Fiber optic (FO) cables are based upon the concept of total internal reflection (TIR), which is achieved when the FO core and cladding have the same refractive indices.

a. True
b. False

Answer:

An Object is positively charged if it has more Positive Electrons in that object

Answer:

C. both a speed and a direction

Answer:

a)  T = 2838.6 N,  b)   W = 1003.2 J,  c) W = 6.22 10⁴ J,  d) W = 2.79 10³ J

e) v_f = 2.65  m / s

Explanation:

a) To find the tension of the cable let's use Newton's second law

        T - W = m a

         T = W + ma

        T = m (g + a)

let's calculate

        T = 285 (-9.8 - 0.160)

        T = 2838.6 N

b) net work is stress work minus weight work

        W = F d

        W = (T-W) d

        W = (m a) d

        W = (285 0.160) 22

        W = 1003.2 J

c) the work done by the cable

         W = T d cos 0

          W = 2838.6 22.0

          W = 6.22 10⁴ J

d) The work done by the weight

the displacement is upwards and the weight points downwards, so the angle is 180º

        W = F. d

         W = F d cos 180

         W = -285 22.0

         W = 2.79 10³ J

e) the final speed of the load. Let's use the relationship between work and the change in kinetic energy

         W = ΔK

as part of rest K₀ = 0

          W = ½ m v_f²

          v_f = [tex]\sqrt{ \frac{2W}{m} }[/tex]

          v_f = [tex]\sqrt{\frac{2 \ 1003.2}{285} }[/tex]

          v_f = 2.65  m / s

Answer:

M = 1.409 10¹⁴ kg

Explanation:

In this exercise we have that the prioress with a minimum speed can escape from the asteroid, therefore we can use the conservation of energy relation.

Starting point. When you drop the stone

         Em₀ = K + U

         Em₀ = ½ m v² - G m M / r

where M and r are the mass and radius of the asteroid

Final point. When the stone is too far from the asteroid

          Em_f = U = - G m M / R_f

as there is no friction, the energy is conserved

          Em₀ = Em_f

          ½ m v² - G m M / r = - G m M / R_f

          ½ v² = G M (1 / r - 1 /R_f)

indicate that for the speed of v = 45.7 m /s, the stone does not return to the asteroid so R_f = ∞

          ½ v² = G M (1 /r)

          M = [tex]\frac{v^2 r}{2G}[/tex]

let's calculate

          M = [tex]\frac{45.7^2 \ 9}{ 2 \ 6.67 \ 10^{-11}}[/tex]

          M = 1.409 10¹⁴ kg

Answer:

Refer to the attachment!~

Answer:

The speed of the combined mass after the collision is 2.1 m/s.

Explanation:

mass of runner, m = 70 kg

speed  of runner, u = 2.7 m/s

mass of shortstop, m' = 85 kg

speed  of shortstop, u' = 1.6 m/s

Let the velocity of combined system is v.

Use conservation of momentum

Momentum before collision = momentum after collision

m u + m' u' = (m + m') v

70 x 2.7 + 85 x 1.6 = (70 + 85) v

189 + 136 = 155 v

v = 2.1 m/s

Answer:

The right solution is "50200 days".

Explanation:

Given:

Calories intake,

= 6000 kcal,

or,

= [tex]2.52\times 10^7 \ J[/tex]

Force,

= 500 N

As we know,

⇒ [tex]Work \ done = Force\times distance[/tex]

Or,

⇒ [tex]distance = \frac{Work \ done}{Force}[/tex]

By putting the values, we get

                  [tex]=\frac{2.52\times 10^7}{500}[/tex]

                  [tex]=0.502\times 10^5[/tex]

                  [tex]=50200 \ m[/tex]

hence,

The number of days will be:

= [tex]\frac{50200}{1}[/tex]

= [tex]50200 \ days[/tex]

Answer:

S = 1/2 gt² = 1/2 × 9.8 × 3² = 4.9×9 = 44.1 m

Explanation:

Answer:

A

Explanation:

I guess not that much confidential!

Answer:

D_y = 414.38m

Explanation:

D_y = D*sin(x)

D_y = 673m*sin(38°)

D_y = 414.38m

Answer:

The correct answer is "4000 J".

