Consider (a) an electron (b) a photon, and (c) a proton, all moving in vacuum. Choose all correct answers for each question. (v) Which move at the speed of light?

Answer:

See below

Explanation:

[tex]\displaystyle {\textsf {Acceleration is the rate of change of velocity with time:}}[/tex]

[tex]\boxed {g = \frac{dv}{dt}}[/tex]

[tex]\sf {where \;dv = \;change \;in \;velocity\; and \;dt \;the \;change \;in \;time}[/tex]

here  g is the gravitational force

When a body is thrown vertically upward, it has an initial speed. As it keeps going up, it speed keeps decreasing until it becomes zero, then the ball starts dropping down

On the upward movement, the change in velocity is negative while the change in time is always positive

So dv/dt = a negative number divided by a positive number which is a negative value for g in the equation for g

The magnitude of the force on the test charge is 1.

Greatness is the quantitative worth of seismic energy. It is a particular worth having no connection with distance and heading of the focal point. The idea of extents traces all the way back to the Ancient Greeks, when stars overhead was sorted into six sizes. Extent alludes to the evaluating of the splendor of stars, the first being the most splendid. It has been moved to different issues since basically the seventeenth 100 years. In Physics, greatness is characterized as the most extreme degree of size and the course of an item. Greatness is utilized as a typical consider vector and scalar amounts. By definition, we realize that scalar amounts are those amounts that have size as it were.

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The distance should be 4m from the wire in order to get the magnetic field of 0.100μ .

It is given that magnetic field 40.0 cm away from a straight wire is  1.00μT having  current 2.00 A .

From equation (1)  magnetic field 40.0 cm = 0.4m away from a straight wire is 1.00μT which is given by    [tex]1.00=\frac{u_0I}{2\pi \times0.4}[/tex]      .....(2)

From equation (1)  magnetic field 'r' m away from a straight wire is 0.100μT which is given by    [tex]0.100=\frac{u_0I}{2\pi \times r}[/tex]       ...(3)

On dividing equation (2) by (3) , we get

             [tex]\frac{1}{0.1} =\frac{r}{0.4} \\\\r=4m[/tex]

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While current is the same in series and parallel, voltage is not. Therefore, because these three resistors are connected in parallel, their voltage is the same, or 12 volts. Using the equation V=IR, 12V=I*10Ω gives you a current of 1.2A.

A resistor is a passive two-terminal electrical component used in circuits to implement electrical resistance. Resistors have a variety of purposes in electronic circuits, reducing current flow, modifying signal levels, dividing voltages, biassing active components, and terminating transmission lines are a few examples. High-power resistors that can generate many watts of heat instead of electrical energy can be utilised as test loads for generators, power distribution systems, and motor controls. Fixed resistors' resistances only sporadically vary as a result of changes in operating voltage, time, or temperature. Variable resistors can be utilised as force sensors, heat sensors, light sensors, volume controls, lamp dimmers, humidity sensors, and chemical activity sensors.

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Answer:

An object at rest remains at rest

Explanation:

An object at rest stays at rest and an object in motion stays in motion with the same speed

The average normal strain in wires A, C and D is found to be 0.035, 0.035 and 0.059 respectively when there is a change in the angle by 2° clockwise.

A physical measurement called strain illustrates how much an object's dimensions vary while it is under stress. The ratio between the object's changed length and its initial length is known as the linear strain.

All wires are stretched when the stiff lever rotates 2° in a clockwise direction around pin B.

Elongation would be R x angle in radian

Elongation in the wire A = 200 mm x 3.142 x 2/180 = 6.982 mm

Elongation in the wire C = 300 x 3.142 x 2/180 = 10.473 mm

Elongation in the wire D = 500 x 3.142 x 2/180 = 17.455 mm

Therefore,

Average normal strain in the wire A = 6.982/200 = 0.03491

Average normal strain in the wire C = 10.473/300 = 0.03491

Average normal strain in the wire D = 17.455/300 = 0.05818

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Answer: partially decomposed

Explanation:

The force on the wire will be

F = 2.88 Newton

We have a current carrying wire in a uniform magnetic field.

We have to determine the magnitude of the magnetic force on this section of wire.

The force on the current carrying wire will be -

F = IBL sinθ

According to the question, we have -

Length (L) = 0.75 m (along +x - axis)

Current (I) = 2.4 A (along + x - axis)

Magnetic Field (B) = 0.39 k T (along +z axis)

Therefore, the angle between Magnetic field and Length = θ = 90°.

