Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen if the spacing between the first and second dark fringes is to be 6.0 mm

Answers

Answer 1

Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

[tex]\lambda = 436.1 \ nm[/tex]

or,

  [tex]=436.1\times 10^{-9} \ m[/tex]

Distance,

[tex]d = 0.31 \ mm[/tex]

or,

  [tex]=0.31\times 10^{-3} \ m[/tex]

Distance between the 1st and 2nd dark fringes,

[tex](y_2-y_1) = 6\times 10^{-3} \ m[/tex]

As we know,

⇒ [tex]\frac{d}{L} (y_2-y_1) = \lambda[/tex]

or,

⇒ [tex]L=\frac{d(y_2-y_1)}{\lambda}[/tex]

By substituting the values, we get

       [tex]=\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}[/tex]

       [tex]=\frac{0.31\times 6\times 10^3}{436.1}[/tex]

       [tex]=\frac{1860}{436.1}[/tex]

       [tex]=4.26 \ m[/tex]


Related Questions

If 56.5 m3 of a gas are collected at a pressure of 455 mm Hg, what volume will the gas occupy if the pressure is changed to 632 mm Hg? *

Answers

Assuming ideal conditions, Boyle's law says that

P₁ V₁ = P₂ V₂

where P₁ and V₁ are the initial pressure and temperature, respectively, and P₂ and V₂ are the final pressure and temperature.

So you have

(455 mm Hg) (56.5 m³) = (632 mm Hg) V₂

==>   V₂ = (455 mm Hg) (56.5 m³) / (632 mm Hg) ≈ 40.7 m³

if C is The vector sum of A and B C = A + B What must be true about The directions and magnitudes of A and B if C=A+B? What must be tre about the directions and magnitudes of A and B if C=0? ​

Answers

Check attached photo

Check attached photo

Answer:

Explanation:

1. If C = A + B then the lines A and B may have the same magnitude or they may not. The direction of A for example may be northwest ↖️ and the direction of B must be south ⬇️ because the arrow of A and the point of B must connect. Then C’s direction is west ⬅️ because it shouldn’t be as equilibrium.

2. If C = 0 t means the force is at equilibrium. That means all forces add up to zero. A’s direction for example may be northeast ↗️ and the direction of B may be south ⬇️ and the direction of C must be west if it has to be at equilibrium.

The magnitude of A and B must be equal

A proton moves perpendicular to a uniform magnetic field at a speed of 1.75 107 m/s and experiences an acceleration of 2.25 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.

Answers

Answer:

B = 0.013(-j) T

Explanation:

Given that,

The speed of a proton, [tex]v=1.75\times 10^7\ m/s[/tex]

Acceleration experienced by the proton,[tex]a=2.25\times 10^3\ m/s[/tex]

We need to find the magnitude and the direction of the magnetic field. At equilibrium,

[tex]ma=qvB\\\\B=\dfrac{ma}{qv}\\\\B=\dfrac{1.67\times 10^{-27}\times 2.25\times 10^{13}}{1.6\times 10^{-19}\times 1.75\times 10^{7}}\\\\B=0.013\ T[/tex]

The velocity is in +z direction, force in +x direction, then the field must be in -y direction.

what happens to the weight of the body when it is falling freely under the action of gravity​

Answers

Answer:

A freely falling object has weight W=mg, where W-weight, m-mass of the object and g-acceleration produced due to the earth's gravity. ... This happens because the normal reaction force exerted on the object in the lift is equal to zero, and normal force equals to mg, which in turn equals the weight of the object

Explanation:

plz mark me as brainliest

Answer:

Gradually increases until the maximum weight reaches the surface of the earth

Explanation:

Find the ratio of the diameter of aluminium to copper wire, if they have the same

resistance per unit length. Take the resistivity values of aluminium and copper to

be 2.65× 10−8 Ω m and 1.72 × 10−8 Ω m respectively​

Answers

Answer:

1.24

Explanation:

The resistivity of copper[tex]\rho_1=2.65\times 10^{-8}\ \Omega-m[/tex]

The resistivity of Aluminum,[tex]\rho_2=1.72\times 10^{-8}\ \Omega-m[/tex]

The wires have same resistance per unit length.

