanation:
As per general postulate of relativity
Lorentz covariance of special relativity becomes a local Lorentz covariance in the presence of matter.So
light speed c is independent of States of matter
A step down transformer with an input voltage of 220V decreases the voltage to half of the input .There is a current flowing of 15A in the primary coil. Find the output current
The output current for the given step down transformer is 30A.
What is the step down transformer?
A step down transformer is a passive device that converts high voltage power to low voltage power, while the output current is higher than the input current. They are used in power adaptors and rectifiers to decrease the voltage to the desired level. It works according to Faraday's law of Electromagnetic induction.
The current in the windings of a step down transformer is inversely proportional to the voltage in windings as:
Input voltage / Output voltage = Output current / Input current
220 / 110 = Outout current / 15
Output current = 30A
Hence, the output current (30A) obtained is higher than the given input current (15A).
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Determine the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is 1/60 of its value at the Earth's surface.
The distance from the Earth's center to the point outside the Earth is 55800 Km
How to determine the distance from the surface of the EarthAcceleration due to gravity of Earth = 9.8 m/s²Acceleration due to gravity of the poin (g) = 1/60 × 9.8 = 0.163 m/s²Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²Mass of the Earth (M) = 5.97×10²⁴ KgDistance from the surface of the Earth (r) =?g = GM / r²
Cross multiply
GM = gr²
Divide both sides by g
r² = GM / g
Take the square root of both sides
r = √(GM / g)
r = √[(6.67×10¯¹¹ × 5.97×10²⁴) / 0.163)]
r = 4.94×10⁷ m
Divide by 1000 to express in Km
r = 4.94×10⁷ / 1000
r = 4.94×10⁴ Km
How to determine the distance from the center of the EarthDistance from the surface of the Earth (r) = 4.94×10⁴ KmRadius of the Earth (R) = 6400 Km Distance from the centre of the Earth =?Distance from the centre of the Earth = R + r
Distance from the centre of the Earth = 6400 + 4.94×10⁴
Distance from the centre of the Earth = 55800 Km
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What does it mean to say that two systems are in thermal equilibrium
Answer:
In simple words, thermal equilibrium means that the two systems are at the same temperature.
A baseball (m=145g) traveling 39 m/s moves a fielder's glove backward 23 cm when the ball is caught.
What was the average force exerted by the ball on the glove?
Express your answer to two significant figures and include the appropriate units.
The average force exerted by the ball on the glove is 480 N.
What is the force exerted by the ball on the glove?
The average force exerted on the glove by the ball is equal in magnitude to the force on the ball.
Force = mass * accelerationmass = 145 g = 0.145 kg
Acceleration of the baseball, a = (v² - u²)/2s
where:
v is final velocity = 0
u is initial velocity = 39 m/s
s is distance = -23 cm = 0.23 m
a = (0 - 39²)/2(-0.23)
a = 3306.52 m/s²
Force = 0.145 * 3306.52
Force = 479.4 N
Average force = 480 N
In conclusion, force is derived from the product of mass and acceleration.
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Q2. A force of 1000N is experienced by a charge in a field of strength of 250NC. Find the value of the charge establishing the field.
Answer:
4 C
Explanation:
The strength of an electric field can be defined as: [tex]E = \frac{f}{q}[/tex] where f=force and q=charge, and e=strength
Since we're given the strength and the force we can simply rearrange the equation so that we solve for Q:
Original Equation:
[tex]E=\frac{f}{q}[/tex]
Multiply both sides by q
[tex]E*q = f[/tex]
Divide both sides by E
[tex]q=\frac{f}{e}[/tex]
So now plug the known values into the equation
[tex]q=\frac{1000 N}{250 N/C}[/tex]
Simplify:
[tex]q = 4 c[/tex]
Three parallel wires each carry current I in the directions shown in (Figure 1). The separation between adjacent wires is d.
a) Calculate the magnitude of the net magnetic force per unit length on the top wire.
b) Calculate the magnitude of the net magnetic force per unit length on the middle wire.
c) Calculate the magnitude of the net magnetic force per unit length on the bottom wire.
