Answer:
a. 78 degree
Explanation:
According to Snell's Law, we have:
(ni)(Sin θi) = (nr)(Sin θr)
where,
ni = Refractive index of medium on which light is incident
ni = Refractive index of ethyl alcohol = 1.361
nr = Refractive index of medium from which light is refracted
nr = Refractive index of ethyl alcohol = 1.333
θi = Angle of Incidence
θr = Angle of refraction
So, the Angle of Incidence is know as the Critical Angle (θc), when the refracted angle becomes 90°. This is the case of total internal reflection. That is:
θi = θc
when, θr = 90°
Therefore, Snell's Law becomes:
(1.361)(Sin θc) = (1.333)(Sin 90°)
Sin θc = 1.333/1.361
θc = Sin⁻¹ (0.9794)
θc = 78.35° = 78° (Approximately)
Therefore, correct answer will be:
a. 78 degree
The angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.
From the information given;
the refractive index of the ethyl alcohol [tex]\mathbf{n_1= 1.361}[/tex]the refractive index of the water [tex]\mathbf{n_2 = 1.333}[/tex] the angle of incidence is the critical angle [tex]\theta_i = \theta_c[/tex] the angle of refraction [tex]\theta _r = 90^0[/tex]According to Snell's Law of refraction;
[tex]\mathbf{n_1 sin \theta _c = n_2 sin \theta_r}[/tex]
[tex]\mathbf{1.361 \times sin \theta _c = 1.333 \times sin 90}[/tex]
[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times sin 90}{1.361}}[/tex]
[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times 1}{1.361}}[/tex]
[tex]\mathbf{ \theta _c = sin^{-1} (0.9794)}[/tex]
[tex]\mathbf{ \theta _c =78.35^0}[/tex]
[tex]\mathbf{ \theta _c \simeq78^0}[/tex]
Therefore, we can conclude that the angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.
Learn more about Snell Law of refraction here:
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The International Space Station is about 90 meters across and about 380 kilometers away. One night t appears to be the same angular size as Jupiter. Jupiter is 143,000 km in size. Use serxa to figure out how far away Jupiter is in AU Note: 1 AU= 1.5 x 10-km
a) 6.0 x 10 Au
b) 4.0 AU
c) 9.1 x 1010 AU
d) 4.0 x 10 AU
Complete Question
The complete question is shown on the first uploaded image
Answer:
The distance is [tex]r_2 = 4 \ AU[/tex]
Explanation:
From the question we are told that
The size of Jupiter is [tex]s_2 = 143,000 \ km[/tex]
The length of the International Space Station is [tex]r_1 = 380\ km[/tex]
The size of the International Space Station is [tex]s_1 = 90 \ m =0.09 \ km[/tex]
The angular size where the same one night and this angular size is mathematically represented as
[tex]\theta = \frac{s}{r}[/tex]
Since [tex]\theta[/tex] is constant
[tex]\frac{s_1}{r_1} = \frac{s_2}{r_2}[/tex]
substituting values
[tex]\frac{0.09}{380} = \frac{143000}{r_2}[/tex]
=> [tex]r_2 = 6.04 * 10^{9} \ km[/tex]
Now we are told to convert to AU and 1 AU [tex]= 1.5 * 10^8 \ km[/tex]
So
[tex]r_2 = \frac{6.04 * 10^8}{1.5*10^{8}}[/tex]
[tex]r_2 = 4 \ AU[/tex]
Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest.
Car A: 500 kg, 10 m/s,
Car B: 2000 kg, 5 m/s,
Car C: 500 kg, 20 m/s,
Car D: 1000 kg, 20 m/s,
Car E: 4000 kg, 5 m/s, and
Car F: 1000 kg, 10 m/s.
(a) Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest.
(b) Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest.
