Answer:
The y-value is z = 0.759 m
Explanation:
From the question we are told that
The position of the first y-axis is [tex]y_1 = 0.300 \ m[/tex]
The current on the first wire is [tex]I_ 1 = 26.0 \ A[/tex]
The force per unit length on each wire is [tex]\frac{F}{l} = 295 \mu N/m = 295 * 10^{-6} \ N/m[/tex]
Generally the force per unit length on first wire is mathematically represented as
[tex]\frac{F}{l} = \frac{\mu_o * I_1 * I_2 }{2*\pi* y_1}[/tex]
Where [tex]\mu _o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting values
[tex]295 *10^{-6} = \frac{ 4\pi * 10^{-7} * 26.0 * I_2 }{2 *3.142* 0.300}[/tex]
[tex]I_2 = \frac{295 *10^{-6 } * 0.300 * 2* 3.142 }{ 4\pi * 10^{-7} * 26 }[/tex]
[tex]I_2 = 17.0 \ A[/tex]
Now the at the point where the magnetic field is zero the magnetic field of each wire are equal , let that point by z meters from the second wire on the y-axis so
[tex]\frac{\mu_o I_2}{2 * \pi * y_1} = \frac{\mu_o I_1}{2 * \pi * (y_1-z)}[/tex]
[tex]I_2 (y_1 - z) = I_1 * y_1[/tex]
substituting values
[tex]17.0 ( 0.300 - z) = 26 * 0.300[/tex]
z = 0.759 m
A current carrying loop of wire lies flat on a table top. When viewed from above, the current moves around the loop in a counterclockwise sense.
(a) For points OUTSIDE the loop, the magnetic field caused by this current:________.
a. points straight up.
b. circles the loop in a clockwise direction.
c. circles the loop in a counterclockwise direction.
d. points straight down.
e. is zero.
(b) For points INSIDE the loop, the magnetic field caused by this current:________.
a. circles the loop in a counterclockwise direction.
b. points straight up.
c. points straight down.
d. circles the loop in a clockwise direction.
e. is zero
Answer:
D &B
Explanation:
Using Fleming right hand rule that States that if the fore-finger, middle finger and the thumb of left hand are stretched mutually perpendicular to each other, such that fore-finger points in the direction of magnetic field, the middle finger points in the direction of the motion of positive charge, then the thumb points to the direction of the force
The tune-up specifications of a car call for the spark plugs to be tightened to a torque of 38N⋅m38N⋅m. You plan to tighten the plugs by pulling on the end of a 25-cm-long wrench. Because of the cramped space under the hood, you'll need to put at an angle of 120∘with respect to the wrench shaft. With what force must you pull?
Answer:
F= 175.5N
Explanation:
Given:
Torque which can also be called moment is defined as rotational equivalent of linear force. It is the product of the external force and perpendicular distance
torque of 38N⋅m
angle of 120∘
Torque(τ): 38Nm
position r relative to its axis of rotation: 25cm , if we convert to metre for consistency we have 0.25m
Angle: 120°
To find the Force, the torque equation will be required which is expressed below
τ = Frsinθ
We need to solve for F, if we rearrange the equation, we have the expression below
F= τ/rsinθ
Note: the torque is maximum when the angle is 90 degrees
But θ= 180-120=60
F= 38/0.25( sin(60) )
F= 175.5N
lock of mass m2 is attached to a spring of force constant k and m1 . m2. If the system is released from rest, and the spring is initially not stretched or com- pressed, find an expres- sion for the maximum displacement d of m2
Answer:
The maximum displacement of the mass m₂ [tex]= \frac{2(m_1-m_2)g}{k}[/tex]
Explanation:
Kinetic Energy (K) = 1/2mv²
Potential Energy (P) = mgh
Law of Conservation of energy states that total energy of the system remains constant.
i.e; Total energy before collision = Total energy after collision
This implies that: the gravitational potential energy lost by m₁ must be equal to sum of gravitational energy gained by m₂ and the elastic potential energy stored in the spring.