Explanation:

Given that,

Force,

= 200 N

Displacement,

= 20 m

Now,

The work done will be:

⇒ [tex]Work=Force\times displacement[/tex]

By putting the values, we get

              [tex]=200\times 20[/tex]

              [tex]=4000 \ J[/tex]

Answer:

ans: 2.25 meter

explanation

use following equations

F = ma

V = U + aT

S = UT + 1/2 aT^2

Answer:

WIND Is what you're looking for

Explanation:

The word is WIND

We call the acceleration of an object moving in uniform circular motion— resulting from a net external force—the centripetal ...

Answer:

Explanation:

Here's what we know and in which dimension:

y dimension:

[tex]v_0=30[/tex] m/s

v = 0 (I'll get to that injust a second)

a = -9.8 m/s/s

The final velocity of 0 is important because that's the velocity of the ball right at the very top of its travels. If we knew how long it takes to get to that max height, we can also use that to find out how long it will take to hit the ground. Therefore, we will find the time it takes to reach its max height and pick up with the investigation of what this means after.

x dimension:

Δx = 70 m

v = ??

Velocity is our unknown.

Solving for the time in the y dimension:

[tex]v=v_0+at[/tex] and filling in:

0 = 30 + (-9.8)t and

-30 = -9.8t so

t = 3.1 seconds

We know it takes 3.1 seconds to get to its max height. In order to determine how long it will take to hit the ground, just double the time. Therefore, it will take 6.2 seconds for the ball to come back to the ground, which is where the persom trying to catch the ball comes in. We will use that time in our x dimension now.

In the x dimension, the equation we need is just a glorified d = rt equation since the acceleration in this dimension is 0.

Δx = vt and

70 = v(6.2) so

v = 11.3 m/s

Answer:

wavelength=75.0

speed of sound(v)=350 .0m/s

frequency(f)=?

we know,

v=f*wavelengh

350.0 =f*750

f. =350/75

=4.667

pls mark me brainlest

Answer: The weight of the object is 29.4 N

Explanation:

To calculate the weight of the object, we use the equation:

[tex]W=m\times g[/tex]

where,

m = mass of the object = 3 kg

g = acceleration due to gravity = [tex]9.8m/s^2[/tex]

Putting values in above equation, we get:

[tex]W=3kg\times 9.8m/s^2\\\\W=29.4N[/tex]

Hence, the weight of the object is 29.4 N

The boy pushed the sled for 16 seconds.

We have a boy who pushes his little brother on a sled.

We have to determine for how long time does boy push the sled.

The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

According to the question -

The sled is initially at rest → initial velocity (u) = 0.

Final velocity (v) = 4 m/s

Mass of boy and sled (M) = 40 kg

Power developed (P) = 20 W = 20 Joules/sec

According to work - energy theorem -

Work done (W) = Δ E(K) = E(f) - E(i)

Therefore -

W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule

Now, Power is defined as the rate of doing work -

P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]

20 = [tex]\frac{320}{t}[/tex]

t = 16 seconds

Hence, the boy pushed the sled for 16 seconds.

To solve more questions on Work, Energy and Power, visit the link below -

https://brainly.com/question/208670

#SPJ2

Answer:

a) [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]

b) [tex]a=30.7[/tex]

c) [tex]a=35.91[/tex]

Explanation:

From the question we are told that:

Initial angular velocity [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]

Initial Length [tex]L_1=0.600m[/tex]

Final angular velocity [tex]\omega _2=6.22rev/s=39rad/s[/tex]

Final Length [tex]L_2=0.900m[/tex]

a)

Generally the rotation with the greater speed is

[tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]

b)

Generally the equation for centripetal acceleration at 8.16 is mathematically given by

[tex]a=\omega_1^2*L_1[/tex]

[tex]a=8.16 rev/s*0.6[/tex]

[tex]a=30.7[/tex]

c)

At 6.35 rev/s

[tex]a=6.35 rev/s*0.9[/tex]

[tex]a=35.91[/tex]

This question is incomplete, the missing diagram is uploaded along this answer below.