Answer:
a)Car E = Car D > (Car F = Car B = Car C) > Car A
b)Car E = Car D > (Car F = Car B = Car C) > Car A
Explanation:
Car A: mass = 500 kg; speed = 10 m/s
Car B: mass = 2000 kg;speed = 5 m/s
Car C:mass = 500 kg; speed = 20 m/s
Car D: mass = 1000 kg; speed = 20 m/s
Car E:mass = 4000 kg; speed = 5 m/s
Car F: mass = 1000 kg; speed = 10 m/s
Part a) Now we know that momentum of each car is product of mass and velocity , so we will have
CarA:
[tex]P_1 = m \times v\\P_1 = (500)(10)\\P_1 = 5 \times 10^3 kg m/s[/tex]
Car B:
[tex]P_2 = m v\\P_2 = (2000)(5)\\P_2 = 10^4 kg m/s[/tex]
Car C:
[tex]P_3 = m v\\P_3 = (500)(20)\\P_3 = 10^4 kg m/s[/tex]
Car D:
[tex]P_4 = m v\\P_4 = (1000)(20)\\P_4 = 2\times 10^4 kg m/s[/tex]
Car E:
[tex]P_5 = m v\\P_5 = (4000)(5)\\P_5 = 2\times 10^4 kg m/s[/tex]
Car F:
[tex]P_6 = m v\\P_6 = (1000)(10)\\P_6 = 10^4 kg m/s[/tex]
So the momentum is given as ,
Car E = Car D > (Car F = Car B = Car C) > Car A
Part b)Impulse is given as change in momentum so here we can say that final momentum of all the cars will be zero as they all stops and hence the impulse is same as initial momentum of the car
so the order of impulse from largest to least is given as
Car E = Car D > (Car F = Car B = Car C) > Car A
One end of an insulated metal rod is maintained at 100c and the other end is maintained at 0.00 c by an ice–water mixture. The rod has a length of 75.0cm and a cross-sectional area of 1.25cm . The heat conducted by the rod melts a mass of 6.15g of ice in a time of 10.0 min .find the thermal conductivity k of the metal?k=............ W/(m.K)
Answer:
The thermal conductivity of the insulated metal rod is [tex]202.92\,\frac{W}{m\cdot K}[/tex].
Explanation:
This is a situation of one-dimensional thermal conduction of a metal rod in a temperature gradient. The heat transfer rate through the metal rod is calculated by this expression:
[tex]\dot Q = \frac{k_{rod}\cdot A_{c, rod}}{L_{rod}}\cdot \Delta T[/tex]
Where:
[tex]\dot Q[/tex] - Heat transfer due to conduction, measured in watts.
[tex]L_{rod}[/tex] - Length of the metal rod, measured in meters.
[tex]A_{c,rod}[/tex] - Cross section area of the metal rod, measured in meters.
[tex]k_{rod}[/tex] - Thermal conductivity, measured in [tex]\frac{W}{m\cdot K}[/tex].
Let assume that heat conducted to melt some ice was transfered at constant rate, so that definition of power can be translated as:
[tex]\dot Q = \frac{Q}{\Delta t}[/tex]
Where Q is the latent heat required to melt the ice, whose formula is:
[tex]Q = m_{ice}\cdot L_{f}[/tex]
Where:
[tex]m_{ice}[/tex] - Mass of ice, measured in kilograms.
[tex]L_{f}[/tex] - Latent heat of fussion, measured in joules per gram.
The latent heat of fussion of water is equal to [tex]330000\,\frac{J}{g}[/tex]. Hence, the total heat received by the ice is:
[tex]Q = (6.15\,g)\cdot \left(330\,\frac{J}{g} \right)[/tex]
[tex]Q = 2029.5\,J[/tex]
Now, the heat transfer rate is:
[tex]\dot Q = \frac{2029.5\,J}{(10\,min)\cdot \left(60\,\frac{s}{min} \right)}[/tex]
[tex]\dot Q = 3.382\,W[/tex]
Turning to the thermal conduction equation, thermal conductivity is cleared and computed after replacing remaining variables: ([tex]L_{rod} = 0.75\,m[/tex], [tex]A_{c,rod} = 1.25\times 10^{-4}\,m^{2}[/tex], [tex]\Delta T = 100\,K[/tex], [tex]\dot Q = 3.382\,W[/tex])
[tex]\dot Q = \frac{k_{rod}\cdot A_{c, rod}}{L_{rod}}\cdot \Delta T[/tex]
[tex]k_{rod} = \frac{\dot Q \cdot L_{rod}}{A_{c,rod}\cdot \Delta T}[/tex]
[tex]k_{rod} = \frac{(3.382\,W)\cdot (0.75\,m)}{(1.25\times 10^{-4}\,m^{2})\cdot (100\,K)}[/tex]
[tex]k_{rod} = 202.92\,\frac{W}{m\cdot K}[/tex]
The thermal conductivity of the insulated metal rod is [tex]202.92\,\frac{W}{m\cdot K}[/tex].
