What is the magnitude of the applied electric field inside an aluminum wire of radius 1.4 mm that carries a 4.5-A current

Answer: 0.0617

Explanation:

Given: The probability of wet weather on any given day in a city of Punjab : p=15%=0.15

Let X be a binomial variable that represents the number of days having wet weather.

Binomial probability formula : [tex]P(X=x)=^nC_xp^x(1-p)^x[/tex], where n= total outcomes, p = probability of success in each outcomes.

Here, n= 7 ( 1 week = 7 days)

The probability that it will take a week for it three wet weather on 3 separate days:

[tex]P(X=3)^=\ ^7C_3(0.15)^3(1-0.15)^{7-3}\\\\=\dfrac{7!}{3!(7-3)!}(0.15)^3(0.85)^4\\\\=\dfrac{7\times6\times5}{3\times2}\times 0.003375\times0.52200625\approx0.0617[/tex]

Hence, the required probability =0.0617

Answer:

the interation blood veins

Explanation:

Answer:

4 x 10¹⁵

Explanation:

Answer:

the analog along the diameter of the acceleration of the particle executing simple harmonic motion is the projection along the diameter of the centripetal acceleration of the particle in the circle

Answer:

Stay the same

Explanation:

Since, friction is negligible:

Initial Momentum = Final Momentum

Initial KE = Final KE

m1 * v1 = m2 * v2

When m increases v decreases.

The momentum and kinetic energy remain the same if you drop paper clips into an open cart rolling along a straight horizontal track with negligible friction.

Between two surfaces that are sliding or attempting to slide over one another, there is a force called friction. For instance, friction makes it challenging to push a book down the floor. Friction always moves an object in a direction that is counter to the direction that it is traveling or attempting to move.

Given:

The paperclips into an open cart rolling along a straight horizontal track with negligible friction,

Calculate the momentum, Since friction is negligible,

Initial Momentum = Final Momentum

Initial Kinetic Energy = Final Kinetic Energy

m₁ × v₁ = m₁  × v₂

When m increases, v decreases,

Thus, momentum will remain the same.

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Answer:

hello the needed diagram is missing attached below is the diagram and the detailed solution

The tension in BC = [tex]\frac{\sqrt{2} }{4} mg[/tex]

Explanation:

ATTACHED BELOW IS THE DETAILED SOLUTION T THE GIVEN PROBLEM

Ma = mg - T/ [tex]\sqrt{2}[/tex]  equation 1

Ma = 3T / [tex]\sqrt{2}[/tex]   equation 2

equate both equations to determine the tension on BC

Answer:

3x10^5m/s

Explanation:

See attached file

Explanation:

The speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].

According to the Doppler effect, the difference between the frequency at which light wave leave a source and reaches an observer is caused by the relative motion of the observer and the wave source.

Given that the difference in the frequency is 0.10 %. The speed of light emitted from the galaxy can be calculated by the Doppler effect.

[tex]\dfrac {\Delta f}{f} = \dfrac {v}{c}[/tex]

Where f is the frequency of the light, v is the speed of light emitted from the galaxy and c is the speed of light emitted from the earth.

[tex]\dfrac {0.10 f}{100 f} = \dfrac {v}{3\times 10^8}[/tex]

[tex]v = 3\times 10^5\;\rm m/s[/tex]

Hence we can conclude that the speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].

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Answer:

Explanation:

Find:

Computation:

F = (9/5)C + 32

3 times that of the Celsius

If C = x

So F = 3x

So,

3x = (9/5)x + 32

15x = 9x +160

6x = 160

x = 26.67

So, C = 26.67° and F = 80°

1/5 times that of the Celsius

If C = x

So F = x/5

So,

x/5 = (9/5)x + 32

x = 9x + 160

x = -20

So, C = -20° and F = -4°

Answer:

6.9kgm²

Explanation:

For an axis through the center of the rectangle, I = m[(w²+L²)/12

Using the parallel axis theorem, the added value of I = mR² = m[(w²/4 + L²/4]

Adding the 2 expressions,

I = (m/3)*(w²+L²)

I =6.95 kg∙m²

Answer

A)184.9J

B)=3.63m/s

C) Zero

Explanation:

A)potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is

U=Mgh

Where m=28kg

g= 9.8m/s

h= difference in height between the initial position and the bottom position

We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical

h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)

=0.674m

Her Potential Energy will now

= 28× 9.8×0.674

=184.9J

B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

E= 0.5mv^2

where

m = 28.0 kg is the mass of the child

v is the speed of the child at the bottom position

Solving the equation for v, we find

V=√2k/m

V=√(2×184.9/28

=3.63m/s

C)we can find work done by the tension in the rope is given using expresion below

W= Tdcosx

where W= work done

T is the tension

d = displacement of the child

x= angle between the directions of T and d

In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. x= 90∘ and cos90∘=0 hence, the work done is zero.