Answer:

- the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N

- the magnitude of compression force at the knee joint is 900 N

Explanation:

Given the data in the question and diagram below;

Net torque = 0

Torque = force × lever arm

so

F[tex]_{ConF[/tex]  × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in

given that F[tex]_{ConF[/tex] = 90 N

90 × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in

90 N × 16.5 in =  T[tex]_{HonL[/tex] × 1.5 in

T[tex]_{HonL[/tex] = ( 90 N × 16.5 in ) / 1.5 in

T[tex]_{HonL[/tex] = 990 N

Therefore, the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N

b) magnitude of compression force at the knee joint;

In equilibrium, net force = 0

along horizontal

F[tex]_{FonB[/tex] - T[tex]_{HonL[/tex] + F[tex]_{ConF[/tex] = 0

we substitute

F[tex]_{FonB[/tex] - 990 + 90 = 0

F[tex]_{FonB[/tex] - 900 = 0

F[tex]_{FonB[/tex] = 900 N

Therefore, the magnitude of compression force at the knee joint is 900 N

Answer:

The net field will be the sum of the fields created by each charge.

where the charge Q in a position r' is given by:

E(r) = k*Q/(r - r')^2

Where k is a constant, and r is the point where we are calculating the electric field.

Then for the charge 3q, in the position r₁ = (-d, 0, 0) the electric field will be:

E₁(r) = k*3q/(r - r₁)^2

While for the other charge of -2q in the position r₂ = (d, 0, 0)

The electric field is:

E₂(r) = -k*2*q/(r - r₂)^2

Then the net field at the point r is:

E(r) = E₁(r) + E₂(r) = k*3q/(r - r₁)^2 + -k*2*q/(r - r₂)^2

E(r) = k*q*( 3/(r - r₁)^2 - 2/(r - r₂)^2)

Then if the we want to find the points r = (x, y, z) such that:

E(r) = 0 = k*q*( 3/(r - r₁)^2 - -k*2*q/(r - r₂)^2)

Then we must have:

0 = ( 3/(r - r₁)^2 - 2/(r - r₂)^2)

Also remember that the distance between two points:

(x, y, z) and (x', y', z') is given by:

D = √( (x - x')^2 + (y - y)^2 + (z -z')^2)

Then we can rewrite:

r - r₁ = √( (x - (-d))^2 + (y - 0 )^2 + (z -0)^2)

       = √( (x + d))^2 + y^2 + z^2)

and

r - r₂ =  √( (x - d)^2 + (y - 0 )^2 + (z -0)^2)

       = √( (x - d))^2 + y^2 + z^2)

Replacing that in our equation we get:

0 = ( 3/(√( (x + d))^2 + y^2 + z^2))^2 - -k*2*q/(√( (x - d))^2 + y^2 + z^2))^2)

0 = (3/((x + d))^2 + y^2 + z^2) - 2/ (x - d))^2 + y^2 + z^2)

We want to find the values of x, y, z such that the above equation is true.

2/ (x - d))^2 + y^2 + z^2) = (3/((x + d))^2 + y^2 + z^2)

2*[((x + d))^2 + y^2 + z^2] = 3*[(x - d))^2 + y^2 + z^2]

2*(x + d)^2  + 2*y^2 + 2*z^2 = 3*(x - d)^2 + 3*y^2 + 3*z^2

2*(x + d)^2 - 3*(x - d)^2 =  3*y^2 + 3*z^2 -  2*y^2 - 2*z^2

2*(x + d)^2 - 3*(x - d)^2  = y^2 + z^2

2*x^2 + 2*2*x*d + 2*d^2 -  3*x^2 + 3*2*x*d - 3*d^2 = y^2 + z^2

-x^2 + 10*x*d - d^2 = y^2 + z^2

we can rewrite this as:

- ( x^2 - 10*x*d + d^2) =  y^2 + z^2

now we can add and subtract 24*d^2 inside the parenthesis to get

- ( x^2 - 10*x*d + d^2 + 24*d^2 - 24*d^2) =  y^2 + z^2

-( x^2 - 2*x*(5d) + 25d^2 - 24d^2) = y^2 + z^2

-(x^2 - 2*x*(5d) + (5*d)^2) + 24d^2 = y^2 + z^2

The thing inside the parenthesis is a perfect square:

-(x - 5d)^2 + 24d^2 = y^2 + z^2

we can rewrite this as:

24d^2 = y^2 + z^2 + (x - 5d)^2

This equation gives us the points (x, y, z) such that the electric field is zero.

Where we need to replace two of these values to find the other, for example, if y = z = 0

24d^2 = (x - 5d)^2

√(24d^2)  = x - 5d

√24*d = x - 5d

√24*d + 5d = x

so in the point (√24*d + 5d, 0, 0) the net field is zero.