The resistance of a wire is given by :

[tex]R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi (\dfrac{d}{2})^2}\\\\\dfrac{R}{l}=\rho \dfrac{1}{\pi (\dfrac{d}{2})^2}[/tex]

According to given condition,

[tex]\rho_1 \dfrac{1}{\pi (\dfrac{d_1}{2})^2}=\rho_2 \dfrac{1}{\pi (\dfrac{d_2}{2})^2}\\\\\rho_1 \dfrac{1}{{d_1}^2}=\rho_2 \dfrac{1}{{d_2}^2}\\\\(\dfrac{d_2}{d_1})^2=\dfrac{\rho_1}{\rho_2}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{2.65\times 10^{-8}}{1.72\times 10^{-8}}}\\\\=1.24[/tex]

So, the required ratio of the diameter of Aluminum to Copper wire is 1.24.

what is threshold frequency?​

Answers

Answer:

"the minimum frequency of radiation that will produce a photoelectric effect."

Explanation:

That answer was derived from gogle cuz my explanations was harder to explain but good luck

You drive 7.5 km in a straight line in a direction east of north.

a. Find the distances you would have to drive straight east and then straight north to arrive at the same point.
b. Show that you still arrive at the same point if the east and north legs are reversed in order.

Answers

Answer:

a)  a = 5.3 km, b) sum fulfills the commutative property

Explanation:

This is a vector exercise, If you drive east from north, we can find the vector using the Pythagorean theorem

              R² = a² + b²

where R is the resultant vector R = 7.5 km and the others are the legs

If we assume that the two legs are equal to = be

             R² = 2 a²

             r = √2 a

             a = r /√2

we calculate

             a = 7.5 /√2

             a = 5.3 km

therefore, you must drive 5.3 km east and then 5.3 km north and you will reach the same point

b) As the sum fulfills the commutative property, the order of the elements does not alter the result

         a + b = b + a

therefore, it does not matter in what order the path is carried out, it always reaches the same end point

3 of 3 : please help got an extra day for a test and i don’t get this (must show work) points and brainliest!

Answers

Explanation:

[tex]qV = \frac{1}{2}mv^2[/tex]

Multiply both sides by 2 and then divide by m to get

[tex]\dfrac{2qV}{m} = v^2[/tex]

Take the square root of both sides to get

[tex]v = \sqrt{\dfrac{2qV}{m}}[/tex]

how many rings does saturn have

Answers

Answer:

From far away, Saturn looks like it has seven large rings. Each large ring is named for a letter of the alphabet. The rings were named in the order they were discovered.

A car starting at rest accelerates at 3m/s² How far has the car travelled after 4s?​

Answers

Answer:

24m

Explanation:

you can use the formula

s=ut+1/2at²

s=0+1/2(3)(4)²

=1/2(3)(8)

=24m

I hope this helps

when 999mm is added to 100m ______ is the result​

Answers

Answer:

what,     100.999m

Explanation:

convert 999 mm into meters, which is 0.999m and add that to a 100 m and that will make the total 100.999 m

The result of the addition of the two values is equal to 100.999 meters.

Given the following data:

Value 1 = 999 millimetersValue 2 = 100 meters

To determine the result of the addition of the two values:

First of all, we would convert the value in millimeter (mm) to meter (m) as follows:

Conversion:

1 millimeter = 0.001 meter

999 millimeter = X meter

Cross-multiplying, we have:

[tex]X = 0.001 \times 999[/tex]

X = 0.999 meter.

For the result:

[tex]Result = 0.999 +100[/tex]

Result = 100.999 meters.

Read more on measurements here: https://brainly.com/question/24842282

Two loudspeakers, 5.5 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.25 m . Assume the speed of sound is 340 m/s.

Required:
a. What is the frequency of the sound?
b. If the frequency is then increased while you remain 0.21 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?
c.

Answers

Solution :

Let [tex]$d_1=\frac{5.5}{2}[/tex]

          = 2.75 m

[tex]d_2 = 0.21 \ m[/tex]

And [tex]$d=|d_1-d_2|$[/tex]

       [tex]$d=(d_1+d_2) - (d_1-d_2)$[/tex]

       [tex]$d=(2.75+0.21) - (2.75-0.21)$[/tex]

       [tex]$d = 2.96-2.54$[/tex]

       [tex]d = 0.42 \ m[/tex]

a). At minimum,

[tex]$d=\frac{\lambda}{2}$[/tex]

[tex]$\lambda = 2d$[/tex]

  = 2 x 0.42

  = 0.84 m

Frequency, [tex]$\nu = \frac{v}{\lambda}$[/tex]

                      [tex]$=\frac{340}{0.84}$[/tex]

                      = 404.76 Hz

Therefore, the frequency of he sound, [tex]$\nu$[/tex] = 404.76 Hz

b). At maximum, λ = d = 0.42 m

Therefore, the frequency, [tex]$\nu = \frac{v}{\lambda}[/tex]

                                             [tex]$=\frac{350}{0.42}$[/tex]

                                             = 809.52 Hz

A car whose tire have radii 50cm travels at 20km/h. what is the angular velocity of the tires?