(a) The magnitude of the net magnetic force per unit length on the top wire is μI²/πd.
(b) The magnitude of the net magnetic force per unit length on the middle wire is zero.
(c) The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.
Force per unit length
The magnitude of the net magnetic force per unit length on the top wire is calculated as follows;
F₁/L = (μI₁/2π) x (I₂/d + I₃/d)
F₁/L = (μI/2π) x (I/d + I/d)
F₁/L = (μI/2π) x (2I/d)
F₁/L = μI²/πd
The magnitude of the net magnetic force per unit length on the middle wire is calculated as follows;
F₂/L = (μI₂/2π) x (I₃/d - I₁/d)
F₂/L = (μI/2π) x (I/d - I/d) = 0
The magnitude of the net magnetic force per unit length on the middle bottom is calculated as follows;
F₃/L = (μI₂/2π) x (I₁/d + I₂/d)
F₃/L = (μI/2π) x (I/2d + I/d)
F₃/L = (μI/2π) x (3I/2d)
F₃/L = 3μI²/4πd
Thus, the magnitude of the net magnetic force per unit length on the top wire is μI²/πd.
The magnitude of the net magnetic force per unit length on the middle wire is zero.
The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.
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An electron gun shoots electrons at a metal plate that is 4.0 mm away in a vacuum. The plate is 5.0 V lower in potential than the gun. How fast must the electrons be moving as they leave the gun if they are to reach the plate?
The speed of the electron is 1.3 * 10^6 m/s
What is the velocity?We know that when the electron gun is shot, the potential energy of the electron is converted into kinetic energy. The mass of the electron is given as 9.11 * 10^-31 Kg.
The energy of the electron is;
eV = 1e * 5V = ev or 8 * 10^-19 J
Given that E = 1/2mv^2
8 * 10^-19 = 0.5 * 9.11 * 10^-31 * v^2
v = √ 8 * 10^-19/0.5 * 9.11 * 10^-31
v = √1.75 * 10^12
v = 1.3 * 10^6 m/s
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Calculate the amount of air in a room 6m long, 5m wide and 3mm high.
Answer:
0.09kg of air
Explanation:
The dimensions of the room are given
change the height to meters by dividing it by thousand.
For the volume multiplying the length,width and height (all should be in the same unit most suitable being meters).
Volume refers to the amount of space inside a box or a object.
The amount of air is equal to the volume.
Answer:
90 m^3
Explanation:
Volume of the room:
6 m * 5 m * 3 m = 90 m^3 <=====( I changed 3mm to 3 m)
if 3mm is not a typo mistake
volume becomes ( 3 mm = .003 m)
6 m * 5 m * .003 m = .09 m^3 ( though unlikely )
A circular loop of wire with radius 0.0410 m and resistance 0.169 Ω is in a region of spatially uniform magnetic field, as shown in the following figure (Figure 1). The magnetic field is directed out of the plane of the figure. The magnetic field has an initial value of 7.78 T and is decreasing at a rate of -0.605 T/s.
a) Is the induced current in the loop clockwise or counterclockwise?
b) What is the rate at which electrical energy is being dissipated by the resistance of the loop?
(a) The induced current in the loop will be counterclockwise.
(b) The rate at which electrical energy is being dissipated by the resistance of the loop is 0.012 W.
Direction of the currentThe induced current in the loop will be counterclockwise to the direction of magnetic field.
Emf induced in the loopemf = -NdФ/dt
emf = -NBA/dt
where;
A is area of the loop
A = πr² = π(0.041)² = 5.28 x 10⁻³ m²
emf = -(-0.605 - 7.78) x 5.28 x 10⁻³
emf = 8.385 x 5.28 x 10⁻³
emf = 0.0442 V
Rate of energy dissipationP = emf²/R
P = (0.0442)²/0.169
P = 0.012 W
Thus, the rate at which electrical energy is being dissipated by the resistance of the loop is 0.012 W.
