If three cars, each of mass 1000 kg , occupy the NE, SE, and SW corners, determine the coordinates of the CM of the loaded ferryboat relative to the center of the raft. Take the origin to be at the CM of the raft.

Answers

Answer 1

Answer:

Using X = sum( x's * mass)/sum( masses)

= [(0)*(6400)+ (-9.5*1000) + (9.5)*(1000) + (9.5)*(1000)]/(6400+1000+1000+1000)

= 1.01 m

Y = sum( y's * mass)/sum( masses)

= [(0)*(6400) + (9.5)*(1000) +(-9.5)*(1000) + (-9.5)*(1000)]/ (6400+1000+1000+1000)

= -1.01 m


Related Questions

Which of the following object is in dynamic equilibrium?

Answers

Answer:

A car driving in a straight line 20 m/s

Explanation:

ayepecks silly

If the absolute pressure inside the bottom of a container open to the atmosphere and filled with an unknown substance is 300 Pa. And the height of the container is 800.0 cm. What is the density of the substance?

Answers

Answer:

2.5 kg/m³

Explanation:

Absolute pressure = gauge pressure + atmospheric pressure

P = Pg + Pa

The gauge pressure caused by the weight of a fluid is called the static pressure.  It is equal to the density of the fluid × acceleration due to gravity × depth of the fluid,

Pg = ρgh

Therefore:

P = ρgh + Pa

300 Pa = ρ (10 m/s²) (8.00 m) + 100 Pa

ρ = 2.5 kg/m³

formula of minimmum pressure​

Answers

Answer:

pressure=force/area

calculate the energy dissipated by an electric iron which draws a current of 5A from a240v power supply for 25minutes​

Answers

Explanation:

Given,

I = 5 A

V = 240 V

T = 25 mins = 1500 sec

Now,

Energy dissipated = IVT= 5×240×1500 = 1800000 J

The total mass of eight identical
building blocks is 31.52 kg. Find the
mass of 1 block.​

Answers

Answer:

3.94

Explanation:

divide total mass by the number of blocks since they are identical

Answer:

3.94

Explanation:

You want to find the mass of one block. Since we know there is 8 blocks with the same mass, you can divide the total mass by 8 since the mass is equally distributed within the 8 blocks

3.) [15 points] A physics teacher is on the west side of a small lake and wants to swim across and up at a point directly across from his starting point. He notices that there is a current in the lake and
that a leaf floating by him travels 4.2m [S] In 5.0s. He is able to swim 1.9 m/s in calm water,
(a) What direction will he have to swim in order to arrive at a point directly across from his position?​

Answers

Answer:

The teacher should swim in a direction 29.24° North of East

Explanation:

Given that the there is a water  current across the lake, and the physics teacher intends to swim directly across the lake, the direction the physics teacher will have to swim is found as follows;

The speed of the water current is given by the speed of the floating leaf traveling with the water current  

Distance traveled by the leaf = 4.2 m South

Time of travel of the leaf = 5.0 s

Speed of leaf = 4.2/5 = 0.84 m/s = Speed of the water current

Swimming peed of the teacher, v = 1.9 m/s

To swim directly across the lake, the teacher has to swim slightly in the opposite direction of the water current, the y-component of the teacher's swimming speed should be equal to and opposite that of the speed of the water current.

Y-component of v = v×sin(θ), where θ is the angle of the direction, the teacher should swim

Therefore;

1.9 × sin(θ) = 0.84

sin(θ) = 0.84/1.9 = 0.44

θ = 26.24°

That is the teacher should swim in a direction 29.24° North of East.

To cross the lake the teacher has to swim in a direction 29.24° North of the East

Finding the direction of speed required:

The speed of the water current can be derived from the speed of the floating leaf :

The distance traveled by the leaf L = 4.2 m South

Time taken T = 5s

So, the speed of the leaf is:

u = 4.2/5

u = 0.84 m/s South

So, the speed of the current is 0.84 m/s South

Now, it is given that the speed of the teacher is, v = 1.9 m/s East

To cross the lake the speed of the teacher must be in a Northeast direction so that the North component of the speed of the teacher cancels out the speed of the current which is directed towards the South.

