A 590-turn solenoid is 12 cm long. The current in it is 36 A . A straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).
What is the magnitude of the force on this wire assuming the solenoid's field points due east?

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:

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:

Explanation:

1.The light reflected from the CD/DVD or soap bubble or oil film forms an interference with the surrounding light. The inference both constructive and destructive making some color appear and some disappear.

2.As light behaves as wave it will interfere differently at different angles. At certain angle it will interfere constructively and at certain angle it will interfere destructively making some color brighter and some disappear. So, at different angles the color are different.

Interference pattern is responsible for the formation of different colour when a light reflected from CD or soap bubble.

We can see colors when we look at reflected light from a CD or DVD disk, or a soap bubble or oil film on water because of the interference pattern. The colors that we see on the CD are created due to the reflection of white light from ridges in the metal. When light passes through something with many small ridges or scratches, we often see rainbow colors and interesting patterns.

These patterns are called interference patterns. White light is made up of 7 colors i.e. red, orange, yellow, green, blue, indigo, violet. The CD converts or separates the white light into 7 colors so we can conclude that interference pattern is responsible for the formation of different colour when a light reflected from CD OR soap bubble.

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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:

[tex]y = 0.0394 \ m[/tex]

Explanation:

From the question we are told that

        The  distance of the screen is  [tex]D = 2.20 \ m[/tex]

       The distance of separation of the slit is  [tex]d = 0.0328 \ mm = 0.0328*10^{-3} \ m[/tex]

        The  wavelength of light is  [tex]\lambda = 588 \ nm = 588 *10^{-9} \ m[/tex]

Generally the condition for constructive interference is

            [tex]dsin\theta = n * \lambda[/tex]

=>        [tex]\theta = sin^{-1} [ \frac{ n * \lambda }{d } ][/tex]

here n = 1 because we are considering the central diffraction peak

=>        [tex]\theta = sin^{-1} [ \frac{ 1 * 588*10^{-9} }{0.0328*10^{-3} } ][/tex]

=>       [tex]\theta = 1.0274 ^o[/tex]

Generally the width of central diffraction peak on a screen is mathematically evaluated as

           [tex]y = D tan (\theta )[/tex]

substituting values

        [tex]y = 2.20 * tan (1.0274)[/tex]

        [tex]y = 0.0394 \ m[/tex]

Answer:

the interation blood veins

Explanation:

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:

A. 110Hz,220Hz, 330 Hz

Explanation:

for organ open at open both ends;

the length of the organ for the fundamental frequency, L = A---->N + N----->A

A---->N  = λ /4 and N----->A = λ /4

L = λ /4 + λ /4 = λ /2

[tex]L = \frac{\lambda}{2} \\\\\lambda = 2L[/tex]

λ  = 2 x 1.5m = 3.0 m

Wave equation is given by;

V = Fλ

Where;

V is the speed of sound

F is the frequency of the wave

F = V/ λ

F₀ = V / 2L

Where;

F₀  is the fundamental frequency

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

the length of the organ for the first overtone, L = A---->N + N----->A + A----->N +  N----->A

L = 4λ /4

L = λ

λ = 1.5 m

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

For open organ at one end

the length of the organ for the fundamental frequency, L = N------A

L = λ /4

λ = 4L

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

the length of the organ for the first overtone, L = N-----N + N-----A

L = λ/2 + λ / 4

L = 3λ /4

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Thus the fundamental frequency for both organs is 110 Hz,

The first overtone for the organ open at both ends is 220 Hz

The first overtone for the organ open at one end is 330 Hz

The correct option is "A. 110Hz,220Hz, 330 Hz"

The correct option is option (A)

the frequencies produced by the pipes are (A) 110Hz,220Hz, 330 Hz

(I) For an organ pipe open at open both ends the frequency of different modes is given by:

F =  nv/2L

where

F is the frequency

L is the length of the organ pipe

v is the speed of the wave

and, n is the mode of frequency

the fundamental frequency corresponds to n = 1, given by:

F₀ = v/2L

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

The first overtone corresponds to n = 2, the second overtone corresponds to n = 3, and so on...