Answer:

4.60 × 10⁻⁸

Explanation:

From the given information;

Assuming that q charges are transferred, then:

[tex]F = \dfrac{kq^2}{d^2}[/tex]

where;

k = 9 ×10⁹

[tex]900000 = \dfrac{9*10^9 \times q^2}{1.2^2}[/tex]

[tex]q = \sqrt{\dfrac{900000\times 1.2^2 }{9*10^9}}[/tex]

q = 0.012 C

No of the electrons transferred is:

[tex]= \dfrac{0.012}{1.6\times 10^{-19}} C[/tex]

[tex]= 7.5 \times 10^{16} \ C[/tex]

Initial number of electrons =  N × 47 × no  of moles

here;

[tex]\text{ no of moles }= \dfrac{6.2}{107.87}[/tex]

no of moles = 0.0575 mol

∴

Initial number of electrons =  [tex]6.023\times 10^{23} \times 47 \times 0.0575 mol[/tex]

= 1.63 × 10²⁴

The fraction of electrons transferred  [tex]=\dfrac{7.5\times 10^{16} }{1.6 3\times 10^{24}}[/tex]

= 4.60 × 10⁻⁸

Answer:

c. kilometers and kilograms

Explanation:

If two units represent different things, like for example grams and joules, where the first one is a unit for mass, and the second one a unit for energy we can not add them together. (

Let's analyze each one of the given options:

a: grams and milligrams

both grams and milligrams are units of mass, so after the appropriate unit conversions, we can add them.

b: miles and kilometers

Both are units for distance, after the appropriate unit conversions, we can add them.

c: kilometers and kilograms

"Kilometer" is a unit for distance, and "kilogram" a unit for mass, we can not add distance and mass, so these two can not be added together.

d: centimeters and yards

Both are units of distance,  after the appropriate unit conversions, we can add them.

Then the only one that we can not add together is option c:  kilometers and kilograms

Answer: The correct option is C (kilometers and kilograms).

Explanation:

UNITS is the quantity of a constant magnitude which is used to measure the magnitudes of other quantities of the same nature. The magnitude of a physical quantity is expressed as, Physical quantity = (numerical value) × (unit). There are six basic units of measurement which includes:

--> metre (m) - unit of length.

--> kilograms (kg) - unit of mass.

--> second (s) - unit of time.

--> ampere (A) - unit of electrical current.

--> kelvin (K) - unit of temperature and

--> mole (mol) - unit of the amount of substance.

Mass is the quantity of matter present in a body which is constant and is measured in KILOGRAMMES or GRAMS. While position is the distance and direction of an object from a particular reference point which is known to vary and is measured in METERS OR KILOMETRES.

From the above definition, KILOGRAMMES is used to measure the magnitude of a different quantity from KILOMETERS and therefore should not be added up after conversation.

Answer:

225,000 kilometers per second

Explanation:

Have a nice day

Answer:

the law of conservation of energy.

Explanation:

An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.

Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.

In Physics, a point where at least three circuit paths (wires) meet is referred to as a junction.

The Kirchhoff’s circuit laws are two(2) equations first published by Gustav Kirchhoff in 1845.

Fundamentally, they address the conservation of energy and charge in the context of electrical circuits.

One of the laws known as Kirchoff's Current Law (KCL) deals with the principle of application of conserved energy in electrical circuits.

Kirchoff's Current Law (KCL) states that the sum of all currents entering a junction must equal the sum of all currents leaving the junction.

This simply means that the algebraic sum of currents in a network of conductors(wires) meeting at a point is equal to zero.

The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.

This ultimately implies that, Kirchhoff's junction rule is a statement of the law of conservation of energy.

Answer:

eqcubuohwehuuc

Explanation:

the the the the the the the the the the the the the the the the dhueirrjhrhjrirjrheh3jeiiruj fnrhfjjjrjrj fjfiirrjejrjejej jrkrjrjrjrjrjdjdjfjrhruriruru rjridjhwjjsjd

Answer:

1) 0.3675

2) 0.367

3) 72.6186

4) 0.5

5) 0.000671

Answer:

1) 367.5 mg = 0.3675 g

2) 367 mL = 0.367 L

3) 28.59 in = 72.61 cm

4) 8 0z = 0.5 lb

5) 0.671 mm = 0.0000671 m

RQ/PQ I think

rise/run