Answers

Radius=r=50cm=0.5mVelocity=20km/h=v

Convert to m/s

[tex]\\ \sf\longmapsto v=20\times 5/18=5.5m/s[/tex]

We know

[tex]\boxed{\sf \omega=\dfrac{rv}{|r|^2}}[/tex]

[tex]\\ \sf\longmapsto \omega=\dfrac{0.5(5.5)}{|0.5|^2}[/tex]

[tex]\\ \sf\longmapsto \omega=\dfrac{2.75}{0.25}[/tex]

[tex]\\ \sf\longmapsto \omega=11rad/s[/tex]

A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r < R) from the center of the sphere the electric field has a value E. If the same charge Q were distributed uniformly throughout a sphere of radius 2R the magnitude of the electric field at a radius r would be equal to:__________

Answers

Answer:  

Hence the answer is E inside [tex]= KQr_{1} /R^{3}[/tex].

Explanation:  

E inside [tex]= KQr_{1} /R^{3}[/tex]  

so if r1 will be the same then  

E  [tex]\begin{bmatrix}Blank Equation\end{bmatrix}[/tex] proportional to 1/R3  

so if R become 2R  

E becomes 1/8 of the initial electric field.

Answer:

The electric field is E/8.

Explanation:

The electric field due to a solid sphere of uniform charge density inside it is given by

[tex]E =\frac{\rho r}{3}[/tex]

where, [tex]\rho[/tex] is the volume charge density and r is the distance from the center.

For case I:

[tex]\rho = \frac{Q}{\frac{4}{3}\pi R^3}[/tex]

So, electric field at a distance r is

[tex]E = \frac { 3 Q r}{3\times 4\pi R^3}\\\\E = \frac{Q r}{4\pi R^3}[/tex]

Case II:

[tex]\rho = \frac{Q}{\frac{4}{3}\pi 8R^3}[/tex]

So, the electric field at a distance r is

[tex]E' = \frac { 3 Q r}{3\times 32\pi R^3}\\\\E' = \frac{Q r}{8\times 4\pi R^3}\\\\E' = \frac{E}{8}[/tex]

what aspect of the US justice system has its roots in Jewish scripture?​

Answers

The aspect of the US justice system that has its roots in Jewish scripture is:

the idea that all people are subject to the same rules and laws.

It is the doctrine of "equality before the law."  Equality before the law means that every individual is equal in the eyes of the law, whether the individual is a lawmaker, a judge, a law enforcement officer, etc.  Equality before the law is also known as equality under the law, equality in the eyes of the law, legal equality, or legal egalitarianism.  It is a legal principle that treats each independent being equally and subjects each to the same laws of justice and due process.

Answer:

answer is C

the idea that all people are subject to the same rules and laws

Explanation:

hope this helps!

The double bond between two oxygen atoms (a molecule of oxygen air) has
two characteristics. What are they?
A. Four valence electrons are shared.
B. A metallic bond is formed.
C. Valence electrons are shared between oxygen atoms.
D. An ionic bond is formed.

Answers

Answer:

valance electrons are shared between oxygen atoms.. making them have eight in the outer most shells.

I hope this helps

write any two physical hazard occuring in the late choldhood​

Answers

Answer:

Hazards during late childhood

Health Problems: Chronic health ailments like T.B., Pneumonia etc will hinder the child's motor abilities.Accidents: School age children are more adventurous in nature, they run fast, play hard, ride bicycles and scooters and engage in a variety of sports.

A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?


PLEASE EXPLAIN HOW YOU GOT THE ANSWER THANK YOU SO MUCH

Answers

Answer:

0

Explanation:

The speed of the ball when it reaches the floor is 0 because when an object is at rest or in uniform motion, it has no speed/velocity

The final speed of the ball when it reaches the floor is 7.10 m/s.

What is the conservation of energy?

The conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but only converted from one form to another or transferred from one system to another. In other words, the total amount of energy in a closed system remains constant over time, even though it may be converted from one form to another.

This principle is based on the first law of thermodynamics, which states that the total energy of a closed system is always conserved, and can only be changed by the transfer of heat, work, or matter into or out of the system. The conservation of energy has important applications in various fields of physics, including mechanics, thermodynamics, and electromagnetism, and is a fundamental principle in the understanding of the natural world.