F₁ =2v/2L

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

The difference between successive overtones is F₀

(II) For an organ pipe open at one end the frequency of different modes is given by:

F =  nv/4L

where

F is the frequency

L is the length of the organ pipe

v is the speed of the wave

and, n is the mode of frequency

the fundamental frequency corresponds to n = 1, given by:

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

For an organ pipe open at one end, only those overtones are present which correspond to odd n, that is n = 3,5,...so:

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

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Answer:

Explanation:

At the point midway between wires

magnetic field due to wire having current 2I₀

= 10⁻⁷ x 2 x2I₀ / r     where 2r is the distance between wires .

magnetic field due to wire having current I₀

= 10⁻⁷ x 4 I₀ / r

magnetic field due to wire having current I₀

= 10⁻⁷ x 2I₀ / r    

= 10⁻⁷ x 2 I₀ / r     where 2r is the distance between wires .

these fields are in opposite direction as direction of current is same in both .

net magnetic field = (4 - 2 )x 10⁻⁷ x I₀ / r

= 2 x 10⁻⁷ x  I₀ / r

At point A net magnetic field = 2 x 10⁻⁷ x  I₀ / r

At point B , we shall calculate magnetic field

magnetic field due to nearer wire having current  2 I₀ = 10⁻⁷ x 4 I₀ / r

magnetic field due to wire far away = 10⁻⁷ x 2 I₀ / 3r

These magnetic fields act in the same direction so they will add up

net magnetic field = [ (4 I₀ / r)  + (2 I₀ / 3r) ] x 10⁻⁷

= (14 I₀ / 3r ) x 10⁻⁷

Magnetic field at point B = (14 I₀ / 3r ) x 10⁻⁷

Ratio of field at A and B

= 3 / 7 . Ans

The ratio of the magnitude of the magnetic field at point A to point B is :

3 / 7

Given data :

Magnitude of the left current is  2I₀

Magnitude of the right current is  I₀

First step : Determine the magnetic field at point A  

The magnetic field due to the left current ( 2I₀ )

10⁻⁷ * 2 * 2I₀ / r       ( 2r = distance between wires )

The magnetic field due to the right current ( I₀ )

10⁻⁷ * 2 I₀ / r

From the expressions above the magnetic fields are in  opposite direction

∴ Net magnetic field = (4 - 2 )* 10⁻⁷ * I₀ / r =   2 * 10⁻⁷ *  I₀ / r

Hence The magnetic field at point A = 2 * 10⁻⁷ *  I₀ / r

Next step : determine the magnetic field at point B

Magnetic field due to the closest wire to point B ( i.e.2I₀ ) = 10⁻⁷ * 4 I₀ / r

Magnetic field due to the wire away from point A = 10⁻⁷ * 2 I₀ / 3r

Since the fields acts in the same directions

The net magnetic field =  (4 I₀ / r)  + (2 I₀ / 3r) ] * 10⁻⁷ = ( 14 I₀ / 3r ) * 10⁻⁷

Hence The magnetic field at point A = ( 14 I₀ / 3r ) * 10⁻⁷

Therefore the ratio of the magnitude of the magnetic field at point A to point B  =  3/ 7

Hence we can conclude that the ratio of the magnitude of the magnetic field at point A to point B  = 3 / 7

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Answer:

2/5 m/s

Explanation:

There are two vectors  v and w . Let θ be angle b/w the two vector.

[tex]cos\theta =\frac{\overleftarrow{v}\cdot \overleftarrow{w}}{\left | v \right |\left | w \right |}\\=\frac{6-4}{\sqrt(2^2+1^2)\sqrt(3^2+4^2)} =\frac{2}{5\sqrt(5)}[/tex]

velocity of the ball in direction of the the wind

[tex]\left | vcos\theta \right |\\\left | v \right |cos\theta\\\sqrt(2^2+1^2)\frac{2}{5\sqrt(5)} = \frac{2}{5}[/tex]

The size of the velocity of the ball in the direction of the wind is 2/5 ms.

Since there are two vectors v and w

Also, here we assume θ be angle b/w the two vector.

So

Cos θ = 6-4 / √(2^2 + 1^2) √(3^2 + 4^2)

= 2/5√5

Now the velocity of the ball should be

= √(2^2 + 1^2) 2 ÷ 5√(5)

= 2 /5

hence, The size of the velocity of the ball in the direction of the wind is 2/5 ms.

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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:

The values is  [tex]m_{max} = 8001 \ bright \ spots[/tex]

Explanation:

From the question we are told that

    The slit distance is  [tex]d = 2 \ mm = 2*10^{-3} \ m[/tex]

    The  wavelength is  [tex]\lambda = 500 \ nm = 500 *10^{-9} \ m[/tex]

At the first half of the screen from the central maxima

   The number of bright spot according to the condition for constructive interference is  

          [tex]n = \frac{d * sin (\theta )}{\lambda}[/tex]

For maximum number of spot [tex]\theta = 90^o[/tex]

So  

       [tex]n = \frac{2*10^{-3} * sin (90 )}{500 *10^{-9}}[/tex]

        [tex]n =4000[/tex]

Now for the both sides plus the central maxima  we have

      [tex]m_{max} = 2 * n + 1[/tex]

substituting values

       [tex]m_{max} = 2 * 4000 + 1[/tex]

       [tex]m_{max} = 8001 \ bright \ spots[/tex]

Answer:

The water pressure on the upper pipe is 92.5 kPa.