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A 269 kg weather balloon is designed to lift a 2910 kg package. What volume should the balloon have after being inflated with helium in order that the total load can be lifted? (Assume standard temperature and pressure, at which the density of air is ρ[tex]_{air}[/tex] = 1.29 kg//[tex]m^{3}[/tex] and the density of helium is ρ[tex]_{He}[/tex] = 0.179 kg/[tex]m^{3}[/tex].)
As, per the buoyancy force,the volume that the balloon should have is 2863[tex]m^{3}[/tex]
What is buoyancy force?
Air buoyancy is the upward force exerted on an object by the air that is displaced by object. Air buoyancy is responsible for the buoyancy created by the displaced air.
[tex]F_{b}[/tex] = -Vρg , where V= volume of the object
ρ = density of the object
g = acceleration due to gravity
[tex]F_{b}[/tex] = buoyant force
The buoyancy force must be equal to the total load lifted
ρ[tex]_{He}[/tex] × V × g + 269 + 2910 = ρ[tex]_{air}[/tex] × V × g
0.179 × V + 3179 = 1.29 V
0.179V + 3179 = 1.29V
0.179V- 1.29V = - 3179
1.11V = 3179
On solving , we get
V = 2863 [tex]m^{3[/tex]
Therefore, the volume that the balloon should have is 2863[tex]m^{3}[/tex].
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A flat circular coil carrying a current of 8.80 A has a magnetic dipole moment of 0.194 A⋅m2 to the left. Its area vector A⃗ is 4.0 cm2 to the left.
a) How many turns does the coil have?
b) An observer is on the coil's axis to the left of the coil and is looking toward the coil. Does the observer see a clockwise or counterclockwise current?
c) If a huge 45.0 T external magnetic field directed out of the paper is applied to the coil, what magnitude of torque results?
d) What direction of torque results?
Hi there!
a)
We can use the equation for the magnetic dipole moment to solve for the number of turns:
[tex]\mu_m = NIA\vec{n}[/tex]
[tex]\mu_m[/tex] = Magnetic dipole moment (0.194 Am²)
N = Number of loops (?)
A = Area of loop (4.0 cm²)
[tex]\vec{n}[/tex] denotes the area vector, or the normal line perpendicular to the area.
We first need to convert cm² to m² using dimensional analysis.
[tex]4.0 cm^2 * \frac{0.01m}{1 cm} * \frac{0.01 m}{1cm} = 0.0004 m^2[/tex]
Rearranging the equation to solve for 'N':
[tex]N = \frac{\mu_m}{IA}\\\\N = \frac{0.194}{(8.8)(0.0004)} = \boxed{55.11 \text{ turns}}[/tex]
**Since we cannot have part of a turn, the coil has about 55 turns.
b)
For this, we can use the Right-Hand-Rule for current. Looking at the coil from the left with your curled fingers going around the coil with the fingertips pointing through and to the left in the direction of the magnetic moment, your thumb points in the COUNTERCLOCKWISE direction.
c)
Now, let's use the equation for the torque produced by a magnetic field:
[tex]\tau = \mu_m \times B[/tex]
This is a cross-product, but since our magnetic field is perpendicular to the magnetic moment, we can disregard it.
Plugging in the values for the magnetic moment and the magnetic field:
[tex]\tau = 0.194 * 45 = \boxed{8.73 Nm}[/tex]
d)
Using the other RHR (current, field, force), the coil will spin about its vertical axis in the field. In more detail, if you look at the coil from the left-hand side with its opening towards you, from this perspective, the left of the coil will come towards you, and the right side of the coil will move away.
What is double-slit experiment?
The double-slit experiment shows that both matter and light can exhibit properties of conventionally defined waves and particles.
The double-slit experiment is a part of a class of "double path" experiments in which a wave is split into two separate waves that later combine to form a single wave (the wave is typically composed of many photons and is better known as a wave front, which should not be confused with the wave properties of the individual photon).
Isaac Newton's corpuscular theory of light, which had previously prevailed as the accepted explanation of light transmission in the 17th and 18th centuries, was defeated by double-slit experiment , which was conducted in the early 1800s.