In each pair, select a substance that is a better heat conductor.
1. copper wire / wood 3. water / iron
2. water / air 4. iron / glass
Answer:
1)copper wire
Explanation:
it is the best electric conductor
I need help physics
Water flows at 0.850 m/s from a hot water heater, through a 450-kPa pressure regulator. The pressure in the pipe supplying an upstairs bathtub 3.70m above the heater is 414-kPa. What's the flow speed in this pipe?
Answer:
The velocity is [tex]v_2= 0.45 \ m/s[/tex]
Explanation:
From the question we are told that
The initial speed of the hot water is [tex]v_1 = 0.85 \ m/s[/tex]
The pressure from the heater [tex]P_1 = 450 \ KPa = 450 *10^{3} \ Pa[/tex]
The height of the hot water before flowing is [tex]h_1 = 0 \ m[/tex]
The height of bathtub above the heater is [tex]h_2 = 3.70 \ m[/tex]
The pressure in the pipe is [tex]P_2 = 414 KPa = 414 *10^{3} \ Pa[/tex]
The density of water is [tex]\rho = 1000 \ kg/m^3[/tex]
Apply Bernoulli equation
[tex]P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2[/tex]
Substituting values
[tex](450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )[/tex]
=> [tex]v_2^2 = \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}[/tex]
=> [tex]v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}[/tex]
=> [tex]v_2= 0.45 \ m/s[/tex]
Which of these charges is experiencing the electric field with the largest magnitude? A 2C charge acted on by a 4 N electric force. A 3C charge acted on by a 5N electric force. A 4C charge acted on by a 6N electric force. A 2C charge acted on by a 6N electric force. A 3C charge acted on by a 3N electric force. A 4C charge acted on by a 2N electric force. All of the above are experiencing electric fields with the same magnitude
Answer:
The highest electric field is experienced by a 2 C charge acted on by a 6 N electric force. Its magnitude is 3 N.
Explanation:
The formula for electric field is given as:
E = F/q
where,
E = Electric field
F = Electric Force
q = Charge Experiencing Force
Now, we apply this formula to all the cases given in question.
A) A 2C charge acted on by a 4 N electric force
F = 4 N
q = 2 C
Therefore,
E = 4 N/2 C = 2 N/C
B) A 3 C charge acted on by a 5 N electric force
F = 5 N
q = 3 C
Therefore,
E = 5 N/3 C = 1.67 N/C
C) A 4 C charge acted on by a 6 N electric force
F = 6 N
q = 4 C
Therefore,
E = 6 N/4 C = 1.5 N/C
D) A 2 C charge acted on by a 6 N electric force
F = 6 N
q = 2 C
Therefore,
E = 6 N/2 C = 3 N/C
E) A 3 C charge acted on by a 3 N electric force
F = 3 N
q = 3 C
Therefore,
E = 3 N/3 C = 1 N/C
F) A 4 C charge acted on by a 2 N electric force
F = 2 N
q = 4 C
Therefore,
E = 2 N/4 C = 0.5 N/C
The highest field is 3 N, which is found in part D.
A 2 C charge acted on by a 6 N electric force
Consider a system of a cliff diver and the earth. The gravitational potential energy of the system decreases by 24,500 J as the diver drops to the water from a height of 44.0 m. Determine her weight in newtons.
Before she jumped from the cliff, her gravitational potential energy was
GPE = (her weight) x (height of the cliff) .
That's exactly the GPE that she lost on the way down to the water. So we can write
24,500 J = (her weight) x (44.0 m)
Divide each side by 44.0 m:
Her weight = 24,500 J / 44 m
Her weight = 556.8 Newtons
(about 125 pounds)
The amount of friction divided by the weight of an object forms a unit less number called the
Answer:
Coefficient of friction.
Explanation:
The amount of friction divided by the weight of an object is equal to the coefficient of friction. It is a dimensional less number. It can be given by :
[tex]F=\mu N[/tex]
N is normal force.