Therefore -

F = IBL sin(90)

F = 2.4 x 1.6 x 0.75 x sin(90)

F = 2.88 x sin(90)

F = 2.88 Newton

Hence, the force on the wire will be

F = 2.88 Newton

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The total voltage across resistor- three in the circuit is 24V

acc to ohm's law

the electrical current flowing through any circuit is directly proportional to the voltage applied across it.

i.e. V ∝ I

V= IR

I = current

v = voltage

R = resistance (proportionality constant)

R in parallel

(1/Rp = 1/R1 + 1/R2 + 1/R3 +1/R4)  

on rearranging the above equation (V=IR)

I = V/R

I = 24/6 = 4A

I is the current in each resistor.

in the parallel circuit voltage across the component is same and total current is the sum of the current  flowing through each component.

the total voltage across resistor-three (vr3) in the circuit is

V= IR₃ = 4×6= 24 volt

the total voltage across the resistor three (vr3) in the circuit is 24volt.

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mass of one sheet of paper is 0.00504 kg or 5.04g.

First we will find the surface area of a single sheet of paper

surface area = length ×width

surface area =0.3×0.21

surface area =0.063m²

now 0.063m² is the surface area of a single sheet of paper .

so now we will find number of papers required to form 1m² of area

which will be given by dividing 1m² by surface area of a single sheet

i.e.

number of sheets required = 1/0.063m²

number of sheets required =15.87

mass of one sheet = mass of 1 m^2 of paper / number of times narrower

mass of one sheet = 0.080 kg / 15.873

mass of one sheet = 0.00504 kg

mass of one sheet of paper is 0.00504 kg or 5.04g.

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By the charge density, The charge throughout te plastic is 0.059 nC.

We need to know about charge density to solve this problem. The charge density can be determined as

λ = Q / V

where λ is charge density, Q is charge and V is volume.

The parameter given is the charge density and the solid cylinder shape which are :

λ = 500nC/m³

r = 2.5 cm = 0.025 m

L = 6.0 cm = 0.06 m

Find the volume of solid cylinder

V = π .  r² . L

V = π .  0.025² . 0.06

V = 1.18 x 10¯⁴ m³

Find the charges

Q = λ x V

Q = 500 x  1.18 x 10¯⁴

Q = 0.059 nC

Hence, the charge throughout the plastic is 0.059 nC.

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The speed of light u is equal to c.

Given,

The motion of a transparent medium influences the speed of light.

The water moves with speed v in a horizontal pipe.

Assume the light travels in the same direction as the water moves.

The speed of light with respect to the water is c /n.

The index of refraction of water, n = 1.33

Let us assume u' be the speed of light in water.

u' is related to the refractive index of water,as u' = c/n

where, c is the speed of light.

Let, u be the speed of light in water in the lab frame.

Now, u and u' are related as : u = ( u'+v) /(1+u'v/c^2)

Here, v is the speed of water in the horizontal pipe.

We know the value of u'. so by substituting the value, we will get ,

u = (c/n+v) /(1+cv/nc^2)

u = c/n (1+nv/c) /(1+v/nc)

u in the limit as the speed of the water approaches c.

Thus,

u = lim v-->c [c/n (1+nv/c) /(1+v/nc) ]

u= c/n (1+n) /(1+1/n)

u= c .

Hence, the speed of light u is equal to c.

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Disclaimer: incomplete question. here is the complete question.

Question:The motion of a transparent medium influences the speed of light. This effect was first observed by Fizeau in1851. Consider a light beam in water. The water moves with speed v in a horizontal pipe. Assume the light travels in the same direction as the water moves. The speed of light with respect to the water is c / n , where n = 1.33 is the index of refraction of water.

(a) Use the velocity transformation equation to show that the speed of the light measured in the laboratory frame isu = c/n (1 + nv/c / 1+ v/nc).

(d) Evaluate u in the limit as the speed of the water approaches c .

Answer:

3.42 + 4 + 5.2 + 12= 24.62
So since we have 24.62 it will be 4 sig figs

Explanation:

The electric flux is 0 N.m²/C.

We need to know about electric flux to solve this problem. Electric flux is the measure of the electric field through a given surface, although an electric field in itself cannot flow. It can be determined as

Φ = E . S = E . Scosθ

where Φ is electric flux, E is electric field, θ is angle of surface and S is surface area.

From the question above, we know that

E = 6.20 × 10⁵ N/C

θ = 90⁰

S = 3.20 m²

By substituting the following parameter, we get

Φ = E . Scosθ

Φ = 6.20 × 10⁵ . 3.20cos(90⁰)

Φ = 0 N.m²/C

Hence, the electric flux is 0 N.m²/C

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The acceleration of the robot is 1.43 m/s2 in magnitude.

The acceleration of an object is defined as the variation of the velocity with respect to time. The acceleration is calculated by finding the ratio of the change in velocity to the change in time.

The acceleration is also calculated by using the force formula. The force is actually defined as the acceleration produced in the body by applying the force on an object of mass m.