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A wire carrying a 25.0 A current bends through a right angle. Consider two 2.00 mm segments of wire, each 3.00 cm from the bend (Figure 1).
a) Find the magnitude of the magnetic field these two segments produce at point P , which is midway between them.
b) Find the direction of the magnetic field at point P
The magnitude of the magnetic field and the direction of the magnetic field at point P is mathematically given as
B=1.9*10^{-5}T
To determine the magnetic field direction, use the right-hand -rule on the page magnetic field is going.
What is the magnitude of the magnetic field these two segments produce at point P, which is midway between them.?Generally, the equation for Biot savant law is mathematically given as
[tex]B=(\frac{u}{4\pi}*{\frac{Ilsin\theta}{r^2})[/tex]
Their net field is
Bn=2B
[tex]Bn=2* B=(\frac{u}{4\pi}*{\frac{Ilsin\theta}{r^2})[/tex]
Hence
[tex]B=(\frac{4*\p *10^{-7}}{4\pi}*{\frac{(30)(2*10^{-3}sin45)}{\sqrt{(3*10^{-2})^2+((3*10^{-2})^2)}/2})[/tex]
B=1.9*10^{-5}T
In conclusion, To determine the magnetic field direction, use the right-hand -rule on the page magnetic field is going.
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Answer:
The magnitude of the magnetic field at the point P is [tex]1.57*10^{-5}T[/tex] and the field is pointing into the page.
Explanation:
The general form of a similar question to this is:
[tex]\vec{B} = \frac{\mu _{0} }{4\pi } * \oint \frac{Id\vec{l} \times \hat{r}}{r^{2} }[/tex]
where [tex]\vec{B}[/tex] is the vector of the Magnetic Field, [tex]\mu _{0}[/tex] is the Free Space Permeability Constant (equal to [tex]4\pi * 10^{-7} \frac{N}{A^2}[/tex]), [tex]I[/tex] is the current, and [tex]r[/tex] is the distance from the segment to the point P. (I will get to the [tex]d\vec{l} \times \hat{r}[/tex] term in a bit)
This equation is fairly complicated. Luckily, it can be simplified by looking at the magnitude and direction separately.
The first thing to simplify is the cross product. Due to the fact that a cross product can be simplified from [tex]\vec{x} \times \vec{y}[/tex] to [tex]xy\sin(\theta)[/tex], where [tex]\theta[/tex] is the angle between the 2 vectors, and [tex]\hat{r}[/tex] is the unit vector of [tex]r[/tex] (i.e. [tex]\hat{r} = \vec{r}/r[/tex]) we can simplify [tex]d\vec{l} \times \hat{r}[/tex] to just [tex]dl \sin(\theta)[/tex].
Next, we will look at the integral. In this scenario, everything will function as a constant, so we can essentially ignore it.
Finally, [tex]\frac{\mu_{0}}{4\pi}[/tex] simplifies down to [tex]10^{-7}[/tex].
This gives us our new equation for the Magnetic Field produced by a single segment at a point:
[tex]B = \frac{Il\sin\theta}{r^{2}}*10^{-7}[/tex]
Now we need to find values for [tex]r[/tex] and [tex]\theta[/tex]. Luckily, we are dealing with a 45-45-90 triangle with sides of [tex]1.5 cm[/tex]. This means the distance [tex]r[/tex] is [tex](1.5\sqrt2)cm[/tex]! Similarly, because it is a 45-45-90 triangle, our [tex]\theta[/tex] is [tex]45\textdegree[/tex]!
Now we can start plugging things in:
[tex]B = \frac{(25A)(2*10^{-3}m)\sin(45\textdegree)}{(1.5\sqrt2*10^{-2}m)^2}*10^{-7}\frac{N}{A^2}[/tex]
[tex]B = 7.86^{-6} \frac{N}{A}[/tex] or [tex]T[/tex]
This is the magnitude due to only one single segment. In order to find the total field, we need to know the direction of the field due to each segment.