Let, the speed of the teacher makes an angle of θ from the EAST.

So, the North component is given by:

v(north) = vsinθ

it must be equal to the speed of the current:

vsinθ = u

1.9 × sinθ = 0.84

sinθ = 0.84/1.9

sinθ = 0.44

θ = 26.24°

The teacher should swim in a direction 29.24° North of East.

Learn more about vector components:

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b. i. In a hydraulic press, a force of 200N is applied to master piston of area 25cm. If
the press is designed to produce a force of 5000N, determine the area of the slave
piston.
(4marks)

Answers

Answer:

625 cm²

Explanation:

The pressure is the same on both pistons.

F/A = F/A

200 N / 25 cm² = 5000 N / A

A = 625 cm²

A very thin film of soap, of thickness 170 nm, in between air seems dark. On the other hand, when placed on top of glass some visible light is seen to shine from the film. How can this happen and what is the smallest visible light that creates constructive interference when we place the film on top of glass

Answers

Answer:

λ₀ = 2 d n

Explanation:

A soap film is a layer where the lus is reflected on the surface and on the inside of the film, these two reflected rays can interfere with each other either constructively or destructively.

Let's analyze the general conditions of this interference,

* When the ray of light reaches the surface of the film it is reflected, as the index of refraction of the air is less than the index of the film, the reflected ray has a phase change of 180º

* When the ray penetrates the film, its wavelength changes due to the refractive index of the film.

          λ = λ₀ / n

where lick is the wavelength in the vacuum or air and n index of refraction of the film, in general this interference is observed perpendicular to the film, so the sine veils 1. the expression for constructive interference taking in what previous remains

         2d = (m + ½) λ

the expression for destructive interference remains

         2d = m λ

          2d = m λ₀ / n

When the film is placed on a glass plate whose index of refraction is greater than the index of refraction of the film, in the reflection in the lower part of the film another phase difference of 180º is created, for which we have a difference of total phase of 180 +180 = 360º, which is equivalent to no phase difference, therefore the two previous equations are interchanged.

Therefore where we had destructive interference now a cosntructive interference happens we can see the reflected light.

Find us the wavelength that this constructive interference creates

           2d n = m λ₀

           λ₀ = 2 d n / m

To find the minimum wavelength, suppose we observe the first interference pattern m = 1

         λ₀ = 2 d n

where d is the thickness of the film and n the index of refraction of the same

Two trains run in the opposite direction with speeds of v1 = 15 m / s and v2 = 20 m / s. A passenger on the first train (the one on v1) notes that train 2 takes 6 s to pass on its side. What is the length of the second train? (The passenger is supposed to be immobile looking through the window)

Answers

Answer:

210 m

Explanation:

The speed of train 2 relative to train 1 is 15 m/s + 20 m/s = 35 m/s.

It takes 6 seconds for the train to pass, so the length of the train is:

(35 m/s) (6 s) = 210 m

When we double the distance between a source of light and the
surface on which it falls, the amount of illumination on the surface
decreases to
(what fraction) of the
original illumination.

Answers

The amount of lighting on a surface drops to (1/4) of the initial illumination when the distance between a light source and the surface it falls on is doubled.

What is illumination?

The amount of light or luminous flux that falls on a surface is known as illumination. It is expressed as lux or lumens per square meter.

The square of the distance has an inverse relationship with the light intensity;

[tex]\rm I = \frac{1}{r^2}[/tex]

Where,

I is the light intensity

r is the distance

Let r is the distance and I is the sound intensity for case 2;

r' = 2r

[tex]\rm I' = \frac{1}{(2r)^2} \\\\ I' = \frac{1}{4r^2} \\\\ I' = \frac{I}{4}[/tex]

When we double the distance between a source of light and the surface on which it falls, the amount of illumination on the surface decreases to(1/2) of the original illumination.

Hence the value of the fraction is 1/4.