[tex]m_1gd = m_2gd+\frac{1}{2}kd^2\\\\m_1g = m_2g+\frac{1}{2}kd\\\\d = \frac{2(m_1-m_2)g}{k}[/tex]
d = maximum displacement of the mass m₂
Which of the following is an element? A. Fire B. Carbon C. Salt D. Water
Answer:
OPTION B is correct
Carbon
Explanation:
element can be defined as a pure substance which cannot be broken down by into smaller units through a chemical method, an element has atoms with identical numbers of protons in their atomic nuclei
Each element is composed of its own type of atom. And this gives the reason why chemical elements are all very different from each other. And all substance on Earth has atoms of at least one of this elements.
There about 118 elements and all arranged in a row and colomn of the periodic table .This elements of the periodic table are arranged by their atomic number, which helps with the chemical properties. Example of elements are; Hydrogen, Oxygeñ, carbon.
Therefore, among the option only carbon is an element because it cannot be broken down into smaller unit unlike water which is made up of oxygen and hydrogen. Also salt is a compound containing more elements.
The substance which represents an element given the following option is carbon (option B)
What is an element?An element is a pure substance that consist of identical atoms.
An element can not be broken down into simple substances by ordinary methods.
The period table consist of a large number of elements. Some of which are:
HydrogenHeliumLithiumBerylliumBoronCarbonNitrogenOxygenFluorineNeonWe must also understand that when two or more elements are chemically combined together it is called a compound and when they are not chemically combined together, it is called a mixture.
Thus, we can conclude that the correct answer to the question is Carbon (option B)
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What are the potential obstacles preventing you from completing your exercises as scheduled? How can you overcome those obstacles?
Answer:
Sleep, behavior patterns, mental state, and job
Explanation:
Answer:
If I exercise right after school, I might be low on energy. I suppose that I could eat a snack and drink something before my workout. If I exercise before school, I might be tired. But, as long as I keep getting eight to nine hours of sleep, I think that my body will adjust to the new schedule after a while. The trick will be getting to bed on time and eating a little breakfast before I work out. I’m kind of worried that the gym won’t be open early in the morning. On the weekends, my friends might keep me from exercising. I suppose I can try to get them to do it with me. We can pick things that we all like to do. I know they like to play tennis sometimes.
Explanation:
A dentist using a dental drill brings it from rest to maximum operating speed of 391,000 rpm in 2.8 s. Assume that the drill accelerates at a constant rate during this time.
(a) What is the angular acceleration of the drill in rev/s2?
rev/s2
(b) Find the number of revolutions the drill bit makes during the 2.8 s time interval.
rev
Answer:
a
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
b
[tex]\theta = 9124.5 \ rev[/tex]
Explanation:
From the question we are told that
The maximum angular speed is [tex]w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s[/tex]
The time taken is [tex]t = 2.8 \ s[/tex]
The minimum angular speed is [tex]w_{min}= 0 \ rad/s[/tex] this is because it started from rest
Apply the first equation of motion to solve for acceleration we have that
[tex]w_{max} = w_{mini} + \alpha * t[/tex]
=> [tex]\alpha = \frac{ w_{max}}{t}[/tex]
substituting values
[tex]\alpha = \frac{40950.73}{2.8}[/tex]
[tex]\alpha = 14625 .3 \ rad/s^2[/tex]
converting to [tex]rev/s^2[/tex]
We have
[tex]\alpha = 14625 .3 * 0.159155 \ rev/s^2[/tex]
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
According to the first equation of motion the angular displacement is mathematically represented as
[tex]\theta = w_{min} * t + \frac{1}{2} * \alpha * t^2[/tex]
substituting values
[tex]\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2[/tex]
[tex]\theta = 57331.2 \ radian[/tex]
converting to revolutions
[tex]revolution = 57331.2 * 0.159155[/tex]
[tex]\theta = 9124.5 \ rev[/tex]
Two identical wooden barrels are fitted with long pipes extending out their tops. The pipe on the first barrel is 1 foot in diameter, and the pipe on the second barrel is only 1/2 inch in diameter. When the larger pipe is filled with water to a height of 20 feet, the barrel bursts. To burst the second barrel, will water have to be added to a height less than, equal to, or greater than 20 feet? Explain.
Answer:
The 1/2 inch barrel will burst at the same height of 20 ft
Explanation:
The pressure on a column of fluid increases with depth, and decreases with height. This means that if you increase the height of the fluid in the column, the pressure at the bottom will increase.
