Air at 1 atm, 158C, and 60 percent relative humidity is first heated to 208C in a heating section and then humidified by introducing water vapor. The air leaves the humidifying section at 258C and 65 percent relative humidity. Determine (a) the amount of steam added to the air, and (b) the amount of heat transfer to the air in the heating section.

Answer:

absorbent    p = S / c

reflective         p = 2S/c

Explanation:

The moment of radiation on a surface is

          p = U / c

where U is the energy and c is the speed of light.

In the case of a fully absorbent object, the energy is completely absorbed. The energy carried by the light is given by the Poynting vector.

           p = S / c

in the case of a completely reflective surface the energy must be absorbed and remitted, therefore there is a 2-fold change in the process

           p = 2S/c

Answer:

a. The current is induced when there is ____3. EITHER DECREASING OR INCREASING___ magnetic flux through a closed loop of wire.

b. If the magnetic flux was constant then there _______WILL BE NO___ induced current regardless of the magnetic flux value.

c. If the magnetic flux was not constant then there __WILL BE______ induced current regardless of the magnetic flux value.

Explanation:

THIS IS BECAUSE IF FARADAY'S LAW OF ELECTROMAGNETIC INDUCTION WHICH STATES THAT WHENEVER THERE IS A CHANGE IN MAGNETIC LINES OF FORCE LINKED WITH A CLOSED CIRCUIT AN EMF IS INDUCED

Answer:

1.8 x 10^-8A

Explanation:

Using the equation for current:

I= Q/t

current = (9X10^-12)/ (0.50X 10^-3)

= 1.8^-8A

Answer:

Explanation:

jjjjjjjjjjjjjjjj

Answer:

B. qualitative

Explanation:

Since in the question it is mentioned that it takes the teacher opinion with respect to the student lunchroom that arranged the opinion in four classifications and according to that the report is also written

So this research represents the quality of the data collected as it does not specify the rating in terms of liking the student lunchroom

Therefore the correct option is B. qualitative

Answer:

e. The speed of the object is a constant 6 m/s

Explanation:

Since the net force is towards the centre , hence there is no tangential acceleration . Only centripetal acceleration is there . Hence point mass is moving with uniform speed . Let it be u .

Centripetal force = m v² / r , r is radius of circular path .

Putting the given values

.10 x v² / .36 = 10

v = 6 m /s

Answer:

0.29D

Explanation:

Given that

F = G M m / r2

F = GM(6m) / (D-r)2

G Mm/r2 = GM(6m) / (D-r)2

1/r2 = 6 / (D-r)2

r = D / (Ö6 + 1)

r = 0.29 D

See diagram in attached file

Answer:

our face is outside the focal length of the concave side of the spoon. We see a virtual inverted image whereas in case of concave mirror we can see a virtual image which is erect.

I bel.ieve the answer is 279Ghz

The required value of pressure-variation amplitude of the given sound wave is 11.84 Pa.

Given data:

The speed of sound in air is, v = 340 m/s.

The density of air is, [tex]\rho = 1.2 \;\rm kg/m^{3}[/tex].

The frequency of sound wave is, f = 330 Hz.

The displacement amplitude of sound wave is, [tex]A = 14 \;\rm \mu m= 14 \times 10^{-6} \;\rm m[/tex].

The standard expression for the pressure variation amplitude for the sound wave propagating in air medium is,

[tex]\Delta P= B \times A \times K[/tex]

Here,

B is the Bulk Modulus and its value is, [tex]B = \rho \times v^{2}[/tex].

K is the wave form constant and its value is, [tex]K = \dfrac{2 \pi f}{v}[/tex].

Solving as,

[tex]\Delta P= (\rho \times v^{2}) \times A \times \dfrac{2 \pi f}{v}\\\\\Delta P= (\rho \times v) \times A \times (2 \pi f)\\\\\Delta P= (1.2 \times 340) \times (14 \times 10^{-6}) \times (2 \pi \times 330)\\\\\Delta P= 11.84 \;\rm Pa[/tex]

Thus, we can conclude that the required value of pressure-variation amplitude of the given sound wave is 11.84 Pa.