To learn more about the illumination refer;

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Please answer this question

Answers

Explanation:

m = kg. v=m/s. g=m/s^2. h= m

>>1/2mv^2=mgh

>>1/2mv^2=mgh>> kg*(m/s)^2= kg*m/s^2*m

>>1/2mv^2=mgh>> kg*(m/s)^2= kg*m/s^2*m>>kg m^2/s^2=kg m^2/s^2 the fraction 1/2 won't be able to make any changes to to the dimensional expression of energy i.e half of energy is still energy therefore you can neglect the number .

>>kg m^2/s^2=kg m^2/s^2

>>J= J

HELP me pleaseeee somebody

an object is placed 30cm from a mirror of focal length 15 cm the object is 7.5cm tall. where is the image located? how tall is the image??

Answers

Explanation:

It is given that,

Object distance from the mirror, u = -30 cm

Focal length of the mirror, f = +15 cm

Size of the object, h = 7.5 cm

We need to find the image distance and the size of the image.

Mirror's formula, [tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]

v is image distance

[tex]\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(15)}-\dfrac{1}{(-30)}\\\\v=10\ cm[/tex]

Let h' is the size of the image. So,

[tex]\dfrac{h'}{h}=\dfrac{-v}{u}\\\\h'=\dfrac{-vh}{u}\\\\h'=\dfrac{-10\times 7.5}{-30}\\\\h'=2.5\ cm[/tex]

So, the image is located at a distance of 10 cm and the size of the image is 2.5 cm.

The following passage has not been edited. There is an error in each line. Write the

incorrect word and the correction in your answer sheet against the correct question

number. The first one has been done as an example. ( 1 x 4 = 4 )

Community service sensitize people to Error: sensitize ; Correction: sensitizes

other‟s needs and supports inclusive (a) Error: _______ ; Correction: ______

development to the underprivileged (b) Error: _______ ; Correction: ______

sections with society. Courses about social (c) Error: ______ ; Correction: _______

work prepares frontline workers to (d) Error: _______ ; Correction: ______​

Answers

Answer:

(a) Error: Other's ; Correction: Others'

(b) Error: to ; Correction: for

(c) Error: with ; Correction: of

(d) Error: prepares ; Correction: prepare

Explanation:

a)  The error is in the word "other's" as the position of apostrophe is wrong, so the correct word will be "others'", it shows plural nouns.

b)  The error is in the word "to", so the correct word will be "for" as for is use to talk about a purpose.

c)  The error is in the word "with"  and the correct word will be "of" as of indicates relationships between other words including things that made of other things.

d)  The error is in the word "preapres" and the correct word will be "prepare".

A ray of light is projected into a glass tube that is surrounded by air. The glass has an index of refraction of 1.50 and air has an index of refraction of 1.00. At what minimum angle will light in the glass tube be totally reflected at the glass/air interface?

Answers

Answer:

θ = 41.8º

Explanation:

This is an internal total reflection exercise, the equation that describes this process is

         sin θ = n₂ / n₁

where n₂ is the index of the incident medium and n₁ the other medium must be met n₁> n₂

        θ = sin⁻¹ n₂ / n₁

let's calculate

       θ = sin⁻¹ (1.00 / 1.50)

       θ = 41.8º

Let us treat a helicopter rotor blade as a long thin
rod, as shown in Fig. 8–49. (a) If each of the three rotor
helicopter blades is 3.75 m long and has a mass of 135 kg,
calculate the moment of inertia of the three rotor blades
about the axis of rotation. (b) How much torque must the
motor apply to bring the blades from rest up to a speed
of 6.0 rev/s in 8.0 s?
Rotor

Answers

Answer:

(a) 1900 kg m²

(b) 8950 Nm

Explanation:

(a) The moment of inertia of a rod about its end is I = ⅓mL².

For 3 rods of mass m = 135 kg and length L = 3.75 m, the total moment of inertia is:

I = 3 (⅓ (135 kg) (3.75 m)²)

I = 1900 kg m²

(b) Net torque = moment of inertia × angular acceleration

∑τ = Iα

First, find the angular acceleration.