From the equation of fluid pressure,
P = ρgh
where
P is the pressure at the bottom of the fluid due to its height
ρ is the density of the fluid in question
h is the height to which the water stand.
You notice how apart from the height 'h' in the equation, all the other parts of the right hand side of the equation cannot be varied; they are a fixed property of the fluid and gravity. And there is no consideration for the horizontal diameter of the water's cross section area.
We can also think of the pressure at the bottom of the fluid to be as a result of an incremental weight of an infinitesimally small vertical section of the water down.
That been said, we can then say that if the barrel with the 1 ft diameter dimension bursts when filled with water up to 20 ft, then, the barrel with the reduced diameter will still burst at the same height as the former pipe.
NB: The only way to stop the pipe from bursting is to increase the thickness of the barrel wall to counteract the pressure forces due to the height.
An air-filled parallel-plate capacitor has plates of area 2.30 cm2 2 separated by 1.50 mm. The capacitor is connected to a 12.0-V battery. Find the value of its capacitance.
Answer:
[tex]1.357\times 10^{-12}[/tex]
Explanation:
Relevant Data provided
Area which indicates A = 2.3 cm^2 = 2.3 x 10^-4 m^2
Distance which indicates d = 1.50 x 10^-3 m
Voltage which indicates V = 12 V
According to the requirement, the computation of value of its capacitance is shown below:-
[tex]Capacitance, C = \frac{\epsilon oA}{D}[/tex]
[tex]= \frac{= 8.854\times 10^{-12}\times 2.3\times 10^{-4}}{(1.5 \times 10^{-3})}[/tex]
= [tex]1.357\times 10^{-12}[/tex]
Therefore for computing the capacitance we simply applied the above formula.
A brick of mass M has been placed on a rubber cushion of mass m. Together they are sliding to the right at constant velocity on an ice-covered parking lot. (a) Draw a free-body diagram of the brick and identify each force acting on it. (b) Draw a free-body diagram of the cushion and identify each force acting on it. (c) Identify all of the action–reaction pairs of forces in the brick–cushion–planet system.
A) The free-body diagram of the forces acting on the brick is attached.
B) The free-body diagram of the forces acting on the rubber cushion is attached.
C) The action and reaction forces of the entire brick–cushion–planet system has been enumerated below.
A) The brick has a Mass M placed on top of a rubber cushion of mass m.
This means that there will be a normal force acting acting upwards on the brick and also a gravitational force acting downward. These forces are denoted as;
Normal force of rubber cushion acting on brick = [tex]n_{cb}[/tex]
Gravitational force acting on brick = Mg
Find attached the free body diagram.
B) The forces acting on the cushion will be;
Normal force of parking lot pavement on rubber cushion = [tex]n_{pc}[/tex]
Gravitational force of earth acting on cushion = mg
Force of brick acting on the rubber cushion = [tex]F_{bc}[/tex]
C) The action pairs of forces are;
i) Force; Normal force of rubber cushion acting on brick = [tex]n_{cb}[/tex]
Reaction Force; Force of brick acting on the rubber cushion = [tex]F_{bc}[/tex]
ii) Action Force; Gravitational force acting on brick = Mg
Reaction; Gravitational force of brick acting on the earth
iii) Action Force; Normal force of parking lot pavement on rubber cushion = [tex]n_{pc}[/tex]
Reaction; Force of rubber cushion on parking lot pavement
iv) Action Force; Gravitational force of earth acting on rubber cushion = mg
Reaction Force; Gravitational force of rubber cushion on the earth.
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The metal bar on the conducting rails is moving to the left. The magnitude of the uniform magnetic field is 0.80 Tesla. The magnetic field is directed out of the page. The length L is 0.95 m. The resistance R of the circuit is 3.60 Ω. (a) If the induced current needs to be 0.54 A, what should be the induced emf
Answer:
Explanation:
induced emf = ?
induced emf = induced current x resistance of the circuit
= .54 x 3.6
= 1.944 V
induced emf required = 1.944 V .