Finding the direction is really easy. Just use the right hand rule. Point your thumb in the direction of the current and curl the rest of your fingers around an imaginary pole. The direction your fingers point is the direction of the field. In this case, the field lines due to the segments point into the page in the 4th quadrant (the origin is the bend). This means that at point P, both segments induce the same field in the same direction. Therefore, we can take our value from before and double it, giving us our final answer:
[tex]B = 1.57*10^{-5} T[/tex]; into the page.
How does an atom of rubidium-85 a rubidium ion with a +1 charge?
A. The atom loses 1 electron to have a total of 47.
B. The atom gains proton to have a total of 38.
C. The atom loses 1 electron to have a total of 36,
D. The atom gains 1 proton to have a total of 86.
Answer:
C. The atom loses 1 electron to have a total of 36
Explanation:
The number of electrons in a Rubidium atom is 37. Since the atom loses 1 electron, it has 36 left.
which object has a weight of about 22.5 n the book the rock the box the fish
Answer: The rock
Explanation:
As shown in (Figure 1), a layer of water covers a slab of material X in a beaker. A ray of light traveling upwards follows the path indicated.
a) Using the information on the figure, find the index of refraction of material X .
b) Find the angle the light makes with the normal in the air.
Answer: the index of refraction of material X will be 1.09 and the angle the light makes with the normal in the air is 81.25°.
Explanation: To find the answer, we need to know the Snell's law.
What is Snell's law of refraction? Using this, how to solve the problem?The Snell's law for refraction can be written as,[tex]\frac{sin (i)}{sin(r)} =\frac{n_r}{n_i}[/tex]
where, i is the incident angle, r is the refracted angle, n is the refractive index.
As we know that the refractive index of water is 1.33For the first case, incident angle from the picture is 65°, and the refracted angle is 48°. Thus, the refractive index of the medium X will be,[tex]\frac{sin 65}{sin48} =\frac{n_w}{n_X} \\n_X=\frac{1.33*sin48}{sin65} =1.09\\[/tex]
In the second case, incident angle is 48° and we have to find the refracted angle r for the air.As we know that the refractive index of air is 1.Thus, the refracted angle will be,[tex]\frac{sin 48}{sin r}=\frac{1}{1.33} \\\\ sin(r)=\frac{1.33*sin 48}{1}=0.988\\\\r=sin^{-1}(0.988)=81.25 degrees.[/tex]
Thus, we can conclude that, the index of refraction of material X will be 1.09 and the angle the light makes with the normal in the air is 81.25°.
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The light makes an angle of 81.25° with the normal in the air and the index of refraction of material X will be 1.09.
In order to determine the solution, we must understand Snell's law.
What is the refraction law of Snell? How can the issue be resolved with this?One way to express Snell's law for refraction is as follows:[tex]\frac{sin(i)}{sin(r)}=\frac{n_r}{n_i}[/tex]
where the refractive index is n and the incidence angle is i. The refracted angle is r.
As is well known, water has a refractive index of 1.33.The incidence angle from the image in the first case is 65°, while the refracted angle is 48°. Consequently, the medium X's refractive index will be,[tex]n_X=\frac{n_w*sin 48}{sin65} =1.09[/tex]
The incident angle in the second scenario is 48°, and we must determine the refracted angle r for the air.Since we now know that air has a refractive index of 1, so that the refracted angle is,[tex]sin(r)=n_w*sin48=0.988\\r=sin^{-1}(0.988)=81.25 degrees.[/tex]
As a result, we can infer that the material X will have an index of refraction of 1.09 and that the angle the light makes with the normal in the air is 81.25°.
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john has 4 apples , is train is 7 minutes early calculate te mass of the sun
Answer:
The mass of Sun doesn't change with respective to the conditions.
Michael has 4 Apples, which may increase his own mass or weight but not the Sun's .
His train is 7 minutes, but this doesn't mean the Sun has been made to change. The train coming late affects the time management and delays work.