Here in the Question,

We can use the conservation of energy to solve this problem. Initially, the ball has kinetic energy due to its motion on the tabletop, but no potential energy since it is at a constant height. When the ball rolls off the edge of the table, it loses some kinetic energy due to friction but gains potential energy as it moves upward. When it reaches the floor, it has gained potential energy but lost kinetic energy due to friction. We can assume that the energy lost due to friction is converted to thermal energy, so the total energy of the system is conserved.

Let's start by calculating the potential energy gained by the ball as it moves from the edge of the table to the floor:

ΔPE = mgh

where ΔPE is the change in potential energy, m is the mass of the ball, g is the acceleration due to gravity, and h is the vertical distance traveled by the ball.

ΔPE = (0.50 kg)(9.81 m/s^2)(1.0 m) = 4.905 J

Now we can use the conservation of energy to find the final kinetic energy of the ball, which will allow us to calculate its final speed:

KEi + ΔPEi = KEf + ΔPEf

where KEi and ΔPEi are the initial kinetic and potential energies of the ball, respectively, and KEf and ΔPEf are the final kinetic and potential energies of the ball, respectively.

Since the ball is not bouncing, we can assume that its initial and final potential energies are zero. Therefore:

KEi = KEf + ΔKE

where ΔKE is the change in kinetic energy due to friction.

We can assume that the coefficient of kinetic friction between the ball and the incline is constant, and use the work-energy principle to find ΔKE:

Wfric = ΔKE

where Wfric is the work done by friction.

The work done by friction can be expressed as:

Wfric = ffricd

where ffric is the force of friction and d is the distance traveled by the ball on the incline.

The force of friction can be expressed as:

ffric = μmg

where μ is the coefficient of kinetic friction, and m and g have their usual meanings.

Putting it all together, we get:

KEi = KEf + ffricd

KEi = KEf + μmgd

(1/2)mv^2 = (1/2)mu^2 + μmgd

v^2 = u^2 + 2gd

where u is the initial speed of the ball on the tabletop, and v is the final speed of the ball on the floor.

Plugging in the given values, we get:

v^2 = (5.0 m/s)^2 + 2(9.81 m/s^2)(1.0 m)

v^2 = 50.405

v = 7.10 m/s

Therefore, the final speed of the ball when it reaches the floor is 7.10 m/s.

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If Vector A is (6, 4) and Vector B is (-2, -1), what is A – B?
A. (8,5)
B. (4,5)
C. (4,3)
D. (8,3)

Answers

Answer:

I think the answer is A...I'm not sure

Explanation:

A=(6,4)

B=(-2,-1)

A-B=(6-(-2)),(4-(-1))

=(6+2),(4+1)

=(8,5)

Answer:

[tex]6-(-2)=[/tex]

[tex]6+2[/tex]

[tex]=8[/tex]

[tex]4-\left(-1\right)[/tex]

[tex]=4+1[/tex]

[tex]=5[/tex]

[tex](8,5)[/tex]

[tex]\textbf{OAmalOHopeO}[/tex]

An observer on Earth sees spaceship 1 fly by at 0.80c. 6.0 years later, the observer on Earth sees spaceship 2 fly by at 0.80c, traveling in the same direction as the first. Both spaceships continue to travel with constant velocities. An observer in spaceship 1 observes Earth to pass spaceship 2 ____ years after Earth passed spaceship 1.

Answers

Answer:

[tex]t_2=10[/tex]

Explanation:

From the question we are told that:

Velocity of both spaceships [tex]V=0.8c[/tex]

Time [tex]t_1=6[/tex]

Generally the equation for time of spaceship 2 through earth is mathematically given by

[tex]t_2=\frac{t_1}{\sqrt{1-v^2}}[/tex]

[tex]t_2=\frac{6}{\sqrt{1-0.8^2}}[/tex]

[tex]t_2=10[/tex]

In first case a mass M is split into two parts with one part being 1/6.334 th of the original mass. In second case M is split into two equal parts. In both the cases the two parts are separated by same distance. What ratio of the magnitude of the gravitational force in first case to the magnitude of the gravitational force in the second case

Answers

Answer:

[tex]F_r=0.132:0.25[/tex]

Explanation:

From the question we are told that:

[tex]M_1=M*\frac{1}{6.334}[/tex]

Therefore

[tex]M_2=M-M*\frac_{1}{6.334}[/tex]