On a non-stick surface, a toy robot and doll are standing side by side. The robot weighs 0. 30 kg, while the doll weighs 0. 20 kg. With a force of 0. 30 n, the robot pushes against the doll.

According to Newton's second law.

F = mass x acceleration

Given that

Force applied = 1N

Acceleration = Force/mass

Substitute the values and get acceleration.

Acceleration = 1/0.7

Acceleration = 1.43m/s²

Hence, the acceleration of the robot is 1.43 m/s2 in magnitude.

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Answer:

he would tap on the "close tab "button and see old the apps he has used in that period

Explanation:

A boat travels 12. 0 m while reduces its velocity from 9. 5 m/s to 5. 5 m/s, while it travels the 12. 0 m, the acceleration of the boat is -2.5 m/s²

The rate at which an object's velocity with respect to the time changes is referred to as acceleration. It is a vector quantity to accelerate (in that they have magnitude and direction). The direction of the net force acting on an object determines the direction of its acceleration. According to Newton's Second Law, the amount of an object's acceleration is the combined result of two causes:

The given parameters;

distance traveled by boat, d = 12 m

initial velocity of boat, u = 9.5 m/s

final velocity of boat, v = 5.5 m/s

The acceleration of boat is calculated as;

[tex]v^{2}=u^{2}+2as\\ a=\frac{v^{2}-u^{2} }{2s}\\ a=\frac{5.5^{2}-9.5^{2} }{2*12}[/tex]

a=-2.5m/[tex]s^{2}[/tex]

Thus, the deceleration of boat is 2.5 m/s².

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The electric field on the axis of the ring at the center of the ring at 100.00cm is 6.64×〖10〗^5N/C

Where a positive charge (q) is given, always note that the charges are pointing away from the electric field. The expression used to calculate the electric field away from the center of the ring is given by the equation below,

E=(K_e×b)/(r^3×Q)

Given that;

E = Electric field

Ke = electric field constant

a = radius of the charge q from the center of the circle

Q = electric charge = 75.0μC

The radius of the charge is a = 10cm= 0.1m

Distance from the electric field is 100cm = 1m

Ke is a constant given as 8.99×〖10〗^(9) Nm^2/c^2

E=(8.99×〖10〗^(9  ) Nm^2/c^2×1)/(〖0.1〗^3×75×〖10〗^(-6) )

E=6.64×〖10〗^5N/C

Therefore, the electric field away from the center of the ring is 6.64×〖10〗^5N/C away from the center of the ring.

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When the variable name has been assigned with the value 33, then the value of the name after the assignment name will be 66.

In the programming language the sign ' = ', is referred to as the assignment operator which is used to assign values to a particular variable, where the variable referred to the name of the memory location. As it is given the assignment name = name * 2, it means that name on the right-hand side is the assignment name, so whatever value it gets is replaced by twice that previous value according to the operation.

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The option that is NOT something that would lead astronomers to believe a black hole might be close-by in space is option C: an object that has height and depth but no width.

There is said to be a wandering black hole that can be seen in an approximately  5,000 light-years away from earth.

It is one that can be seen in the Carina-Sagittarius spiral aspect of our galaxy.

Note that, its discovery gives room for astronomers to state that the nearest form of isolated stellar-mass black hole to Earth can be as close as about 80 light-years away.

Therefore, The option that is NOT something that would lead astronomers to believe a black hole might be close-by in space is option C: an object that has height and depth but no width because it has no width, no depth and no height.

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The average energy density of the solar component of electromagnetic radiation at this location is determined as 2.58 x 10⁻⁶ J.

The average energy density of the solar component of electromagnetic radiation at this location is calculated as follows;

U(avg) = (Brms)²/μ₀

U(avg) = (1.8 x 10⁻⁶)² / (4π x 10⁻⁷)

U(avg) = 2.58 x 10⁻⁶ J

Thus, the average energy density of the solar component of electromagnetic radiation at this location is determined as 2.58 x 10⁻⁶ J.

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Answer: 1) D & 2) H

Explanation:

D. Tropical rainy &

H. Long-term average temperature  and precipitation for the area

The electric flux through the surface is 5.65×〖10〗^2 Nm^2/C ≈5.6×〖10〗^2  Nm^2/C

The electric flux through a surface describes the number of electric field lines that cross that surface. It is expressed as the surface integral of the electric field through that surface.

Gauss's law states that for a closed surface, the electric flux through that surface depends only on the total charge present inside that surface. Hence, for a closed surface, the electric flux can be computed without evaluating any integrals.

The charges are: q1=3.0nC, q2=−2.0nC, q3=−7.0nC, and q4=1.0nC

Gauss's Law provides that the net electric flux through a closed surface depends only on the net charge enclosed by the surface, by the equation:

ϕ=Qnet÷ϵ0

where,

Φ is the net electric flux through the closed surface.