[tex]\mu[/tex] = coefficient of friction
[tex]\mu=\dfrac{F}{N}[/tex]
Now consider a different electromagnetic wave, also described by: Ex(z,t) = Eocos(kz - ω t + φ) In this equation, k = 2π/λ is the wavenumber and ω = 2π f is the angular frequency. In this case, though, assume φ = +30o and Eo = 1 kV/m. What is the value of Ex(z,t) when z/λ = 0.25 and ft = 0.125?
Answer:
Explanation:
Ex(z,t) = Eocos(kz - ω t + φ)
k = 2π/λ , ω = 2π f
φ = +30° , E₀ = 10³ V .
z/λ = 0.25 , ft = 0.125
Ex(z,t) = Eocos(2πz/λ - 2πf t + φ)
Putting the values given above
Ex(z,t) = 10³ cos ( 2π / 4 - 2π x .125 + 30⁰ )
= 1000cos (90⁰ - 45+30)
= 1000 cos 75
=258.8 V .
Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which would allow astronauts to experience a sort of artificial gravity when walking along the inner wall of the station's outer rim. (a) Imagine one such station with a diameter of 104 m, where the apparent gravity is 2.20 m/s2 at the outer rim. How fast is the station rotating in revolutions per minute
Answer:
f = 1.96 revolutions per minute
Explanation:
The formula for the the frequency of revolution of a satellite, to develop an artificial gravity, with the help of centripetal acceleration is given as follows:
f = (1/2π)√(ac/r)
where,
f = frequency of rotation = ?
ac = centripetal acceleration= apparent gravity or artificial gravity = 2.2 m/s²
r = radius of station or satellite = diameter/2 = 104 m/2 = 52 m
Therefore,
f = (1/2π)√[(2.2 m/s²)/(52 m)]
f = (0.032 rev/s)(60 s/min)
f = 1.96 revolutions per minute
Which of the following statements is true of a gas?
It has a fixed volume, but not a fixed shape
It has closely packed molecules
It can change into a liquid by adding heat
It takes the shape and size of a container
Answer:
it takes the shape and size of the container that it is in
Explanation:
Answer:
it takes the shape and size of a container
The first antiparticle, the positron or antielectron, was discovered in 1932. It had been predicted by Paul Dirac in 1928, though the nature of the prediction was not fully understood until the experimental discovery. Today, it is well accepted that all fundamental particles have antiparticles.
Suppose that an electron and a positron collide head-on. Both have kinetic energy of 3.58 MeV and rest energy of 0.511 MeV. They produce two photons, which by conservation of momentum must have equal energy and move in opposite directions. What is the energy Eloton of one of these photons?
Answer:
4.09 MeV
Explanation:
Find the given attachment
"A trooper is moving due south along the freeway at a speed of 28 m/s. At time t = 0, a red car passes the trooper. The red car moves with constant velocity of 40 m/s southward. At the instant the trooper's car is passed, the trooper begins to speed up at a constant rate of 2.9 m/s2. What is the maximum distance ahead of the trooper that is reached by the red car?"
Answer:
24.83 m
Explanation:
Applying the equation of motion;
d = vt + 0.5at^2 ......1
Where;
d = distance
v = velocity
t = time
a = acceleration
For the trooper;
v = 28 m/s
a = 2.9 m/s^2
Substituting into equation 1;
d1 = 28t + 0.5(2.9t^2)
d1 = 28t + 1.45t^2
For the red car;
v = 40 m/s
a = 0
Substituting into equation 1
d2 = 40t
The difference in distance is;
d = d2 - d1
d = 40t - (28t + 1.45t^2)
d = 12t - 1.45t^2
The maximum distance is at d(d)/dt = 0
differentiating d;
d' = 12 - 2.9t = 0
2.9t = 12
t = 12/2.9 = 4.137931034482
t = 4.138 s
Substituting t into function d;
d(max) = 12(4.138) - 1.45(4.138^2)
d(max) = 24.8275862 = 24.83 m
the maximum distance ahead of the trooper that is reached by the red car is 24.83 m
A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 m/s. (a) Determine the magnitude of her acceleration, assumed to be constant. (b) If the ramp is inclined at 32.6 ° with respect to the ground, what is the component of her acceleration that is parallel to the ground?