[tex]M_2=M*\frac{5.334}{6.334}[/tex]

Generally the equation for Gravitational force of attraction is mathematically given by

For Unequal split

[tex]F=\frac{GM_1M_2}{d^2}[/tex]

[tex]F=\frac{G(M*\frac_{1}{6.334})(M*\frac{5.334}{6.334})}{d^2}[/tex]

[tex]F=\frac{GM^2}{d^2}*(0.132)[/tex]

For equal split

[tex]F=\frac{GM_1M_2}{d^2}[/tex]

[tex]F=\frac{G(\frac{M}{2})((\frac{M}{2}}{d^2}[/tex]

[tex]F=0.25 \frac{GM^2}{d^2}[/tex]

Therefore the ratio of the gravitational force is

[tex]F_r=0.132:0.25[/tex]

In part A of the lab we see that the magnetic field of a long straight wire __. a. increases with distance in a linear relationship b. increases with distance in a non-linear relationship c. decreases with distance in an inverse (1/r) relationship d. decreases with distance in an inverse-square (1/r2) relationship

Answers

Explanation:

a long straight wire __. a. increases with distance in a linear relationship b. increases with distance in a non-linear relationship c. decreases with distance in an inverse (1/r) relationship d. decreases with distance in an inverse-square (1/r2) relationship

In part A of the lab, we see the magnetic field of a long straight wire decreases with distance in an inverse (1/r) relationship, therefore the correct option is C.

What is a magnetic field?

A magnetic field could be understood as an area around a magnet, magnetic material, or an electric charge in which magnetic force is exerted. The SI unit of the magnetic field is tesla.

For a  long straight wire carrying the current, the relation with the distance as given below

B = μI/(2πr)

where B is the magnetic field

μ is the permeability of the free space

r is the distance from the wire

As we can see from the above relation

the magnetic field of a long straight wire decreases with distance in an inverse (1/r) relationship, therefore the correct answer is option C.

Learn more about the magnetic fields from here

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Which circuit has the larger equivalent resistance: a circuit with two 10 ohm resistors connected in parallel or a circuit with two 10 ohm resistors connected in series?

Answers

Answer:

A circuit with two 10 ohm resistors connected in series.

Explanation:

The formula for the equivalent resistance for resistors in parallel is

[tex]\frac{1}{Rt} = \frac{1}{R1} + \frac{1}{R2}[/tex]   So if R1=R2= 10  [tex]\frac{1}{Rt} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} <=> Rt =\frac{10}{2} =5 ohm[/tex]

The formula for the equivalent resistance for resistors in series is

Rt = R1 + R2  So Rt= 10 + 10 = 20

you happen to visit the moon when some people on earth see a total solar eclipse. who has a better experience of this event, you or the friends you left behind back on earth

Answers

The friends left on earth because they can see the total eclipse, where as you are on the moon witnessing sections get dark rather than the whole picture

Your friend would have a better experience of this event, than you .

What is an eclipse?

An eclipse is produced when a planetary body moves in front of another planetary body and is visible from a third planetary body. Considering the sun, moon, and earth's locations in relation to one another during the time of the eclipse,

there are various types of eclipses in our solar system. For instance, a lunar eclipse occurs when the earth passes between the moon and the sun.

For the solar eclipse to happen the light from the sun is obstructed by the moon observing from the earth.

The buddies left Earth because they could view the whole eclipse, but you were on the moon and only saw parts of the eclipse turn black.

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Given that the temperature of a body is 527K determine the value in degree C

Answers

Answer:

253.85°C

Explanation:

Here is the formula for converting K to °C

527K − 273.15 = 253.85°C

A charged particle having mass 6.64 x 10-27 kg (that of a helium atom) moving at 8.70 x 105 m/s perpendicular to a 1.30-T magnetic field travels in a circular path of radius 18.0 mm. What is the charge of the particle

Answers

Answer:

the charge of the particle is 2.47 x 10⁻¹⁹ C

Explanation:

Given;

mass of the particle, m = 6.64 x 10⁻²⁷ kg

velocity of the particle, v = 8.7 x 10⁵ m/s

strength of the magnetic field, B = 1.3 T

radius of the circle, r = 18 mm = 1.8 x 10⁻³ m

The magnetic force experienced by the charge is calculated as;

F = ma = qvB

where;

q is the charge of the particle

a is the acceleration of the charge in the circular path

[tex]a = \frac{v^2}{r} \\\\ma = qvB\\\\q = \frac{ma}{vB} \\\\q = \frac{mv^2}{rvB} = \frac{mv}{rB} \\\\q = \frac{(6.64\times 10^{-27} ) \times (8.7\times 10^5)}{(1.8\times 10^{-2}) \times (1.3)} \\\\q = 2.47 \ \times 10^{-19} \ C[/tex]

Therefore, the charge of the particle is 2.47 x 10⁻¹⁹ C

Cho các máy cắt sử dụng trong công nghiệp có ký hiệu trên nhãn thiết bị: C350; B500. Hãy tính dòng điện bảo vệ ngắn mạch và dòng điện bảo vệ quá tải của từng thiết bị?