ω₀ = 0 rad/s

ω = 6.0 rev/s (2π rad/rev) = 37.7 rad/s

t = 8.0 s

α = (37.7 rad/s − 0 rad/s) / 8.0 rad/s = 4.71 rad/s²

∑τ = Iα

∑τ = (1900 kg m²) (4.71 rad/s²)

∑τ = 8950 kg m² / s²

∑τ = 8950 Nm

Two kilograms of nitrogen (N2) at 25°C is contained in a 0.62 m3 rigid tank. This tank is connected by a valve to a 0.16 m3 rigid tank containing 0.8 kg of oxygen (O2) at 127°C. The valve is opened, and the gases are allowed to mix, achieving an equilibrium state at 87°C.
initial pressures of N2 is 5.7293 bar and O2 is 5.2 bar.
the final pressure is 6.44 bar.
the magnitude of the heat transfer for the process is 162.8 kJ, and the direction of energy flow is going in.
What is the entropy change for the mixing process, in kJ/K?

Answers

Answer:

Explanation:

For entropy change the formula is

ΔS = ΔQ / T

ΔQ = Δ H

ΔS = Δ H / T

Given

Δ H = + 162.8 kJ

We can take equilibrium temperature as average temperature of the whole process

So, T = 273 + 87 = 360 K

ΔS = Δ H / T

=  162.8 kJ  / 360

= +  0.508 kJ / K .

When the magnitude of the heat transfer for the process is 162.8 kJ, Then the entropy change for the mixing process, in kJ/K is = + 0.508 kJ / K

What is Entropy change?

For The entropy change, the formula is

Then ΔS = ΔQ / T

After that ΔQ = Δ H

Then ΔS = Δ H / T

Given as per question are:

Then Δ H = + 162.8 kJ

Now We can take equilibrium temperature as average temperature of the whole process are:

So, T is = 273 + 87 = 360 K

Then ΔS = Δ H / T

After that = 162.8 kJ / 360

Therefore, = + 0.508 kJ / K.

Find more information about Entropy change here:

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Allocate birr 5000 among the three workers in the ratio 1/3 :1/6 and 5/12.​

Answers

Answer:

1666.7 ETB (birr)

833.3 ETB (birr)

2083.3 ETB (birr)

Explanation:

The first worker

5000*1/3=1666.7

The second worker

5000*1/6=833.3

The third worker

5000*5/12=2083.3

Hope this helps :) ❤❤❤

The large-scale distribution of galaxies in the universe reveals Group of answer choices a smooth, continuous, and homogenous arrangement of clusters large voids, with most of the galaxies lying in filaments and sheets a large supercluster at the center of universe a central void with walls of galaxies at the edge of the universe

Answers

Question

 

The large-scale distribution of galaxies in the universe reveals

A) a smooth, continuous, and homogenous arrangement of clusters

B) large voids, with most of the galaxies lying in filaments and sheets a large supercluster at the center of the universe

C) a central void with walls of galaxies at the edge of the universe

Group of answer choices

Answer:

The correct answer is B)

Explanation:

The universe is arranged in a filamentary structure. Filamentary structures are very large. They are the largest kind of structures in the universe and comprise mostly of galaxies that are held together by gravity.

The structures found within Galaxy filaments have thread-like qualities spanning 52 to 78.7 megaparsecs h⁻¹ in lenght.

Other phenomena associated with the nature fo the universe is the existence of void spaces.

Cheers!

a 6 letter word a way of explaining an object or event using a set of facts

Answers

Explanation:

A theory is a way of explaining an object or event using a set of facts.

A plane drops a package for delivery. The plane is flying horizontally at a speed of 120m/s,and the package travels 255 m horizontally during the drop. We can ignore air resistance.What is the package's vertical displacement during the drop?