Answer:
a) a = 3.09 m/s²
b) aₓ = 2.60 m/s²
Explanation:
a) The magnitude of her acceleration can be calculated using the following equation:
[tex] V_{f}^{2} = V_{0}^{2} + 2ad [/tex]
Where:
[tex]V_{f}[/tex]: is the final speed = 8.89 m/s
[tex]V_{0}[/tex]: is the initial speed = 0 (since she starts from rest)
a: is the acceleration
d: is the distance = 12.8 m
[tex] a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2} [/tex]
Therefore, the magnitude of her acceleration is 3.09 m/s².
b) The component of her acceleration that is parallel to the ground is given by:
[tex] a_{x} = a*cos(\theta) [/tex]
Where:
θ: is the angle respect to the ground = 32.6 °
[tex] a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2} [/tex]
Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².
I hope it helps you!
A skateboarder, starting from rest, rolls down a 12.8-m ramp the magnitude of the skateboarder's acceleration is approximately 3.07 [tex]m/s^2[/tex], the component of her acceleration that is parallel to the ground is approximately 1.66 [tex]m/s^2[/tex].
(a) The following kinematic equation can be used to calculate the skateboarder's acceleration:
[tex]v^2 = u^2 + 2as[/tex]
[tex](8.89)^2 = (0)^2 + 2a(12.8)[/tex]
78.72 = 25.6a
a = 78.72 / 25.6
a = 3.07 [tex]m/s^2[/tex]
(b) Trigonometry can be used to calculate the part of her acceleration that is parallel to the ground. We are aware that the ramp's angle with the ground is 32.6°.
[tex]a_{parallel }= a * sin(\theta)[/tex]
Plugging in the values:
[tex]a_{parallel[/tex] = 3.07 [tex]m/s^2[/tex]* sin(32.6°)
[tex]a_{parallel[/tex]≈ 1.66 [tex]m/s^2[/tex]
Therefore, the component of her acceleration that is parallel to the ground is approximately 1.66 [tex]m/s^2[/tex].
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Of one of the planets becomes a black hole , what would the escape speed be?
Answer:
If, instead, that rocket was on a planet with the same mass as Earth but half the diameter, the escape velocity would be 15.8 km/s Any object that is smaller than its Schwarzschild radius is a black hole – in other words, anything with an escape velocity greater than the speed of light is a black hole.
Explanation:
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A small block with a mass of 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. 6.34). The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is observed to be 2.80 m/s.
(a) What is the tension in the cord in the original situation when the block has speed v = 0.70 m/s? (b) What is the tension in the cord in the final situation when the block has speed v = 2.80 m/s? (c) How much work was done by the person who pulled on the cord?
Answer:
a) 0.147 N
b) 9.408 N
c) 9.261 N
Explanation:
The tension on the cord is the only force keeping the block in circular motion, thus representing the entirety of its centripetal force [tex]\frac{mv^{2} }{r}[/tex]. Plugging in values for initial and final states and we get answers for a and b. The work done by the person causes the centripetal force to increase, and thus is the difference between the final tension and the initial tension.
In an RC-circuit, a resistance of R=1.0 "Giga Ohms" is connected to an air-filled circular-parallel-plate capacitor of diameter 12.0 mm with a separation distance of 1.0 mm. What is the time constant of the system?
Answer:
[tex]\tau = 1\ ms[/tex]
Explanation:
First we need to find the capacitance of the capacitor.
The capacitance is given by:
[tex]C = \epsilon_0 * area / distance[/tex]
Where [tex]\epsilon_0[/tex] is the air permittivity, which is approximately 8.85 * 10^(-12)
The radius is 12/2 = 6 mm = 0.006 m, so the area of the capacitor is:
[tex]Area = \pi * radius^{2}\\Area = \pi * 0.006^2\\Area = 113.1 * 10^{-6}\ m^2[/tex]
So the capacitance is:
[tex]C = \frac{8.85 * 10^{-12} * 113.1 * 10^{-6}}{0.001}[/tex]
[tex]C = 10^{-12}\ F = 1\ pF[/tex]
The time constant of a rc-circuit is given by:
[tex]\tau = RC[/tex]
So we have that:
[tex]\tau = 10^{9} * 10^{-12} = 10^{-3}\ s = 1\ ms[/tex]
A conducting bar with mass m and length L slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current I in the rails and bar, and a constant, uniform, vertical magnetic field B fills the region between the rails . Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance and electrical resistance :
A. v2m / ILB to yhe right
B. 3v2m /2 ILB to yhe left
C. 5v2m/ 2ILB to the right
D. v2m / 2ILB to the left
Answer:
F = ILB
Explanation:
To find the net force on the conducting bar you take into account the following expression:
[tex]\vec{F}=I( \vec{L}X \vec{B})[/tex]
I: current in the conducting bar
L: length of the bar
B: magnitude of the magnetic field
In this case the direction of the magnetic field and the motion of the bar are perpendicular between them. The direction of the bar is + i, and the magnetic field poits upward + k. The cross product of these vector give us the direction of the net force:
+i X +k = +j
The direction of the force is to the right and its magnitude is F = ILB
On a brisk walk, a person burns about 331 Cal/h. If the brisk walk were done at 3.0 mi/h, how far would a person have to walk
to burn off 1 lb of body fat? (A pound of body fat stores an amount of chemical energy equivalent to 3,500 Cal.)