Explanation:

Given that,

Pressure in lower pipe= 120 kPa

Speed of water in lower pipe= 1 m/s

Acceleration due to gravity = 10 m/s²

Density of water = 1000 kg/m³

Radius of lower pipe = 12 m

Radius of uppes pipe = 6 m

Height of upper pipe = 2 m

We need to calculate the velocity in upper pipe

Using continuity equation

[tex]A_{1}v_{1}=A_{2}v_{1}[/tex]

[tex]\pi r_{1}^2\times v_{1}=\pi r_{2}^2\times v_{2}[/tex]

[tex]v_{2}=\dfrac{r_{1}^2\times v_{1}}{r_{2}^2}[/tex]

Put the value into the formula

[tex]v_{2}=\dfrac{12^2\times1}{6^2}[/tex]

[tex]v_{2}=4\ m/s[/tex]

We need to calculate the water pressure on the upper pipe

Using bernoulli equation

[tex]P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}[/tex]

Put the value into the formula

[tex]120\times10^{3}+\dfrac{1}{2}\times1000\times1^2+1000\times10\times0=P_{2}+\dfrac{1}{2}\times1000\times(4)^2+1000\times10\times2[/tex]

[tex]120500=P_{2}+28000[/tex]

[tex]P_{2}=120500-28000[/tex]

[tex]P_{2}=92500\ Pa[/tex]

[tex]P_{2}=92.5\ kPa[/tex]

Hence, The water pressure on the upper pipe is 92.5 kPa.

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:

4 / 7

Explanation:

1/total resistance = 1/1 + 1/2 + 1/4

= 1¾

total resistance = 1 ÷ 1¾

= 4/7

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:

2500 kg/m³

Explanation:

P = P

ρgh = ρgh

ρh = ρh

(1000 kg/m³) (8.9 cm) = ρ (3.5 cm)

ρ ≈ 2500 kg/m³

Answer:

The net force is directed downwards.

Explanation:

Since the magnet is falling much more faster than it would unaided, then there is a net force that is accelerating the magnet downwards. We know that acceleration is due to a force acting on a mass, and in this case, the magnet is the mass. Also, the acceleration is always in the direction of the force producing it, which means that the net force on the magnet is vertically downwards.

Answer:

The number of windings is 1.

Explanation:

The radius of the solenoid = 8.0 cm = 0.08 m

Length of the solenoid = 45.0 cm = 0.45 m

number of turn = ?

circumference of each winding = 2πr = 2 x 3.142 x 0.08 = 0.503 m

The number of windings = (Length of the solenoid)/(circumference of each winding)

==> 0.45/0.503 = 0.89 ≅ 1

Answer:

A) E(r) = 1.3957 × 10^(5) N/C

B) E(r) = 9.8864 × 10⁴ N/C

C) E(r) = 1.13 × 10^(5) N/C

Explanation:

We are given;

q = 6 nc = 6 × 10^(-9) C

L = 10 cm = 0.1 m

d = 4.4 cm = 0.044 m

r1 = 1 cm = 0.01 m

r2 = 2 cm = 0.02 m

r3 = 3 cm = 0.03 m

Formula for the electric field strength in this question is given as;

E(r) = q/(2π(ε_o)rL) + q/(2π(ε_o)(d - r)L)

When factorized, we have;

E(r) = q/(2π(ε_o)L) × [(1/r) + (1/(d - r))]

Plugging in the relevant values for q/(2π(ε_o)L)

We know that (ε_o) has a constant value of 8.854 × 10^(−12) C²/N².m

Thus; q/(2π(ε_o)L) = (6 × 10^(-9))/(2π(8.854 × 10^(−12)0.1) = 1078.53

Thus;

E(r) = 1078.52 [1/r + 1/(d - r)]

A) E1 is at r = 1 cm = 0.01m

Thus;

E(r) = 1078.52 (1/0.01 + (1/(0.044 - 0.01))

E(r) = 1.3957 × 10^(5) N/C

B) E2 is at r = 2 cm = 0.02 m

Thus;

E(r) = 1078.52 (1/0.02 + (1/(0.044 - 0.02))

E(r) = 9.8864 × 10⁴ N/C

C) E2 is at r = 3 cm = 0.03 m

Thus;

E(r) = 1078.52 (1/0.03 + (1/(0.044 - 0.03))

E(r) = 1.13 × 10^(5) N/C

Given that,

Energy [tex]H=2.7\times10^{31}\ W[/tex]

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

[tex]H=Ae\sigma T^4[/tex]

[tex]A=\dfrac{H}{e\sigma T^4}[/tex]

[tex]4\pi R^2=\dfrac{H}{e\sigma T^4}[/tex]

[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]

Put the value into the formula

[tex]R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}[/tex]

[tex]R=5.0\times10^{10}\ m[/tex]

(b). Given that,

Radiates energy [tex] H=2.1\times10^{23}\ W[/tex]

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]

Put the value into the formula

[tex]R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}[/tex]

[tex]R=5.42\times10^{6}\ m[/tex]

Hence, (a). The radius of the star is [tex]5.0\times10^{10}\ m[/tex]

(b). The radius of the star is [tex]5.42\times10^{6}\ m[/tex]

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:

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:

Explanation:

When plastic balloon is rubbed with wool , charges are created on both balloon and silk in equal amount . Rubber balloon will acquire negative charge and silk will acquire positive charge .