Answers

Answer:

Package's vertical displacement(s) = 22.12 meter

Explanation:

Given:

Speed of plane = 120 m/s

Total distance = 255 m

Find:

Package's vertical displacement(s)

Computation:

Time taken = Distance / Speed

Time taken = Total distance / Speed of plane

Time taken = 255 / 120

Time taken = 2.125 s

Acceleration due to gravity(g) = 9.8 m/s²

Initial velocity (u) = 0

So,

Package's vertical displacement(s) = ut + (1/2)gt²

Package's vertical displacement(s) = (0)(2.125) + (1/2)(9.8)(2.125)²

Package's vertical displacement(s) = 22.12 meter

Answer: -22.1

Explanation:

I just did the Khan Academy and that was the answer, not the one provided by that one person. :)))

A load of 500N is carried by 200N effort in a simple machine having load distance 3m Calculate effort distance.​

Answers

Answer:

2.5 m

Explanation:

Load ( L ) = 500 N

Effort ( E ) = 200 N

Load distance ( LD ) = 3 m

Effort distance ( ED ) = ?

Now, Let's find the Effort distance ( ED )

We know that,

Output work = Input work

i.e L × LD = E × ED

plug the values

[tex]500 \times 3 = 200 \times ED[/tex]

multiply the numbers

[tex]1500 = 200 \times ED[/tex]

Swipe the sides of the equation

[tex]200 \: ED \: = 500[/tex]

Divide both sides of the equation by 200

[tex] \frac{200 \: ED}{200} = \frac{500}{200} [/tex]

Calculate

[tex]ED\: = 2.5 \: m[/tex]

Hope this helps..

best regards!!

The graph shows a wave that oscillates with a frequency of 60 Hz. Based on the information given in the diagram, what is the speed of the wave?

Answers

Answer:

900 cm/s or 9 m/s.

Explanation:

Data obtained from the question include the following:

Length (L) = 30 cm

frequency (f) = 60 Hz

Velocity (v) =.?

Next, we shall determine the wavelength (λ).

This is illustrated below:

Since the wave have 4 node, the wavelength of the wave will be:

λ = 2L/4

Length (L) = 30 cm

wavelength (λ) =.?

λ = 2L/4

λ = 2×30/4

λ = 60/4

λ = 15 cm

Therefore, the wavelength (λ) is 15 cm

Now, we can obtain the speed of the wave as follow:

wavelength (λ) = 15 cm

frequency (f) = 60 Hz

Velocity (v) =.?

v = λf

v = 15 × 60

v = 900 cm/s

Thus, converting 900 cm/s to m/s

We have:

100 cm/s = 1 m/s

900 cm/s = 900/100 = 9 m/s

Therefore, the speed of the wave is 900 cm/s or 9 m/s.

The angle of incidence of a ray of light striking an equilateral triangular prisms ABC of refracting angle 60o is 40o. Calculate:
(i)the angle of refraction at the first face
(ii)the angle of emergence

Answers

Answer:

1: the refracted angle in the first face is equal to the incident angle that is 60degrees

2. Emergence Angle is 42degrees

Explanation:

Pls see attached file

A 24 cm radius aluminum ball is immersed in water. Calculate the thrust you suffer and the force. Knowing that the density of aluminum is 2698.4 kg / m3

Answers

Answer:

W =1562.53 N

Explanation:

It is given that,

Radius of the aluminium ball, r = 24 cm = 0.24 m

The density of Aluminium, [tex]d=2698.4\ kg/m^3[/tex]

We need to find the thrust and the force. The mass of the liquid displaced is given by :

[tex]m=dV[/tex]

V is volume

Weight of the displaced liquid

W = mg

[tex]W=dVg[/tex]

So,

[tex]W=dg\times \dfrac{4}{3}\pi r^3\\\\W=2698.4\times 10\times \dfrac{4}{3}\times \pi \times (0.24)^3\\\\W=1562.53\ N[/tex]

So, the thrust and the force is 1562.53 N.

Iron man wears an awesome ironsuit.He is flying over high current carrying wire. Will he be affected?

Answers

Answer:

According to super hero logic , nothing will happen to him.

But according to science , yes he will get current shock but good news is that he wouldn't get elected until he is in contact with the wires.

He may / may not be affected but his suit will be damaged for sure as it is made of metal.

HOPE THIS HLEP AND PLSSSSS MARK AS BRAINLIEST AND THNXX :)

A carpenter measured the lengeth of a small piece of timber as 24.6cm .Calculate the relative error in the measurement if the true length is 24.5cm​

Answers

ANSWER:

0.4081%

Explanation:

Difference=24.6-24.5=0.1

Relative error = 0.1/24.5*100=0.4081%

Relative error is equal to the = difference between both the values/The true value *100

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickness is 7 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 17 m2 and 0.20 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window

Answers

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

Find the mass. 10 points. Will give brainliest.