mi?
Answer:
32mi
Explanation:
If 1lb contains 3,500 Cal
It means the number of hours required to burn 3500cal would be;
3500/331 = 10.57hours
But a brisk walk is 3.0 mi/h,
It means a distance of 3.0 × 10.57 mi would be covered = 31.71 miles
32miles{ approximated to the nearest whole}
Note Distance = speed × time
Michelson and Morley's experiment is widely considered to have been:______
a. a success because it detected a shift in the interference pattern.
b. a failure because it detected a shift in the interference pattern.
c. a success because it did not detect a shift in the interference pattern.
d. a failure because it did not detect a shift in the interference pattern.
e. lacking the necessary precision to determine a shift in the interference pattern.
Answer:
The correct answer is option (c) a success because it did not detect a shift in the interference pattern.
Explanation:
In Michelson and Morley experiment it was considered to be successful.
They both found out that the experiment that was carried out was not a failure since it did not detect any shift in the interference pattern.
With this findings it was widely regarded as correct and precise.
someone please help me with this thanks
A point charge is located at the center of a thin spherical conducting shell of inner and outer radii r1 and r2, respectively. A second thin spherical conducting shell of inner and outer radii R1 and R2, respectively, is concentric with the first shell. The flux is as follows for the different regions of this arrangement. Ф -10.3 103 N-m2/C for
0 for r<2 4:
-36.8 x 10נ N-m2/c
0 for r > R2
36.8 x 10נ N-m2/c
Determine the magnitude ond sign of the point chorge ond the charge on the surface of the two shels point charge inner shell outer shel.
Answer:
the magnitude is 7 and sign of the point charge on the surface shell is -13
Explanation:
please help
Complete the first and second sentences, choosing the correct answer from the given ones.
1. A temperature of 100 K corresponds on a Celsius scale to 100 ° C / 0 ° C / 173 ° C / –173 ° C.
2. At 50 ° C, it corresponds to a Kelvin scale of 150 K / 323 K / 273 K / 223 K.
1) 100 ° C
2) 323 K
hope it helps youuuuuu
This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
Answer:
rod end A is strongly attracted towards the balls
rod end B is weakly repelled by the ball as it is at a greater distance
Explanation:
When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.
Therefore rod end A is strongly attracted towards the balls and
rod end B is weakly repelled by the ball as it is at a greater distance
an object's resistance to any change in motion is the_________ of the object.
An object's resistance to any change in motion is the Inertia of the object.
What is the highest point at which weather will generally occur?
Answer:
At thestratosphere: it 20- 25km
a vector has components x=6 m and y=8 m. what is its magnitude and direction?
Answer: 10m
Explanation:
The magnitude of the vector would be 10
[tex]\sqrt{6^{2}+8^{2} } =10[/tex]
A) In the figure below, a cylinder is compressed by means of a wedge against an elastic constant spring = 12 /. If = 500 , determine what the minimum compression in the spring will be so that the pad does not move. Disregard the weight of the blocks and . The coefficient of friction between and the pad and between the floor and the pad is s = 0.4. Consider that the friction between the cylinder and the vertical walls is negligible
Answer: 4.08 cm.
B) Determine the lowest force required to lift the weight of 750 . The static coefficient of friction between and and between and is s= 0.25, and between and is 's = 0.5. Disregard the weight of the shims and .