The current in a series circuit is 15.0 A. When an additional 8.00-% resistor is inserted in series, the current drops to 12.0 A. What is the resistance in the original circuit
Answer:
The resistance of the original circuit is [tex]32\,\,\Omega[/tex]
Explanation:
In the original circuit, we have an unknown resistor that we call R, an unknown power supply that we call V, and the current is 15 Amps. in the second circuit with an added 8 Ω resistor in series, which gives an equivalent resistance of R+8 Ω, using the same power supply V, the current is 12 Amps. SO, we can write a system of two equations with two unknowns as follows:
[tex]V=R\,(15)\\V=(R+8)\,(12)\\then\\15\,R=12\,R+92\\3\,R=96\\R=\frac{96}{3} \,\Omega\\R=32\,\,\Omega[/tex]
A cube of metal has a mass of 11 grams and a volume of 1 cm . When fully submerged in water this metal cube hanging from an accurate spring scale will weigh what amount?
Answer:
0.098 N
Explanation:
From the question,
Spring scale reading = W-U............... Equation 1
Where W = weight of the cube, U = upthrust.
W = mg
Where m = mass of the cube, g = acceleration due to gravity.
Given: m = 11 g = 0.011 kg, g = 9.8 m/s².
W = 0.011(9.8)
W = 0.1078 N.
From Archimedes principle,
Upthrust = weight of water displaced.
U = (Density of water×volume of metal cube)×acceleration due to gravity.
U = (D×V)g
Given: D = 1000 kg/m², V = 1 cm³ = (1/1000000) = 1×10⁻⁶ m³, g - 9.8 m/s²
U = 1000(9.8)(10⁻⁶)
U = 0.0098 N.
Substitute the value of W and U into equation 1
Reading of the spring scale = 0.1078-0.0098
Reading of the spring scale = 0.098 N
When separated by distance d, identically charged point-like objects A and B exert a force of magnitude F on each other. If you reduce the charge of A to one-half its original value, and the charge of B to one-tenth, and reduce the distance between the objects by half, what will be the new force that they exert on each other in terms of force F
Answer:
F = F₀ 0.2
Explanation:
For this exercise we apply Coulomb's law with the initial data
F₀ = k q_A q_B / d²
indicate several changes
q_A ’= ½ q_A
q_B ’= 1/10 q_B
d ’= ½ d
let's substitute these new values in the Coulomb equation
F = k q_A ’q_B’ / d’²
F = k ½ q_A 1/10 q_B / (1/2 d)²
F = (k q_A q_B / d2) ½ 1/10 2²
F = F₀ 0.2
The core of an optical fiber has an index of refraction of 1.35 , while the index of refraction of the cladding surrounding the core is 1.21 . What is the critical angle θc for total internal reflection at the core‑cladding interface?
Answer:
The critical angle is [tex]\theta_c = \ 63.68^o[/tex]
Explanation:
From the question we are told that
The refractive index of the core is [tex]n_c = 1.35[/tex]
The refractive index of the cladding is [tex]n_s = 1.21[/tex]
Generally according to Snell's law
[tex]\frac{sin i }{sin r } = \frac{n_s}{n_c }[/tex]
Here for total internal reflection the refractive angle is [tex]r = 90^o[/tex] and the critical angle is equal to the critical angle so [tex]i = \theta_c[/tex]
[tex]\frac{sin \theta_c }{sin (90) } = \frac{n_s}{n_c }[/tex]
substituting values
[tex]\frac{sin \theta_c }{sin (90) } = \frac{1.21}{1.35 }[/tex]
[tex]\theta_c = sin^{-1} [\frac{1.21}{1.35} ][/tex]
[tex]\theta_c = \ 63.68^o[/tex]
The operator of a space station observes a space vehicle approaching at a constant speed v. The operator sends a light signal at speed c toward the space vehicle. What is the speed of the light signal as viewed from the space vehicle
Answer:
The speed of the light signal as viewed from the observer is c.
Explanation:
Recall the basic postulate of the theory of relativity that the speed of light is the same in ALL inertial frames. Based on this, the speed of light is independent of the motion of the observer.
In the circuit shown, the galvanometer shows zero current. The value of resistance R is :
A) 1 W
B) 2 W
C) 4 W
D) 9 W
Answer:
its supposed to be (a) 1W
A 46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this causeway?