Now when balloon is brought near a wall , there is induction of charge on the wall due to charge on the balloon . On the near surface of wall positive charge is produced and on the surface deep inside the wall negative charge is produced . The charge deep inside goes inside the earth but the positive charge near the surface of wall can not escape . It remains trapped by negative charge on the balloon .

hence there is mutual attraction between balloon and surface of wall is just like attraction between opposite charges . But once the ballon due to mutual attraction comes in contact with the wall , the charge on balloon and on wall neutralises each other and hence after some time the balloon falls off from the wall on the ground . It does not remain attracted to wall for ever . It happens due to neutralisation of charges on balloon and wall .

Answer:

a

    [tex]z= 2.5 \ m[/tex]

b

   [tex]z = (1 \ m , 4 \ m )[/tex]

Explanation:

From the question we are told that

     Their distance apart is  [tex]d = 5.00 \ m[/tex]

      The  wavelength of each source wave [tex]\lambda = 6.0 \ m[/tex]

Let the distance from source A  where the construct interference occurred be z

Generally the path difference for constructive interference is

              [tex]z - (d-z) = m \lambda[/tex]

Now given that we are considering just the straight line (i.e  points along the line connecting the two sources ) then the order of the maxima m =  0

  so

        [tex]z - (5-z) = 0[/tex]

=>     [tex]2 z - 5 = 0[/tex]

=>     [tex]z= 2.5 \ m[/tex]

Generally the path difference for destructive  interference is

           [tex]|z-(d-z)| = (2m + 1)\frac{\lambda}{2}[/tex]

=>         [tex]|2z - d |= (0 + 1)\frac{\lambda}{2}[/tex]

=>        [tex]|2z - d| =\frac{\lambda}{2}[/tex]

substituting values

          [tex]|2z - 5| =\frac{6}{2}[/tex]

=>      [tex]z = \frac{5 \pm 3}{2}[/tex]

So  

      [tex]z = \frac{5 + 3}{2}[/tex]

      [tex]z = 4\ m[/tex]

and

      [tex]z = \frac{ 5 -3 }{2}[/tex]

=>   [tex]z = 1 \ m[/tex]

=>    [tex]z = (1 \ m , 4 \ m )[/tex]

Answer:

The maximum temperature increase is [tex]\Delta T = 0.0497 \ ^oC[/tex]

Explanation:

From the question we are told that

    The mass of the bullet is [tex]m = 17.0 \ g =0.017 \ kg[/tex]

     The  speed is  [tex]v_1 = 785 \ m/s[/tex]

     The mass of the water is  [tex]m_w = 13.5 \ kg[/tex]

     The velocity it emerged with is  [tex]v_2 = 534 \ m/s[/tex]

Generally due to the fact that energy can nether be created nor destroyed but transferred from one form to another then  

the change in kinetic energy of the bullet =  the heat gained by the water

 So

 The change in kinetic energy of the water is  

          [tex]\Delta KE = \frac{1}{2} m (v_1^2 - v_2 ^2 )[/tex]

substituting values  

        [tex]\Delta KE =0.5 * 0.017 * (( 785)^2 - (534) ^2 )[/tex]

        [tex]\Delta KE = 2814.1 \ J[/tex]

Now the heat gained by the water is

     [tex]Q = m_w* c_w * \Delta T[/tex]

Here [tex]c_w[/tex] is the specific heat of water which has a value  [tex]c_w = 4190 J/kg \cdot K[/tex]

So  since   [tex]\Delta KE = Q[/tex]  

we have that

          [tex]2814.1 = 13.5 * 4190 * \Delta T[/tex]

          [tex]\Delta T = 0.0497 \ ^oC[/tex]

Answer:

Answer:

A. Increasing the number of lines per length.

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:

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:

d. unchanged.

Explanation:

The frequency of a wave is dependent on the speed of the wave and the wavelength of the wave. The frequency is characteristic for a wave, and does not change with distance. This is unlike the amplitude which determines the intensity, which decreases with distance.

In a wave, the velocity of propagation of a wave is the product of its wavelength and its frequency. The speed of sound does not change with distance, except when entering from one medium to another, and we can see from

v = fλ

that the frequency is tied to the wave, and does not change throughout the waveform.

where v is the speed of the sound wave

f is the frequency

λ is the wavelength of the sound wave.