Answers

Answer:

3.94 kg

Explanation:

Given,

Force ( f ) = 30 N

Acceleration(a) = 7.6 m/s

Now, Let's find the mass of the ball

Using the Newton's second law of motion:

We get:

[tex]force \: = mass \: \times acceleration[/tex]

plug the value

[tex]30 \: = m \: \times 7.6[/tex]

Use the commutative property to reorder the terms

[tex] 30 = 7.6 \: m[/tex]

Swap the sides of the equation

[tex]7.6m = 30[/tex]

Divide both sides of the equation by 7.6

[tex] \frac{7.6 \: m}{7.6} = \frac{30}{7.6} [/tex]

Calculate

[tex]m = 3.94 \: kg[/tex]

Hope this helps..

Best regards!!

Answer:

[tex]\displaystyle \boxed{\mathrm{3.95 \: kg }}[/tex]

Explanation:

[tex]\mathrm{force \: (N) = mass \: (kg) \times acceleration \: (m/s^2)}[/tex]

[tex]\mathrm{force = 30N}[/tex]

[tex]\mathrm{acceleration = 7.6 \: m/s^2 }[/tex]

[tex]\mathrm{Find \: the \: mass.}[/tex]

[tex]\mathrm{30 = m \times 7.6}[/tex]

[tex]\displaystyle \mathrm{m =\frac{30}{7.6} }[/tex]

[tex]\displaystyle \mathrm{m = 3.947... }[/tex]

1. Analogies exist between rotational and translational physical quantities. Identify the rotational term analogous to each of the following: acceleration, force, mass, work, translational kinetic energy, linear momentum, impulse.
2. Explain why centripetal acceleration changes the direction of velocity in circular motion but not its magnitude.

Answers

Answer:

1) a    α,  m   I,  W=F.d    W =τ . θ,

2)  a = v²/r

Explanation:

1) The amounts of rotational and translational motion are related

acceleration is

        a = d²x / dt²

    linear displacement is equivalent to angular rotation, therefore angular acceleration is

      α = d²θ / dt²

force in linear motion is equivalent to moment in endowment motion

       F = m a

       τ = I α

the mass is the inertia of the translation, in rotational motion the moment of inertia is the rotational inertia

          I = m r²

Work is defined by W = F. d

in rotation it is defined by W = τ . θ

The linear momentum is p = mv

the angular momentum L = I w

momentum the linear motion is I = F dt

in the rotation it is I = τ dt

 

2) The velocity is a vector therefore it has modulus and direction, linear acceleration changes the modulus of velocity, whereas circular motion changes the direction (the other element of the vector).

      [tex]a_{c}[/tex]Ac = v²/r

Two long, straight wires are parallel and 10 cm apart. One carries a current of 2.0 A, the other a current of 5.0 A. If the two currents flow in opposite directions, what is the magnitude and direction of the force per unit length of one wire on the other

Answers

Answer:

The magnitude and direction of the force per unit length of one wire on the other is 2 x 10⁻⁵ N/m, attractive force.

Explanation:

Given;

distance between the two parallel wires, r = 10 cm = 0.1 m

current in the first wire, I₁ = 2A

current in the second wire, I₂ = 5 A

The force per unit length on each wire can be calculated as;

[tex]\frac{F}{L} = \frac{\mu_oI_1I_2}{2\pi r}[/tex]

where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

[tex]\frac{F}{L} = \frac{\mu_oI_1I_2}{2\pi r} \\\\\frac{F}{L} = \frac{4\pi*10^{-7}*2*5}{2\pi *0.1} \\\\\frac{F}{L} = 2 *10^{-5} \ N[/tex]/ m

The direction of the force between the two wires is attractive since the current in the two wires are in opposite direction.

Therefore, the magnitude and direction of the force per unit length of one wire on the other is 2 x 10⁻⁵ N/m, attractive force.

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