Answer:

in an accident, when the body collides with the air bags, the collision time of impact between the two bodies will increase due to the presence of air bags in the car. Larger is the impact time smaller is the transformation of energy between the body and air bag. That is why air bags greatly reduce the chance of injury in a car accident.

It depends on what stage of the mission you're talking about.

==>  While it's sitting on the pad before launch, the forces on the rocket are balanced, so there's no net force on it.

==>  When the engines ignite, their thrust (d) is greater than the force of gravity.  So the net force on the rocket is upward, and the spacecraft accelerates upward.

==>  After the engines shut down, the net force acting on the rocket is due to Gravity (c).

. . . If the rocket has enough vertical speed, it escapes the Earth completely, and just keeps going.  

. . . If it has enough horizontal speed, it enters Earth orbit.  

. . . If it doesn't have enough vertical or horizontal speed, it falls back to Earth.    

A rocket will preserve to speed up so long as there's a resultant pressure upwards resulting from the thrust of the rocket engine.

A rocket launches whilst the pressure of thrust pushing it upwards is greater than the burden force because of gravity downwards. This unbalanced pressure reasons a rocket to accelerate upwards. A rocket will maintain to hurry up so long as there's a resultant force upwards resulting from the thrust of the rocket engine.

What's the net pressure of unbalanced?

If the forces on an item are balanced, the net pressure is zero. If the forces are unbalanced forces, the results do not cancel each difference. Any time the forces acting on an object are unbalanced, the net pressure is not 0, and the movement of the item modifications.

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Answer:

The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]

Explanation:

Given that,

A beam of light from a laser illuminates a glass.

Suppose, the length of piece is [tex]L=25.21\times10^{-2}\ m[/tex]

Index of refraction is 2.83.

We need to calculate the speed of light pulse in glass

Using formula of speed

[tex]v=\dfrac{c}{\mu}[/tex]

Put the value into the formula

[tex]v=\dfrac{3\times10^{8}}{2.83}[/tex]

[tex]v=1.06\times10^{8}\ m/s[/tex]

We need to calculate the time of short pulse of light beam

Using formula of velocity

[tex]v=\dfrac{d}{t}[/tex]

[tex]t=\dfrac{d}{v}[/tex]

Put the value into the formula

[tex]t=\dfrac{25.21\times10^{-2}}{1.06\times10^{8}}[/tex]

[tex]t=2.37\times10^{-9}\ sec[/tex]

Hence, The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]

Answer:

 v_{1fy} = - 0.4549 m / s

Explanation:

This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved

initial. Before the crash

      p₀ = m v₁₀

final. After the crash

      [tex]p_{f}[/tex] = m [tex]v_{1f}[/tex] + m v_{2f}

Recall that velocities are a vector so it has x and y components

       p₀ = p_{f}

we write this equation for each axis

X axis

       m v₁₀ = m v_{1fx} + m v_{2fx}

Y Axis  

       0 = -m v_{1fy} + m v_{2fy}

the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components

      sin 23.3 = v_{2fy} / v_{2f}

      cos 23.3 = v_{2fx} / v_{2f}

      v_{2fy} = v_{2f} sin 23.3

      v_{2fx} = v_{2f} cos 23.3

we substitute in the momentum conservation equation

       m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3

       0 = - m v_{1f} sin θ + m v_{2f} sin 23.3

      1.83 = v_{1f} cos θ + 1.15 cos 23.3

       0 = - v_{1f} sin θ + 1.15 sin 23.3

      1.83 = v_{1f} cos θ + 1.0562

        0 = - v_{1f} sin θ + 0.4549

     v_{1f} sin θ = 0.4549

     v_{1f}  cos θ = -0.7738

we divide these two equations

      tan θ = - 0.5878

      θ = tan-1 (-0.5878)

       θ = -30.45º

we substitute in one of the two and find the final velocity of the incident ball

        v_{1f} cos (-30.45) = - 0.7738

        v_{1f} = -0.7738 / cos 30.45

        v_{1f} = -0.8976 m / s

the component and this speed is

       v_{1fy} = v1f sin θ

       v_{1fy} = 0.8976 sin (30.45)

       v_{1fy} = - 0.4549 m / s

Answer:

the large amplitude wave keeps moving to the right

the small amplitude wave continues to move to the left.