Answer : 1095.4 N.
Explanation:
A) Draw free body diagrams of both blocks.
Force P is pushing right on block A, which will cause it to move right along the incline. Therefore, friction forces will oppose the motion and point to the left.
There are 5 forces acting on block A:
Applied force P pushing to the right,
Normal force N pushing up and left 10° from the vertical,
Friction force Nμ pushing down and left 10° from the horizontal,
Reaction force Fab pushing down,
and friction force Fab μ pushing left.
There are 2 forces acting on block B:
Reaction force Fab pushing up,
And elastic force kx pushing down.
(There are also horizontal forces on B, but I am ignoring them.)
Sum of forces on A in the x direction:
∑F = ma
P − N sin 10° − Nμ cos 10° − Fab μ = 0
Solve for N:
P − Fab μ = N sin 10° + Nμ cos 10°
P − Fab μ = N (sin 10° + μ cos 10°)
N = (P − Fab μ) / (sin 10° + μ cos 10°)
Sum of forces on A in the y direction:
N cos 10° − Nμ sin 10° − Fab = 0
Solve for N:
N cos 10° − Nμ sin 10° = Fab
N (cos 10° − μ sin 10°) = Fab
N = Fab / (cos 10° − μ sin 10°)
Set the expressions equal:
(P − Fab μ) / (sin 10° + μ cos 10°) = Fab / (cos 10° − μ sin 10°)
Cross multiply:
(P − Fab μ) (cos 10° − μ sin 10°) = Fab (sin 10° + μ cos 10°)
Distribute and solve for Fab:
P (cos 10° − μ sin 10°) − Fab (μ cos 10° − μ² sin 10°) = Fab (sin 10° + μ cos 10°)
P (cos 10° − μ sin 10°) = Fab (sin 10° + 2μ cos 10° − μ² sin 10°)
Fab = P (cos 10° − μ sin 10°) / (sin 10° + 2μ cos 10° − μ² sin 10°)
Sum of forces on B in the y direction:
∑F = ma
Fab − kx = 0
kx = Fab
x = Fab / k
x = P (cos 10° − μ sin 10°) / (k (sin 10° + 2μ cos 10° − μ² sin 10°))
Plug in values and solve.
x = 500 N (cos 10° − 0.4 sin 10°) / (12000 (sin 10° + 0.8 cos 10° − 0.16 sin 10°))
x = 0.0408 m
x = 4.08 cm
B) Draw free body diagrams of both blocks.
Force P is pushing block A to the right relative to the ground C, so friction force points to the left.
Block A moves right relative to block B, so friction force on A will point left. Block B moves left relative to block A, so friction force on B will point right (opposite and equal).
Block B moves up relative to the wall D, so friction force on B will point down.
There are 5 forces acting on block A:
Applied force P pushing to the right,
Normal force Fc pushing up,
Friction force Fc μ₁ pushing left,
Reaction force Fab pushing down and left 15° from the vertical,
and friction force Fab μ₂ pushing up and left 15° from the horizontal.
There are 5 forces acting on block B:
Weight force 750 n pushing down,
Normal force Fd pushing left,
Friction force Fd μ₁ pushing down,
Reaction force Fab pushing up and right 15° from the vertical,
and friction force Fab μ₂ pushing down and right 15° from the horizontal.
Sum of forces on B in the x direction:
∑F = ma
Fab μ₂ cos 15° + Fab sin 10° − Fd = 0
Fd = Fab μ₂ cos 15° + Fab sin 15°
Sum of forces on B in the y direction:
∑F = ma
-Fab μ₂ sin 15° + Fab cos 10° − 750 − Fd μ₁ = 0
Fd μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750
Substitute:
(Fab μ₂ cos 15° + Fab sin 15°) μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750
Fab μ₁ μ₂ cos 15° + Fab μ₁ sin 15° = -Fab μ₂ sin 15° + Fab cos 15° − 750
Fab (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°) = -750
Fab = -750 / (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°)
Sum of forces on A in the y direction:
∑F = ma
Fc + Fab μ₂ sin 15° − Fab cos 15° = 0
Fc = Fab cos 15° − Fab μ₂ sin 15°
Sum of forces on A in the x direction:
∑F = ma
P − Fab sin 15° − Fab μ₂ cos 15° − Fc μ₁ = 0
P = Fab sin 15° + Fab μ₂ cos 15° + Fc μ₁
Substitute:
P = Fab sin 15° + Fab μ₂ cos 15° + (Fab cos 15° − Fab μ₂ sin 15°) μ₁
P = Fab sin 15° + Fab μ₂ cos 15° + Fab μ₁ cos 15° − Fab μ₁ μ₂ sin 15°
P = Fab (sin 15° + (μ₁ + μ₂) cos 15° − μ₁ μ₂ sin 15°)
First, find Fab using the given values.