Learn more about the pressure-variation amplitude here:

https://brainly.com/question/13091615

Answer:

the right answer is true

Answer:

True

Explanation:

Answer:

344.68 m

Explanation:

The computation of the far does the car travel before stopping is

Data provided in the question

Force = F = 8,868 N

mass = m = 2,500 kg

So,

accleration = a is

[tex]= \frac{-F}{m}\\\\\= \frac{8868}{2500}[/tex]

a = -3.54 m/s^2

The initial speed = u = 49.4 m / s

final speed = v = 0

Based on the above information

Now applying the following formula

v^ 2- u^ 2= 2aS

Therefore

[tex]S = \frac{v^ 2- u^ 2}{2a}\\\\\ = \frac{0- 49.4^ 2}{2\times -3.54}[/tex]

= 344.68 m

Answer:

The wave will be travelling in the y-axis

Explanation:

An e-m wave has a spatially varying electric field that is always associated with a magnetic field that changes over time and vice versa. The electric field and the magnetic field oscillates perpendicularly to each other, and together form a wave that travels in a perpendicular direction to the magnetic and the electric field in space. The movement of the e-m wave through space is usually away from the source where it is generated. So, if the electric field travels in the z-axis, and the magnetic field travels through along the x-axis, then the e-m wave generated will travel in the y-axis direction.

Answer:

Vi = 2 m/s

Explanation:

First we find the force applied to the car by wall to stop it. We use Hooke's Law:

F = kx

where,

F = Force = ?

k = spring constant = 4 x 10⁶ N/m

x = compression = 3.18 cm = 0.0318 m

Therefore,

F = (4 x 10⁶ N/m)(0.0318 m)

F = 127200 N

but, from Newton's Second Law:

F = ma

a = F/m

where,

m = mass of car = 2010 kg

a = deceleration = ?

Therefore,

a = 127200 N/2010 kg

a = 63.28 m/s²

a = - 63.28 m/s²

negative sign due to deceleration.

Now, we use 3rd equation of motion:

2as = Vf² - Vi²

where,

s = distance traveled = 3.18 cm = 0.0318 m

Vf = Final Speed = 0 m/s

Vi = Initial Speed = ?

Therefore,

2(- 63.28 m/s²)(0.0318 m) = (0 m/s)² - Vi²

Vi = √4.02 m²/s²

Vi = 2 m/s

Answer:

B. [tex]8sin(6.2x-12t)[/tex]

Explanation:

The general equation of a wave is expressed as [tex]y = Asin(kx-\omega t)[/tex]

A is the amplitude of the wave

k is the wave number and it is expressed as [tex]k =\frac{2\pi}{\lambda}[/tex]

[tex]\omega[/tex] is the angular frequency expressed as [tex]\omega = 2\pi f[/tex]

[tex]\lambda[/tex] is the wavelength and f is the angular frequency

Given k = 6.2rad/m and [tex]\omega = 12rad/s[/tex]

On substituting this value into the general wave equation;

[tex]y = Asin(kx-\omega t)\\y = Asin(6.2x-12t)[/tex]

From the expression gotten, the only equation that could be the equation of the wave is [tex]y = 8sin(6.2x-12t)[/tex]

Answer:

50k/h is the answer to iy

Answer:

Yes. Something similar occurs here on Earth.

Explanation:

Gravity tends to pull objects perpendicularly to the ground. In space, the absence of this force means there is no compression on the spine due to gravity trying to pull it down. This means that astronauts undergo an increase in height in space.

Here on Earth, we experience gravity pull on our spine during the day. At night when we sleep, we lie down with our spine parallel to the ground, which means that our spine is no longer under compression from gravity force. The result is that we are a few centimetres taller in the morning when we wake up, than we are before going to bed at night. The increase is not much pronounced here on Earth because there is a repeated cycle of compression and decompression of our spine due to gravity, unlike when compared to that of astronauts that spend long duration in space, all the while without gravity forces on their spine

Answer:

Explanation:

Using the law of conservation of momentum to solve the problem. According to the law, the sum of momentum of the bodies before collision is equal to the sum of the bodies after collision. The bodies move with the same velocity after collision.

Mathematically.

mu + MU = (m+M)v

m and M are the masses of the bullet and the block respectively

u and U are their respective velocities

v is their common velocity

from the question, the following parameters are given;

m = 20g = 0.02kg

u = 960m/s

M = 4.5kg

U =0m/s (block is at rest)

Substituting this values into the formula above to get v;

0.02(960)+4.5(0) = (0.02+4.5)v

19.2+0 = 4.52v

4.52v = 19.2

Dividing both sides by 4.52

4.52v/4.52 = 19.2/4.52

v = 4.25m/s

Since they have the same velocity after collision, then the speed of the block immediately after the collision is also 4.25m/s

Answer:

The melting point of indium is 157.436 degrees Celsius.

Explanation:

The resistance of the platinum wire, R1 = 2

The temperature at R1 is, T1 = 20 degrees Celsius.