Qnet is the net charge enclosed by the surface.

∈_0=8.85×〖10〗^(-12) C^2/Nm^2 is the permittivity of free space.

The net charge inside the box is:

Qnet = q1+q2+q3+q4

=3.0nC−2.0nC−7.0nC+1.0nC

= −5.0nC=−5.0×10−9C

The net electric flux through the box is:

Φ=Q_net/∈_0

Φ=(-5.0×〖10〗^(-9) C)/(8.85×〖10〗^(-12) C^2/Nm^2 )

Φ=5.65×〖10〗^2 Nm^2/C ≈5.6×〖10〗^2  Nm^2/C

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The ball is on the 50-yard line. The ball travels west 5 yards. The distance ball traveled was 60 yards.

Distance can be defined as the length of any path between any two locations. Displacement is the direct distance between any two points when calculated via the shortest path between them. When computing distance, the direction is disregarded. The displacement computation takes the direction into consideration.

The 5 yard gain would bring the ball to the 45 yard line (ball was tackled  at the 50 yard line, so 50 - 5 = 45), but the 15 yards lost due to the tackle and push back would bring ball to the 60 yard line (45 + 15 = 60).

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For a constant volume, the pressure of a gas is directly proportional to the temperature in Kelvin is the conclusion which can be made from Gay-Lussac’s law and is denoted as option D.

This is referred to as the perpendicular force applied on a body per unit area and the unit is Pascal.

Gay-Lussac was a scientist who discovered through numerous experiments and observations that at a constant volume, the pressure of a gas is directly proportional to the temperature in Kelvin.

This is denoted as the following equation below:

P ∝ T

P = kT where p is pressure, k is constant and t is temperature.

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The options are:

Given:

[tex]speed,\;v=449\;m/s[/tex]

We know that,

[tex]adiabatic index, \gamma = 1.667[/tex]

[tex]gas constant,\;R=8.314\;J[/tex]

The form speed of the sound in a monoatomic gas is,

[tex]v=\sqrt{\frac{\gamma\;R\;T}{M} }[/tex]

Substitute the known values in the above equation,

[tex]449=\sqrt{x} \frac{1.667 \times 8.314 \times 293}{M}[/tex]

[tex]M=\sqrt{x} \frac{1.667 \times 8.314 \times 293}{449}[/tex]

[tex]M=0.02014 \times (\frac{1000}{1})[/tex]

[tex]M=20.14\;g/mol[/tex]

Therefore, monoatomic gas is Neon gas (Ne)

Monatomic gas, which differs from diatomic, triatomic, or generally polyatomic gases in that it contains particles (molecules) made up of just one atom, includes gases like helium or sodium vapor. Because a monatomic gas lacks the rotational and vibrational energy components that characterize polyatomic gases, its thermodynamic behavior in the normal temperature range is incredibly straightforward. As a result, its heat capacity is independent of temperature and molecular weight (in this case, atomic weight), and its entropy (a measure of disorder) only depends on temperature and molecular weight.

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When thermal energy is removed from particles, then the particles' kinetic energy decreases (Option D).

The expression particles' kinetic energy makes reference to the amount of motion energy (i.e. energy in movement) that contain the particles of an object, which depends on the temperature of the material.

In conclusion, when thermal energy is removed from particles, then the particles' kinetic energy decreases (Option D).

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The distance should be 1.27 m from the wire in order to get magnetic field one-tenth as large .

Let [tex]B_1[/tex] be inside the plane and [tex]B_2[/tex] be outside the plane . It is required to calculate magnetic field at point A .

Direction of magnetic field at A is calculated using Right hand rule.

[tex]B_{net}=B_2-B_1\\\\B_{net}=\frac{u_0I}{2\pi (0.4-\frac{3\times10^{-3}}{2}) } -\frac{u_0I}{(0.4+\frac{3\times10^{-3}}{2})} \\\B_{net}=7.5\times 10^{-9}T[/tex]....(1)

Let  a distance "R"  from the wire so that the magnetic field is [tex]\frac{B_{net}}{10}[/tex]

Using the magnetic field equation , we get

[tex]\frac{B_{net}}{10} =B_2-B_1\\\\\\frac{7.5\times 10^{-9}}{10} =\frac{u_0I}{2\pi (R-\frac{3\times10^{-3}}{2}) } -\frac{u_0I}{(R+\frac{3\times10^{-3}}{2})} \\\\\R^2-2.25\TIMES10^{-6}=1.6\\\\R=1.27m[/tex]

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Answer:

the energy of photon will be half.

Explanation:

If you need a deeper explanation please let me know. Hope I helped!

Explanation:

E = hc / wavelength

Given that:

h = 6.626 × 10-34 Js

c = 2.998 x 10^8 m/s

wavelength = 263 x 10^-9 m