Answer:
2.9tons
Explanation:
Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:
parallel ("tangential"): F_t = F sin a
perpendicular ("normal"): F_n = F cos a
At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:
F sin a ~~ (46 tons)*0.064 ~~ 2.9tons
Also see attached file
The required force parallel to the incline to hold the monolith on this causeway will be "2.9 tons".
Angle and ForceAccording to the question,
Angle, a = 3.7 degrees or,
Sin a = 0.064
Force, F = 46 tons
We know the relation,
Parallel (tangential), [tex]F_t[/tex] = F Sin a
By substituting the values,
= 46 × 0.064
= 2.9 tons
Thus the response above is appropriate answer.
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Air at 1 atm, 158C, and 60 percent relative humidity is first heated to 208C in a heating section and then humidified by introducing water vapor. The air leaves the humidifying section at 258C and 65 percent relative humidity. Determine (a) the amount of steam added to the air, and (b) the amount of heat transfer to the air in the heating section.
Answer:
A. the amount of steam added to the air s 0.065kgH20/kg dry air
B.the amount of heat transfer to the air in the heating section is 5.1KJ/Kg of dry air
Explanation:
Pls see attached file
Answer: right side B
Explanation:
Four equal masses m are located at the corners of a square of side L, connected by essentially massless rods. Find the rotational inertia of this system about an axis (a) that coincides with one side and (b) that bisects two opposite sides.
Answer:
Explanation:
a )
Moment of inertial of four masses about axis that coincides with one side :
Out of four masses . location of two masses will lie on the axis so their moment of inertia will be zero .
Moment of inertia of the two remaining masses
= m L² + m L²
= 2 mL²
b )
Axis that bisects two opposite sides
Each of the four masses will lie at a distance of L / 2 from this axis so moment of inertia of the four masses
= 4 x m x ( L/2 )²
= 4 x mL² / 4
= m L² .
Two protons are released from rest, with only the electrostatic force acting. Which of the following statements must be true about them as they move apart? (There could be more than one correct choice.)
A. Their electric potential energy keeps decreasing.
B. Their acceleration keeps decreasing.
C. Their kinetic energy keeps increasing.
D. Their kinetic energy keeps decreasing.
E. Their electric potential energy keeps increasing.
Answer:
(A)
Explanation:
We know , electric potential energy between two charge particles of charges "q" and "Q" respectively is given by kqQ/r where r is the distance between them.
Since the two charged particles are moving apart, the distance between them (r) increases and thus electrical potential energy decreases.
A valuable statuette from a Greek shipwreck lies at the bottom of the Mediterranean Sea. The statuette has a mass of 10,566 g and a volume of 4,064 cm3. The density of seawater is 1.03 g/mL.
a. What is the weight of the statuette?
b. What is the mass of displaced water?
c. What is the weight of displaced water?
d. What is the buoyant force on the statuette?
e. What is the net force on the statuette?
f. How much force would be required to lift the statuette?
Answer:
A) W = 103.55 N
B) mass of displaced water = 4186 g
C) W_displaced water = 41.06 N
D) Buoyant force = 41.06 N.
E) ZERO
F) 62.54 N
Explanation:
We are given;
mass of statuette;m = 10,566 g = 10.566 kg
volume = 4,064 cm³
Density of seawater;ρ = 1.03 g/mL = 1.03 g/cm³
A) The dry weight of the statuette can be calculated as;
W = mg
So;
W = 10.556 × 9.81
W = 103.55 N
B) Mass of displaced water is calculated from;
Density = mass/volume
So, mass = Density × Volume
m = 1.03 × 4,064 = 4186 g
C) Weight of displaced water is given by;
W_displaced water = (m_displaced water) × g
W_displaced water = 4.186 kg × 9.81 m/s^2 = 41.06 N
D) The buoyant force is the same as the weight of the displaced water.
Thus, Buoyant force = 41.06 N.
E) The apparent weight of the statuette is calculated from;
Apparent weight = Dry weight - Weight of displaced water
Apparent weight = 103.6 N - 41.06 N = 62.54 N. It is sitting on the bottom of the sea, so the sea floor is providing an opposite force that is equal but opposite the weight so that the net force on the statuette is zero. Since It has zero acceleration, in any direction, hence the net force on it is zero.