Answers

Answer:

ask in the English then I can help you

Explanation:

please mark me as brain list

26. A square loop whose sides are 6.0-cm long is made with copper wire of radius 1.0 mm. If a magnetic field perpendicular to the loop is changing at a rate of 5.0 mT/s, what is the current in the loop?

Answers

Answer:

Explanation:

The formula for determining the Emf induced in a loop is:

[tex]\varepsilon = \dfrac{d \phi}{dt}[/tex]

[tex]\varepsilon = \dfrac{d (B*A)}{dt}[/tex]

[tex]\varepsilon = A \times \dfrac{dB}{dt}[/tex]

[tex]\varepsilon = (side (l))^2 \times \dfrac{dB}{dt}[/tex]

where;

square area A = ( l²)

l² = 6.0 cm = 6.0 × 10⁻²

[tex]\varepsilon = ( 6.0 \times 10^{-2})^2 \times 5.0 \times 10^{-3} \ T/S[/tex]

[tex]\varepsilon =18 \times 10^{6} \ V[/tex]

Recall that:

The resistivity of copper = [tex]1.68 \times 10^{-8}[/tex] ohm m

We can as well say that the length of the copper wire = perimeter of the square loop;

The perimeter of the square loop = 4L

Thus, the length of the copper wire  = 4 (6.0 × 10⁻² )m

= 24× 10⁻² m

Finally, the current in the loop is determined from the formula:

V = IR

where,

V = voltage

I = current and R = resistance of the wire

Making "I" the subject:

I = V/R

where;

[tex]R = \dfrac{\rho \times l}{A}[/tex]

[tex]R = \dfrac{\rho \times l}{\pi * r^2}[/tex]

[tex]R = \dfrac{1.68 *10^{-8} \times 24*10^{-2}}{\pi * (1*10^{-3})^2}[/tex]

[tex]R = 0.001283 \ ohms[/tex]

[tex]I = \dfrac{18*10^{-6}}{0.001283}[/tex]

I = 14.029 mA

A nylon string on a tennis racket is under a tension of 285 N . If its diameter is 1.10 mm , by how much is it lengthened from its untensioned length of 29.0 cm ? Use ENylon=5.00×109N/m2.

Answers

Answer:

1.74×10⁻³ m

Explanation:

Applying,

ε = Stress/strain............. Equation 1

Where ε = Young's modulus

But,

Stress = F/A.............. Equation 2

Where F = Force, A = Area

Strain = e/L.............. Equation 3

e = extension, L = Length.

Substitute equation 2 and 3 into equation 1

ε = (F/A)/(e/L) = FL/eA............. Equation 4

From the question,

Given: F = 285 N, L = 29 cm = 0.29 m, ε = 5.00×10⁹ N/m²,

A = πd²/4 = 3.14(0.0011²)/4 = 9.4985×10⁻⁶ m²

Substitute these values into equation 4

5.00×10⁹ = (285×0.29)/(9.4985×10⁻⁶×e)

Solve for e

e = (285×0.29)/(5.00×10⁹×9.4985×10⁻⁶)

e = 82.65/4.74925×10⁴

e = 1.74×10⁻³ m

A 2.0 kg frictionless puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. How far does the puck move from rest in 2.5 s?

Answers

Answer:

the distance moved by the puck after 2.5 s is 7.8 m

Explanation:

Given;

mass of the puck, m = 2 kg

initial velocity of the puck, u = 0

applied force, F = 5 N

time of motion, t = 1.5 s

Acceleration of the puck is calculated from Newton's second law of motion;

F = ma

a = F/m

a = 5/2

a = 2.5 m/s²

The distance moved by the puck after 2.5 s is calculated as;

s = ut + ¹/₂at

s = 0 + ¹/₂at²

s = ¹/₂at²

s = 0.5 x 2.5 x (2.5)²

s = 7.8 m

Therefore, the distance moved by the puck after 2.5 s is 7.8 m

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