So, As the per the question, It is evident that Sun's Mass is still the same irrespective of conditions .
Hence, The required answer Sun's Mass is 2*10^30 kg
Explanation:
37 POINTS PLEASE HELP MEE. The distance from A to B consists of an uphill section and a downhill section. A cyclist rides from A to B and then returns to A for a period of 4 hours and 30 minutes. When traveling as well as when returning the uphill speed is 15 km / h and the speed at the downhill is 20 km / h. Do you calculate the length of the distance from A to B?
The distance from A to B is 39.375 Km.
What is the total distance travelled by the cyclist?
The total distance travelled by the cyclist is given by the formula below:
Total distance = average speed * total time takenTotal distance = (15 + 20)/2 * 4.5
Total distance = 78.75 Km
Thus, distance from A to B = 78.75/2 = 39.375 Km
In conclusion, the total distance travelled is determined from the average speed and total time taken.
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What do an electron and a neutron have in common?
Answer:
To give light yo people and the neutron are to give you health care
If the mass of an object is 10 kg and the velocity is 8 m/s, what is the momentum?
A. 8 kgm/s
B. 120 kgm/s
C. 80 kgm/s
D. 40 kgm/s
Answer:
80 kgm/s
Explanation:
Momentum = Mass x Velocity
It can be expressed as [tex]\displaystyle{p = mv}[/tex] where p is momentum, m is mass and v is velocity.
We know that mass is 10 kg and velocity is 8 m/s - therefore, substitute the given information in formula:
[tex]\displaystyle{p=10 \ \times \ 8}\\\\\displaystyle{p=80 \ \ kgm/s}[/tex]
Hence, the momentum is 80 kgm/s.
A 269 kg weather balloon is designed to lift a 2910 kg package. What volume should the balloon have after being inflated with helium in order that the total load can be lifted? (Assume standard temperature and pressure, at which the density of air is ρair = 1.29 kg/m3 and the density of helium is ρHe = 0.179 kg/m3.)
The volume of the helium balloon in order to lift the weight is 17,760m³.
To find the answer, we need to know about the buoyant force.
What's the buoyant force?When a lighter object is kept in a higher density medium, it experiences a force along upward by that medium. This is buoyant force.Mathematically, buoyant force= density × volume of the object×gWhat's the volume of helium required to lift the 269kg weather balloon and 2910kg package?To lift the weight, the buoyant force must equal to the weight.If V is the volume of helium, buoyant force= V×0.179×gSo, V×0.179×g = (269+2910)g=> V= 3179/0.179 = 17,760m³
Thus, we can conclude that the volume of the helium balloon in order to lift the weight is 17,760m³.
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If the range of a projectile is and 256√3 m in the maximum height reached is 64 m. calculate the angle of projection
The angle of projection given that the range is 256√3 m and the maximum height reached is 64 m is 30°
Data obtained from the questionRange (R) = 256√3 mMaximum height (H) = 64 mAcceleration due to gravity (g) = 9.8 m/s²Angle of projection (θ) = ?How to determine the angle of projectionR = u²Sine(2θ) / g
256√3 = u²Sine(2θ) / 9.8
Cross multiply
256√3 × 9.8 = u²Sine(2θ)
Divide both sides by Sine(2θ)
u² = 256√3 × 9.8 / Sine(2θ)
H = u²Sine²θ / 2g
64 = [256√3 × 9.8 / Sine(2θ)] × [Sine²θ / 2 × 9.8]
64 = [256√3 / Sine(2θ)] × [Sine²θ / 2]
Recall
Sine²θ = SineθSineθ
Sine2θ = 2SineθCosθ
Thus,
64 = [256√3 / 2SineθCosθ] × [SineθSineθ / 2]
64 = 256√3 × Sineθ / 4Cosθ
Recall
Sineθ / Cosθ = Tanθ
Thus,
64 = 256√3 / 4 × Tanθ
Divide both side by 256√3 / 4
Tanθ = 64 ÷ 256√3 / 4
Tanθ = 0.5774
Take the inverse of Tan
θ = Tan⁻¹ 0.5774
θ = 30°
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An astronaut on a spacewalk 200km above Earth drops a hammer (mass 2kg), which goes into orbit about the Earth (radius 6,400km) How long does it take the hammer to orbit the Earth?