F. From E above, The Force required to lift the statuette = 62.54 N
Consider a block of mass equal to 10kg sliding on an inclined plane of 30°, as shown in the figure below. The coefficient of kinetic friction between the block and the plane surface is c = 0.4 (a) Determine the value of the horizontal and vertical acceleration of the block. (b) If the block starts from rest in t=0s and when it is in the X=0 and Y=5m position, calculate what its horizontal and vertical position will be at the instant t=1s. (C) How long does the LM block take to reach the base of the tilted plane?
Answer:
(a) aₓ = 1.33 m/s² and aᵧ = -0.770 m/s²
(b) x = 0.665 m and y = 4.62 m
(c) 3.61 s
Explanation:
(a) There are two ways we can solve this. The first way is to sum the forces in the x and y direction, then use the relation tan 30° = -aᵧ/aₓ, where aᵧ is the acceleration in the +y direction (up) and aₓ is the acceleration in the +x direction (right).
The second way is to sum the forces in the parallel and perpendicular directions to find the acceleration parallel to the incline, a. Then, use the relations aᵧ = -a sin 30° and aₓ = a cos 30°.
Let's try the first method. Sum of forces in the +y direction:
∑F = ma
N cos 30° + Nμ sin 30° − mg = maᵧ
N cos 30° + Nμ sin 30° − mg = -maₓ tan 30°
Sum of forces in the +x direction:
∑F = ma
N sin 30° − Nμ cos 30° = maₓ
Substituting:
N cos 30° + Nμ sin 30° − mg = -(N sin 30° − Nμ cos 30°) tan 30°
N cos 30° + Nμ sin 30° − mg = -N sin 30° tan 30° + Nμ sin 30°
N cos 30° − mg = -N sin 30° tan 30°
N (cos 30° + sin 30° tan 30°) = mg
N = mg / (cos 30° + sin 30° tan 30°)
N = (10 kg) (10 m/s²) / (cos 30° + sin 30° tan 30°)
N = 86.6 N
Now, solving for the accelerations:
N sin 30° − Nμ cos 30° = maₓ
aₓ = N (sin 30° − μ cos 30°) / m
aₓ = (86.6 N) (sin 30° − 0.4 cos 30°) / 10 kg
aₓ = 1.33 m/s²
N cos 30° + Nμ sin 30° − mg = maᵧ
aᵧ = N (cos 30° + μ sin 30°) / m − g
aᵧ = (86.6 N) (cos 30° + 0.4 sin 30°) / 10 kg − 10 m/s²
aᵧ = -0.770 m/s²
Now let's try the second method.
Sum of forces in the perpendicular direction:
∑F = ma
N − mg cos 30° = 0
N = mg cos 30°
Sum of forces in the parallel direction:
∑F = ma
mg sin 30° − Nμ = ma
mg sin 30° − mgμ cos 30° = ma
a = g (sin 30° − μ cos 30°)
a = (10 m/s²) (sin 30° − 0.4 cos 30°)
a = 1.536 m/s²
Solving for the accelerations:
aₓ = a cos 30°
aₓ = 1.33 m/s²
aᵧ = -a sin 30°
aᵧ = -0.770 m/s²
As you can see, the second method is faster and easier, but both methods will give you the same answer.
(b) In the x direction:
Given:
x₀ = 0 m
v₀ = 0 m/s
aₓ = 1.33 m/s²
t = 1 s
Find: x
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (1 s) + ½ (1.33 m/s²) (1 s)²
x = 0.665 m
In the y direction:
Given:
y₀ = 5 m
v₀ = 0 m/s
aᵧ = -0.770 m/s²
t = 1 s
Find: y
y = y₀ + v₀ t + ½ at²
y = 5 m + (0 m/s) (1 s) + ½ (-0.770 m/s²) (1 s)²
y = 4.62 m
(c) In the y direction:
Given:
y₀ = 5 m
y = 0 m
v₀ = 0 m/s
aᵧ = -0.770 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
0 m = 5 m + (0 m/s) t + ½ (-0.770 m/s²) t²
t = 3.61 s
Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light by total internal reflection at the critical angle for the interface between the core (ncore = 1.482 ) and the cladding (ncladding = 1.44).