The increased resistance, R2 = 3.072

Let the temperature at 3.072 = T2

Now find the temperature at which the indium starts melting.

We know that α = ( R2 - R1 ) / [ R1 × ( T2 - T1 ) ]

Given, α = 3.9 x 10^-3/ degrees Celsius.

T2- T1   = ( R2 - R1 ) / R1 α

T2 – T1 = (3.072 – 2) / (2 × 3.9 x 10^-3)

T2 – T1 = 137.436

T2 = T1 + 137.436

T2 =  20 + 137.436

T2 = 157.436 degree Celsius

Answer:

1,772  C

Explanation:

Answer:

Energy, [tex]E=4.67\times 10^{-19}\ J[/tex]

Explanation:

It is given that,

Frequency of blue shade is, [tex]f=7.06\times 10^{14}\ Hz[/tex]

We need to find the energy of exactly one photon of this light. The formula that is used to find the energy of photon is given by :

[tex]E=nhf[/tex]

Here, n is number of photon, n = 1

h is Planck's constant

So,

[tex]E=1\times 6.626\times 10^{-34}\times 7.06\times 10^{14}\\\\E=4.67\times 10^{-19}\ J[/tex]

So, the energy of this light is [tex]4.67\times 10^{-19}\ J[/tex].

Answer:

Yes it produces current

Explanation:

Because If this enclosed field is somehow changed, then following the law of electromagnetic induction, a pulse of current will be produced in the loop. Dues to a change is produced in the electric field when the iron parts of a car pass over it, momentarily increasing the strength of the field.

Answer:

a)  v = 2,9992 10⁸ m / s , b)  Eo = 375 V / m ,  B = 1.25 10⁻⁶ T,

c)     λ = 3,157 10⁻⁷ m,   f = 9.50 10¹⁴ Hz ,  T = 1.05 10⁻¹⁵ s , UV

Explanation:

In this problem they give us the equation of the traveling wave

        E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]

a) what the wave velocity

all waves must meet

        v = λ f

In this case, because of an electromagnetic wave, the speed must be the speed of light.

        k = 2π / λ

        λ = 2π / k

        λ = 2π / 1.99 10⁷

        λ = 3,157 10⁻⁷ m

        w = 2π f

        f = w / 2 π

        f = 5.97 10¹⁵ / 2π

        f = 9.50 10¹⁴ Hz

the wave speed is

        v = 3,157 10⁻⁷   9.50 10¹⁴

        v = 2,9992 10⁸ m / s

b) The electric field is

           Eo = 375 V / m

to find the magnetic field we use

           E / B = c

           B = E / c

            B = 375 / 2,9992 10⁸

            B = 1.25 10⁻⁶ T

c) The period is

           T = 1 / f

            T = 1 / 9.50 10¹⁴

            T = 1.05 10⁻¹⁵ s

the wavelength value is

          λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm

this wavelength corresponds to the ultraviolet

Answer:

Explanation:

A.the direction of induced current will be clockwise

B: Changing 18cm and 6.8cm into 0.18m and 0.68

2.5

Divide them both by 2 to find the radius . Now we have 0.09 and .034m.

Now use Φ=(π*0.09^2)(.75 T)cos0 and the 0.019wb

(π*0.034^2)(.75 T)cos0 and the 0.00272wb

ow use ε=-N(ΔΦ/Δt)

For ΔΦ, 0.091-0.0027=0.0883

C.

To find the current, use I=ε/R

0.0883/2.5= 0.035A

Answer:

Explanation:

Range of projectile R = 20 m

formula of range

R = u² sin2θ / g

u is initial velocity , θ is angle of projectile

putting the values

20 = u² sin2x 40 / 9.8

u² = 199

u = 14.10 m /s

At the initial point

vertical component of u

= u sin40 = 14.1 x sin 40

= 9.06 m/s

Horizontal component

= u cos 30

At the final point where the ball strikes the ground after falling , its speed remains the same as that in the beginning .

Horizontal component of velocity

u cos 30

Vertical component

= - u sin 30

= - 9.06 m /s

So its horizontal component remains unchanged .

change in vertical component = 9.06 - ( - 9.06 )

= 18.12 m /s

change in momentum

mass x change in velocity

= .050 x 18.12

= .906 N.s

Impulse = change in momentum

= .906 N.s .

Answer:

The magnitude  and direction of the  magnetic field is 2.7 x 10⁻⁵ T upwards

Explanation:

Given;

current in the eastern wire, [tex]I_e[/tex] = 15 A

current in the western wire, [tex]I_w[/tex] = 6 A

distance between the wires, d = 30 cm = 0.3 m

The magnetic field at a distance R from a line current I, is given as;

[tex]B = \frac{\mu_o I }{2 \pi R}[/tex]

The magnetic field between the wires, are in opposite directions, and since the currents are also in opposite directions, the magnetic fields of the wires will be added.