Fab = -750 / (0.25 × 0.5 cos 15° + 0.25 sin 15° + 0.5 sin 15° − cos 15°)
Fab = 1151.9 N
Now, find P.
P = 1151.9 N (sin 15° + (0.25 + 0.5) cos 15° − 0.25 × 0.5 sin 15°)
P = 1095.4 N
PLS HELP,WILL GIVE BRAINLIEST + 30 POINTS
Describe how fractional distillation and cracking are used so that sufficient petrol is produced from crude oil to meet demand.
Answer:
☟
Explanation:
Fuels made from oil mixtures containing large hydrocarbon molecules are not efficient as they do not flow easily and are difficult to ignite. Crude oil often contains too many large hydrocarbon molecules and not enough small hydrocarbon molecules to meet demand. This is where cracking comes in.
Cracking allows large hydrocarbon molecules to be broken down into smaller, more useful hydrocarbon molecules. Fractions containing large hydrocarbon molecules are heated to vaporise them. They are then either:
heated to 600-700°C
passed over a catalyst of silica or alumina
These processes break covalent bonds in the molecules, causing thermal decompositionreactions. Cracking produces smaller alkanesand alkenes (hydrocarbons that contain carbon-carbon double bonds). For example:
hexane → butane + ethene
C6H14 → C4H10 + C2H4
Some of the smaller hydrocarbons formed by cracking are used as fuels, and the alkenes are used to make polymers in plastics manufacture. Sometimes, hydrogen is also produced during cracking.
Fractional distillation of crude oil
Fractional distillation separates a mixture into a number of different parts, called fractions.
A tall fractionating column is fitted above the mixture, with several condensers coming off at different heights. The column is hot at the bottom and cool at the top. Substances with high boiling points condense at the bottom and substances with lower boiling points condense on the way to the top.
Crude oil is a mixture of hydrocarbons. The crude oil is evaporated and its vapours condense at different temperatures in the fractionating column. Each fraction contains hydrocarbon molecules with a similar number of carbon atoms and a similar range of boiling points.
Oil fractions
The diagram below summarises the main fractions from crude oil and their uses, and the trends in properties. Note that the gases leave at the top of the column, the liquids condense in the middle and the solids stay at the bottom.
As you go up the fractionating column, the hydrocarbons have:
lower boiling points
lower viscosity (they flow more easily)
higher flammability (they ignite more easily).
Other fossil fuels
Crude oil is not the only fossil fuel.
Natural gas mainly consists of methane. It is used in domestic boilers, cookers and Bunsen burners, as well as in some power stations.
Coal was formed from the remains of ancient forests. It can be burned in power stations. Coal is mainly carbon but it may also contain sulfur compounds, which produce sulfur dioxide when the coal is burned. This gas is a cause of acid rain. Also, as all fossil fuels contain carbon, the burning of any fossil fuel will contribute to global warming due to the production of carbon dioxide.
In fractional distillation, the crude oil is added to the chamber and heated. The components with the highest boiling point will condense in the lower part of the column and the components with the lower boiling point will condense at the top of the column. Petrol with a low boiling point is collected from the top of the column.
What is fractional distillation?Fractional distillation can be described as the separation of a mixture into its component fractions. The chemical compound is separated by heating them to a temperature at which fractions of the mixture will vaporize.
Generally, the components have boiling points that differ by less than 25 °C from each other under one atmosphere. When the mixture is heated, the component with the lower boiling point boils and changes to vapours.
The more volatile component remains in a vapour state and repeated distillations are used in the process, and the mixture is separated into component parts.
Therefore, the petrol from the crude oil can easily be separated as it has a boiling point of about 25-60°C.
Learn more about fractional distillation, here:
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