Suppose you wanted the largest angle at which total internal reflection occurred to be θmax = 5 degrees. What index of refraction does the cladding need if the core is unchanged?
Answer:
n_cladding = 1.4764
Explanation:
We are told that θ_max = 5 °
Thus;
θ_max + θ_c = 90°
θ_c = 90° - θ_max
θ_c = 90° - 5°
θ_c = 85°
Now, critical angle is given by;
θ_c = sin^(-1) (n_cladding/n_core)
sin θ_c = (n_cladding/n_core)
n_cladding = (n_core) × sin θ_c
Plugging in the relevant values, we have;
n_cladding = 1.482 × sin 85
n_cladding = 1.4764
A turntable A is built into a stage for use in a theatrical production. It is observed during a rehearsal that a trunk B starts to slide on the turntable 15 s after the turntable begins to rotate. Knowing that the trunk undergoes a constant tangential acceleration of 0.3 m/s^2 , determine the coefficient of static friction between the trunk and the turntable
Answer:
μ = 0.03
Explanation:
In order for the trunk not to slide the frictional force between the turntable and the trunk must be equal to the unbalanced force applied on the trunk by the motion of the turntable. Therefore,
Unbalanced Force = Frictional Force
but,
Unbalanced Force = ma (Newton's second law of motion)
Frictional Force = μN = μW = μmg
Therefore,
ma = μmg
a = μg
μ = a/g
where,
μ = coefficient of static friction between the trunk and the turntable = ?
a = tangential acceleration of trunk = 0.3 m/s²
g = 9.8 m/s²
Therefore,
μ = (0.3 m/s²)/(9.8 m/s²)
μ = 0.03
As light shines from air to another medium, i = 26.0 º. The light bends toward the normal and refracts at 32.0 º. What is the index of refraction? A. 1.06 B. 0.944 C. 0.827 D. 1.21
Explanation:
It is given that,
Angle of incidence from air to another medium, i = 26°
The angle of reflection, r = 32°
We need to find the refractive index of the medium. The ratio of sine of angle of incidence to the sine of angle of reflection is called refractive index. It can be given by :
[tex]n=\dfrac{\sin i}{\sin r}\\\\n=\dfrac{\sin (26)}{\sin (32)}\\\\n=0.82[/tex]
So, the index of refraction is 0.82. Hence, the correct option is C.
Two protons are released from rest, with only the electrostatic force acting. Which of the following statements must be true about them as they move apart? (There could be more than one correct choice.)a. Their electrical potential energy keeps decreasing.b. Their acceleration keeps decreasing.c. Their kinetic energy keeps increasing.d. Their kinetic energy keeps decreasing.e. Their electric potential energy keeps increasing.
Answer:
Explanation:
correct options
a ) Their electrical potential energy keeps decreasing
Actually as they move apart , their electrical potential energy decreases due to increase of distance between them and kinetic energy increases
so a ) option is correct
b ) Their acceleration keeps decreasing
As they move apart , their mutual force of repulsion decreases due to increase of distance between them so the acceleration decreases .
c ) c. Their kinetic energy keeps increasing
Their kinetic energy increases because their electrical potential energy decreases . Conservation of energy law will apply .
The moving apart should be true statements:
a. The electrical potential energy should be reduced.
b. The acceleration should be reduced.
c. The kinetic energy should be increased.
True statements related to moving apart:At the time when the moving part, there is the reduction of the electric potential energy because there is a rise in the distance due to which the increment of the kinetic energy. The reduction of the mutual force of repulsion because of increment in the distance due to this the acceleration should be reduced. There is the increase in the kinetic energy due to the reduction of the electrical potential energy. here the law of conversation of energy should be applied.
Learn more about energy here: https://brainly.com/question/10658188
Two metal sphere each of radius 2.0 cm, have a center-to-center separation of 3.30 m. Sphere 1 has a chrage of +1.10 10^-8 C. Sphere 2 has charge of -3.60 10^-8C. Assume that the separation is large enough for us to assume that the charge on each sphere iss uniformly distribuuted.
A) Calculate the potential at the point halfway between the centers.
B) Calculate the potential on the surface of sphere 1.
C) Calculate the potential on the surface of sphere 2.