The total field = magnetic field (east) + magnetic field (west);

[tex]B = \frac{\mu_o I_e}{2 \pi R_e} + \frac{\mu_0 I_w}{2 \pi R_w} \\\\B = \frac{\mu_o}{2\pi} (\frac{I_e}{R_e} + \frac{I_w}{R_w})[/tex]

where;

[tex]R_w[/tex] is the distance of the field from west = 10cm = 0.1 m

[tex]R_e[/tex] is the distance of the field on east from west = d - 10cm = 30cm - 10cm = 20cm = 0.2 m

The total magnetic field is;

[tex]B = \frac{\mu_o}{2\pi} (\frac{I_e}{R_e} + \frac{I_w}{R_w})\\\\B = \frac{4\pi *10^{-7}}{2\pi} (\frac{15}{0.2} + \frac{6}{0.1})\\\\B = 2*10^{-7}(75 + 60)\\\\B = 2*10^{-7}(135)\\\\B = 2.7*10^{-5} \ T[/tex]

Since total magnetic field is positive, the direction of the field is upwards (positive y direction)

Therefore, the magnitude  and direction of the  magnetic field is 2.7 x 10⁻⁵ T upwards

Answer:

There are 10 cubes in the box

Explanation:

Let the total number of cubes be x and the total number of pyramids be y

Since there are 17 total objects;

Then;

x + y = 17 •••••••••(i)

Also, the total number of sides of cubes is 6 * x = 6x ( a single cube has 6 sides)

For the pyramid we have 5 * y = 5y

adding both gives the total number of sides

6x + 5y = 95 •••••• (ii)

From i, we cabs say y = 17-x

plug this in ii

6x + 5(17-x) = 95

6x + 85 -5x = 95

6x-5x = 95-85

x = 10

Given that,

Rank in order from largest to smallest the magnitude of the electric field at block dot.

Electric field :

Electric field is proportional to the charge divided by square of distance.

In mathematically,

[tex]E\propto\dfrac{q}{r^2}[/tex]

Where, q = charge

r = distance

If the charge is greater then electric field will be greater.

If the distance is greater then electric field will be smaller.

We need to find the electric field at black dot

According to figure,

(I). The electric field at black dot due to positive charge point q to left. the distance is r.

The electric field will be

[tex]E=\dfrac{kq}{r^2}[/tex]

The electric field will be largest.

(II). The electric field at black dot due to positive charge point 2q to left. The distance is 2r.

Then, the electric field will be

[tex]E=\dfrac{k2q}{(2r)^2}[/tex]

[tex]E=\dfrac{kq}{2r^2}[/tex]

The electric field will be smallest.

(III).  The electric field at black dot due to positive charge point 2q to left. The distance is r.

Then, the electric field will be

[tex]E=\dfrac{k2q}{(r)^2}[/tex]

The electric field will be very largest.

(IV). The electric field at black dot due to positive charge point q to left. The distance is 2r.

Then, the electric field will be

[tex]E=\dfrac{kq}{(2r)^2}[/tex]

[tex]E=\dfrac{kq}{4r^2}[/tex]

The electric field will be very smallest.

So, The electric field from largest to smallest will be

[tex]E_{3}>E_{1}>E_{2}>E_{4}[/tex]

Hence, The ranking will be 3, 1, 2, 4.

(D) is correct option.

Answer:

he correct answer is V = ER

Explanation:

In this exercise they give us the electric field on the surface of the sphere and ask us about the electric potential, the two quantities are related

                ΔV = ∫ E.ds

where E is the elective field and normal displacement vector.

Since E is radial in a spray the displacement vector is also radial, the dot product e reduces to the algebraic product.

                 ΔV = ∫ E ds

                 ΔV = E s

since s is in the direction of the radii its value on the surface of the spheres s = R

                  ΔV = E R

checking the correct answer is V = ER

Answer:

It is the carrier of genetic information.

Will occur if you were able to place a rotating magnet near a coil of wire is continually rotating magnetic field would continuously induce current.

When the magnet moves close to the coil or rapidly increasing until the magnet is within the flux of the coil. As it passes through the coil, the magnetic flux through the coil starts to decrease. Consequently, an induced EMF is inverted.

Whenever there is a relative movement between the loop and the magnet, regardless of who moves, an electric current, called induced current, appears in the loop.

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