When checking the answers, the correct ones are 1, 2

Explanation:

The waves fulfill the principle of superposition, which states that the value of the function at a point is the algebraic sum of the waves at a given instant.

The two waves in this exercise travel in the opposite direction, so when they are close, the resulting wave is the sum of the two waves, having a complicated shape. But when the waves follow their movement, they give in the same way as the initial a,

the large amplitude wave keeps moving to the right

the small amplitude wave continues to move to the left.

When checking the answers, the correct ones are 1, 2

Answer:

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Explanation:

The capacitance of a parallel plate capacitor is given by the following formula:

C = ε₀A/d

where,

C = Capacitance

ε₀ = Permeability of free space

A = Area of plates

d = Distance between plates

FOR CAPACITOR A:

C = CA = 17.8 nF = 17.8 x 10⁻⁹ F

A = A₁

d = d₁

Therefore,

CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F   ----------------- equation 1

FOR CAPACITOR B:

C = CB = ?

A = A₁/2

d = 2 d₁

Therefore,

CB = ε₀(A₁/2)/2d₁

CB = (1/4)(ε₀A₁/d₁)

using equation 1:

CB = (1/4)(17.8 X 10⁻⁹ F)

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Answer:

1.1micro meter

Explanation:

Given that

Constructive interference is

ma = alpha x sin theta

Alpha = 1 x 594 x10^ -9/ sin 32.8°

= 1.1 x 10^ -6m

Explanation:

Answer:

3x10⁴v

Explanation:

Using

Wavelength= h/ √(2m.Ke)

880nm = 6.6E-34/√ 2.9.1E-31 x me

Ke= 6.6E-34/880nm x 18.2E -31.

5.6E-27/18.2E-31

= 3 x 10⁴ Volts

Answer:

R = 36.885 km

Explanation:

In order to distinguish the two planes we must use the Rayleigh criterion that establishes two distinguishable objects if in their diffraction the central maximum of one coincides with the first minimum of the other

The diffraction equation for slits is

            a sin θ = m λ

the first minimum occurs for m = 1

             sin θ = λ a

as the diffraction experiments the angles are very small, we approximate

             sin θ = θ

             θ = λ / a

This expression is for a slit, in the case of circular objects, when solving the system in polar coordinates, a numerical constant appears, leaving the expression of the form

            θ = 1.22 λ / a

In this problem they give us the frequency, let's find the wavelength with the relation

           c = λ f

           λ = c / f

           θ = 1.22 c/ f a

since they ask us for the distance between the planes, we can use the definition of radians

          θ = s / R

if we assume that the distance is large, we can approximate the arc to the horizontal distance

          s = x

we substitute

             x / R = 1.22 c / fa

             R = x f a / 1.22c

Let's reduce the magnitudes to the SI system

            f = 9000 MHz = 9 109 Hz

            a = 15 m

           x = 100 m

let's calculate

            R = 100 10⁹ 15 / (1.22 3 108)

            R = 3.6885 10⁴ m

let's reduce to km

            R = 3.6885 10¹ km

            R = 36.885 km

Answer:

 P = 2923.89 W  

Explanation:

Power is

     P = F v

for which we must calculate the force, let's use Newton's second law, let's set a coordinate system with a flat parallel axis and the other axis (y) perpendicular to the plane

X Axis  

         F - Wₓ = 0

         F = Wₓ

Y Axis

         N -  [tex]W_{y}[/tex] = 0

let's use trigonometry for the components of the weight

         sin 6 = Wₓ / W

         cos 6 = W_{y} / W

         Wₓ = W sin 6

         W_{y} = W cos 6

          F = mg cos 6

          F = 75 9.8 cos 6

          F = 730.97 N

let's calculate the power

        P = F v

        P = 730.97 4.0

        P = 2923.89 W

Answer:

Number of turns of wire(N) = 3,036 turns (Approx)

Explanation:

Given:

Diameter = 13 Cm

emf = 5.6 v

Note:

The given question is incomplete, unknown information is as follow.