Answer:
A) V = -136.36 V , B) V = 4.85 10³ V , C) V = 1.62 10⁴ V
Explanation:
To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is
V = k ∑ [tex]q_{i} / r_{i}[/tex]
where [tex]q_{i}[/tex] and [tex]r_{i}[/tex] are the loads and the point distances.
A) We apply this equation to our case
V = k (q₁ / r₁ + q₂ / r₂)
They ask us for the potential at the midpoint of separation
r = 3.30 / 2 = 1.65 m
this distance is much greater than the radius of the spheres
let's calculate
V = 9 10⁹ (1.1 10⁻⁸ / 1.65 + (-3.6 10⁻⁸) / 1.65)
V = 9 10¹ / 1.65 (1.10 - 3.60)
V = -136.36 V
B) The potential at the surface sphere A
r₂ is the distance of sphere B above the surface of sphere A
r₂ = 3.30 -0.02 = 3.28 m
r₁ = 0.02 m
we calculate
V = 9 10⁹ (1.1 10⁻⁸ / 0.02 - 3.6 10⁻⁸ / 3.28)
V = 9 10¹ (55 - 1,098)
V = 4.85 10³ V
C) The potential on the surface of sphere B
r₂ = 0.02 m
r₁ = 3.3 -0.02 = 3.28 m
V = 9 10⁹ (1.10 10⁻⁸ / 3.28 - 3.6 10⁻⁸ / 0.02)
V = 9 10¹ (0.335 - 180)
V = 1.62 10⁴ V
As a wheel turns, the angle through which it has turned varies with time as β(t)=Ct + Bt3 where C=0.400rad/s and B=0.0120rad/s3. Calculate the angular velocity w(t) as a function of time.
Answer:
ω(t) = 0.4 + 0.036 t²
Explanation:
The angular displacement of the disk is given as the function of time:
β(t) = Ct + B t³
where,
C = 0.4 rad/s
B = 0.012 rad/s³
Therefore,
β(t) = 0.4 t + 0.012 t³
Now, for angular velocity ω(t), we must take derivative of angular displacement with respect to t:
ω(t) = dβ/dt = (d/dt)(0.4 t + 0.012 t³)
ω(t) = 0.4 + 0.036 t²
Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If of the light passes through this combination, what is the angle between the transmission axes of the two filters
Answer:
The angle between the transmission axes of the filters is 65°
Explanation:
The complete question is
Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If 18% of the light passes through this combination, what is the angle between the transmission axes of the two filters.
From Malus law,
[tex]I = I_{0} cos^{2} \beta[/tex] ....1
where [tex]I[/tex] is the intensity of the polarized light,
[tex]I_{o}[/tex] is the intensity of the incident light
β the angle between the transmission axes of the two filters
Since the intensity is reduced to 18% or 0.18 of its initial value, this means that
[tex]cos^{2} \beta[/tex] = 0.18
substituting into the equation above, we have
[tex]I = 0.18I_{0}[/tex] ....2
equating the two equations, we have
[tex]I_{0}cos^{2} \beta[/tex] = [tex]0.18I_{0}[/tex]
[tex]cos^{2}\beta[/tex] = [tex]\frac{0.81I_{0} }{I_{0} }[/tex] = 0.18
[tex]cos \beta[/tex] = [tex]\sqrt{0.18}[/tex] = 0.424
[tex]\beta[/tex] = [tex]cos^{-1} 0.424[/tex] = 64.9 ≅ 65°
A car starts from Hither, goes 50 km in a straight line to Yon, immediately turns around, and returns to Hither. The time for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip is:
A. 0
B. 50 km/hr
C. 100 km/hr
D. 200 km/hr
E. cannot be calculated without knowing the acceleration
Answer:
The average velocity for this trip is 0 km/hr
Explanation:
We know that average velocity = total displacement/total time.
Now, its displacement is d = final position - initial position.
Since the car starts and ends at its initial position at Hither, if we assume its initial position is 0 km, then its final position is also 0 km.
So, its displacement is d = 0 km - 0 km = 0 km.
Since the total time for the round trip is 2 hours, the average velocity is
total displacement/ total time = 0 km/2 hr = 0 km/hr.
So the average velocity for this trip is 0 km/hr