Answer:
See below
Explanation:
T = period of orbit = sqrt ( 4 pi^2 r^3 / (G Me) )
G = gravitational constant 6.6743 x 10^-11 m^3 / (kg-s^2)
Me = mass of earth = 6 x 10^24 kg
r = radius = 6600 km = 6,600,000 m
plug in the values to find T = 5323.75 seconds
(check my math)
When a 2.50 - kg object is hung vertically on a certain light spring described by Hooke’s law, the
spring stretches 2.76 cm. (a) What is the force constant of the spring? (b) If the 2.50 - kg object is
removed, how far will the spring stretch if a 1.25 - kg block is hung on it? (c) How much work must
an external agent do to stretch the same spring 8.00 cm from its unstretched position?
The work done in the spring is calculated to be 2.8 J
What is Hooke's law?Hooke's law states that, the extension of a given material is directly proportional to the applied force as long as the elastic limit is not exceeded . First, we must bear in mind that the material must remain within the elastic limit for us to apply the Hooke's law in solving the problem.
Now;
From Hooke's law;
F = Ke
F = force applied
K = force constant
e = extension
F = W = mg = 2.50 - kg * 9.8 m/s^2 = 24.5 N
K = 24.5 N/ 2.76 * 10^-2
K = 888 N/m
e = F/K
F = W = 1.25 - kg * 9.8 m/s^2 = 12.25 N
e = 12.25 N/ 888 N/m = 0.014 m or 1.4 cm
Work done by an external agent = 1/2 Kx^2
= 0.5 * 888 * (8 * 10^-2)^2
= 2.8 J
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A computer monitor accelerates electrons between two plates and sends them at high speed to form an image on the screen. If the elec- trons gain 4.1 * 10-15 J of kinetic energy as they go from one accelerat- ing plate to the other, what is the voltage between the plates?
The voltage between the plates is 3.9 × [tex]10^-^3[/tex] V.
The work-energy theorem states that the delta in the equation equals the change in kinetic energy plus the change in potential energy. Here, a charge's potential energy is expressed as qV, where V is the position's electric potential. The greater the change in voltage per unit distance, the greater the electric field.
The kinetic energy of the electrons = 4.1 × [tex]10^-^1^5[/tex] J
Charge of the electron = 1.602 × [tex]10^-^1^9[/tex] coulomb
Using,
ΔU = q × ΔV
4.1 × [tex]10^-^1^5[/tex] = 1.602 × [tex]10^-^1^9[/tex] × ΔV
ΔV = 3.9 × [tex]10^-^3[/tex] V
Therefore, the voltage between the plates is 3.9 × [tex]10^-^3[/tex] V.
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In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.430 m (1.41 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 15.00 millibar. The atmospheric pressure outside is 940 millibar.
1. Calculate the force required to pull the two hemispheres apart. [Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
2. Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1450N (i.e., about 326 lbs), what is the minimum number of horses required?
The force required to pull the two hemispheres apart is 4.2×10⁴ N and 29 number of horses are needed to pull these hemispheres apart.
What's the expression of force in terms of pressure?Mathematically, force = pressure/areaTotal area of the two hemispheres = 4π×(0.43)²= 2.3 m²Total pressure on the hemispheres= 15 milibar (directed inward) + 940 milibar (atmospheric pressure) = 955 milibar=955×100 N/m²= 9.55×10⁴ N/m²
Force on the hemispheres= 9.55×10⁴/2.3 = 4.2×10⁴ NWhat's the minimum number of horses required to get 4.2×10⁴ N of force, if each horse can pull with a force of 1450N?No. of horses required to separate the hemispheres = 4.2×10⁴/1450 = 29
Thus, we can conclude that the 29 horses are needed to pull the two hemispheres with a force of 4.2×10⁴ N.
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