Magnetic field increases = 0.25 T in 1.8 (Second)

Find:

Number of turns of wire(N)

Computation:

radius (r) = 13 / 2 = 6.5 cm = 0.065 m

Area = πr²

Area = (22/7)(0.065)(0.065)

Area = 0.013278 m²

So,

emf = (N)(A)(dB / dt)

5.6 = (N)(0.013278)(0.25 / 1.8)

5.6 = (N)(0.013278)(0.1389)

N = 3,036.35899

Number of turns of wire(N) = 3,036 turns (Approx)

Answer:

From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases

Explanation:

I cannot  find any attached photo, but we can proceed anyways theoretically.

The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point

i.e

[tex]E=\frac{F}{Q}[/tex]

But the force F

[tex]F= \frac{kQ1Q2}{r^2}[/tex]

But the electric field intensity due to a point charge Q at a distance r meters away is given by

[tex]E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }[/tex]

From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases

Answer:

10.8rev

Explanation:

Using

Wf²-wf = 2 alpha x theta

0²- 56.36x56.36/ 2(-20.13) x theta

Theta = 68.09 rad

But 68.09/2π

>= 10.8 revolutions

Explanation:

Answer:

The electric field strength needed is 4 x 10⁵ N/C

Explanation:

Given;

magnitude of magnetic field, B = 0.1 T

velocity of the charge, v = 4 x 10⁶ m/s

The velocity of the charge when there is a balance in the magnetic and electric force is given by;

[tex]v = \frac{E}{B}[/tex]

where;

v is the velocity of the charge

E is the electric field strength

B is the magnetic field strength

The electric field strength needed is calculated as;

E = vB

E = 4 x 10⁶ x 0.1

E = 4 x 10⁵ N/C

Therefore, the electric field strength needed is 4 x 10⁵ N/C

Answer:

81 N

7.1 m/s²

Explanation:

Draw a free body diagram of the sphere.  There are two forces:

Weight force mg pulling straight down,

and tension force T pulling up along the rope.

Sum of forces in the centripetal direction:

∑F = ma

T − mg sin 45° = m v² / r

T = m (g sin 45° + v² / r)

T = (7 kg) (10 m/s² sin 45° + (3 m/s)² / 2 m)

T = 81 N

Sum of forces in the tangential direction:

mg cos 45° = ma

a = g cos 45°

a = (10 m/s²) cos 45°

a = 7.1 m/s²

Answer:

364days

Explanation:

Pls see attached file

Explanation:

The moon will take 112.7 days to make one revolution around the planet.

The period of the satellite around any planet only depends upon the distance between the planet's center and satellite and also depends upon the planet's mass.

Given, the distance from the moon's center to the planet's surface,

h = 2.165 × 10⁵ km,

The radius of the planet, r = 4175 km  

The mass of the planet = 6.70 × 10²² kg

The total distance between the moon's center to the planet's center:

a = r +h = 2.165 × 10⁵ + 4175

a = 216500 + 4175

a = 220675

a = 2.26750 × 10⁸ m

The period of the planet can be calculated as:

[tex]T =2\pi \sqrt{\frac{a^3}{Gm} }[/tex]

[tex]T =2\3\times 3.14 \sqrt{\frac{(2.20675 \times 10^8)^3}{(6.67\times 10^{-11}).(6.70\times 10^{22})} }[/tex]

T = 9738253.26 s

T = 112.7 days

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Answer:D

Explanation:I got it right

Answer:

They needed to be able to unite even though they spoke different languages.

Explanation:

Answer:

1.37 ×10^-3 T

Explanation:

From;

B= μnI

μ = 4π x 10-7 N/A2

n= number of turns /length of wire = 1700/0.75 = 2266.67

I= 0.48 A

Hence;

B= 4π x 10^-7 × 2266.67 ×0.48

B= 1.37 ×10^-3 T

Answer:

Option (c) : 20°C

Explanation:

[tex]t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2} [/tex]

T(final) = 500* 10 + 100*70/600 = 20°C

Answer:

51. 7m/s

Explanation:

Take speed of sound in air = 340 m/s

fp = fs (V + Vp)/(V + Vs)

128 = 123 (340 + Vp)/(340 + 36.4)

Vp = 51.7m/s

Explanation: