Write the equation in slope-intercept form
x-2y=4

Answer:

The acceleration of the refrigerator together with the objects decreases.

Explanation:

If the mass of the refrigerator is increased by stacking more masses (objects) on it,

and the force applied remains constant, then we know from

F = ma

where

F is the applied force

m is the total mass of the refrigerator and the objects

a is the acceleration of the masses.

If F is constant, and m is increased, the acceleration will decrease

Answer:

The acceleration decreases.

Explanation:

its right

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:

The energy contained in a single pulse is 90,200 J.

Explanation:

Given;

power of the laser, P = 82 TW = 82 x 10¹² W

time taken by the laser to provide the power, t = 1.1 ns = 1.1 x 10⁻⁹ s

the wavelength of the laser, λ = 0.24 μm = 0.24 x 10⁻⁶ m

The energy contained in a single pulse is calculated as;

E = Pt

where;

P is the power of each laser

t is the time to generate the power

E = (82 x 10¹²)(1.1 x 10⁻⁹)

E = 90,200 J

Therefore, the energy contained in a single pulse is 90,200 J

Answer:

1. Energy: ultraviolet>> visible> infrared

2. Frequency: ultraviolet>> visible > infrared

3. Wavelength: infrared >> visible > ultraviolet

Explanation:

Electromagnetic waves are a class of waves that do not require material medium for their propagation, and travel at the same speed. They are arranged with respect to either their decreasing wavelength or increasing frequency to form a spectrum called an electromagnetic spectrum.

Comparing the energy, frequency and wavelength of ultraviolet, infrared and visible regions, it can be deduced that:

1. Energy: ultraviolet has the highest energy, then followed by visible, and infrared has the lowest energy.

i.e energy: ultraviolet>> visible> infrared

2. Frequency: ultraviolet radiation has the highest frequency, visible region has a greater frequency than that of infrared.

i.e frequency: ultraviolet>> visible > infrared

3. Wavelength: infrared radiation has the highest wavelength, followed by visible region, and ultraviolet radiation has the lowest wavelength.

i.e  wavelength: infrared >> visible > ultraviolet

In terms of lowest to the highest energy,frequency, and wavelength is;

Energy: infrared > visible light > ultraviolet

Frequency: infrared > visible light > ultraviolet

Wavelength: ultraviolet > visible light > infrared

The electromagnetic spectrum is made up of all the electromagnetic waves (ultraviolet, infrared, and visible) arranged according to their energy,frequency, and wavelength.

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Answer: Magnitude of electrical force stays the same.

Explanation:

Equation:

[tex]F_{e} =k\frac{Q_{1}Qx_{2} }{r^{2} }[/tex]

Since the magnitude of each charge is halved.

and

the separation is halved.

[tex]F_{e} =k\frac{(.5Q_{1}*.5Q_{2} }{(.5r)^{2} }[/tex]

[tex]F_{e} =k\frac{.25*Q_{1}Qx_{2} }{.25*r^{2} }[/tex]

Cancel out .25 on the numerator and denominator. Leaving the original equation.

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

Assuming constant speeds, the P-wave covers a distance d in time t such that

9000 m/s = d/(60 t)

while the S-wave covers the same distance after 1 more minute so that

5000 m/s = d/(60(t + 1))

Now,

d = 540,000 t

d = 300,000(t + 1) = 300,000 t + 300,000

Solve for t in the first equation and substitute it into the second equation, then solve for d :

t = d/540,000

d = 300,000/540,000 d + 300,000

4/9 d = 300,000

d = 675,000

So the earthquake center is 675,000 m away from the seismic station.

Answer:

24.34 m/s

Explanation:

recall that one of the equations of motions takes the form:

v = u + at

where,

v = final velocity

u = initial velocity (given as 22.2 m/s)

a = acceleration (given as 2.68m/s²)

t = time elapsed during acceleration (given as 0.80s)

since we are told that the the acceleration is in the direction of the intial velocity, we can simply substitute the known values into the equation above:

v = u + at

v = 22.2 + (2.68) (0.8)

v = 24.34 m/s

Answer:

λ = 667.85 nm

Explanation:

Let f be the frequency detected by the observer

Let v be the speed at which the observer is moving.

Now, when the direction at which the observer is moving is away from the source, we have the frequency as;

f = f_o√((1 - β)/(1 + β))

From wave equations, we know that the wavelength is inversely proportional to the frequency. Thus, wavelength is now;

λ = λ_o√((1 + β)/(1 - β))

Where, β = Hr/c

H is hubbles constant which has a value of 0.0218 m/s • ly

c is speed of light = 3 × 10^(8) m/s

r is given as 2.40 x 10^(8) ly

Thus,

β = (0.0218 × 2.4 x 10^(8))/(3 × 10^(8))

β = 0.01744

Since we are given λ_o = 656.3 nm

Then;

λ = 656.3√((1 + 0.01744)/(1 - 0.01744))

λ = 667.85 nm

Explanation:

It is given that,

The velocity of football is 18 m/s

It is projected at an angle of 20 degrees

We need to find the ball's acceleration in the horizontal direction as it flies through the air.

When it is projected with some velocity, it has two rectangular components i.e. horizontal and vertical.

In vertical direction, it will move under the action of gravity. There is no change in velocity in horizontal direction. So, ball's acceleration in the horizontal direction is equal to 0.

Answer:

a)  Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F ,  V = 3.4375 V ,

c)  U = 54.7 nJ ,  d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

        C = Q / V

        Q = C V

can also be calculated using geometry consideration

        C = e or A / d

we reduce to the SI system

       A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

       d = 1.53 cm = 1.53 10⁻² m

we substitute

         Q = eo A / d V

         Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275

         Q = 3.9757 10⁻¹⁰ C

let's reduce to pC

         Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

          Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

           Q = k Q₀

           Q = 80 397.57

           Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

           C = k C₀

           C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

           C = 1.157 10⁻¹⁰ F

           V = Vo / k

            V = 275/80

            V = 3.4375 V

c) The stored energy is

             U = ½ C V²

              U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²     275²

             U = 5.47 10⁻⁸ J

let's reduce to nJ

              109 nJ = 1 J

               U = 54.7 nJ

d) energy after submerging

             U = ½ (kCo) (Vo / k) 2

             U = ½ Co Vo2 / k

             U = U₀ / k

             U = 54.7 / 80 nJ

              U = 0.68375 nJ

the energy change is

         ΔU = U₀ -U

          ΔU = 54.7 - 0.687375

(a) Charge on the plate before immersion, Qi is 5.258 x 10⁻³ pC and the charge after, Qf is 0.421 pC.

(b) The capacitance and potential difference after immersion is 1.157 x 10⁻¹⁰ F and 3.44 V respectively.

(c) The change in energy of the capacitor is 54.02 nJ.

The charge on the plate is calculated as follows;

[tex]Q =\frac{\varepsilon _o A}{Vd} \\\\Q_i = \frac{8.85 \times 10^{-12} \times (25 \times 10^{-4}) }{275\times 0.0153} \\\\Q_i = 5.258 \times 10^{-15} \ C\\\\Q_i = 5.258 \times 10^{-3} pC[/tex]

[tex]Q_f = k Q_i\\\\Q_f = 80 \times 5.258 \times 10^{-3} \ pC= 0.421 \ pC[/tex]

[tex]C = \frac{k\varepsilon _o A}{d} \\\\C = \frac{80 \times 8.85 \times 10^{-12} \times (25\times 10^{-4} )}{0.0153} \\\\C = 1.157 \times 10^{-10} \ F[/tex]

[tex]V = \frac{V_0}{k}\\\\V = \frac{275}{80} \\\\V = 3.44 \ V[/tex]

The initial energy of the capacitor is calculated as follows;

[tex]U_i = \frac{1}{2} CV^2\\\\U_ i = \frac{1}{2} \times (\frac{\varepsilon _o A}{d} )V^2\\\\U_i = \frac{1}{2} \times (\frac{8.85\times 10^{-12} \times 25 \times 10^{-4}}{0.0153} )\times 275^2\\\\U_i = 5.47 \times 10^{-8} \ J\\\\U_i = 54.7 \ nJ[/tex]

The final energy of the capacitor is calculated as follows;

[tex]U_f = \frac{1}{2} (kC) \times (\frac{V}{k} )^2\\\\U_f = \frac{1}{2} C\times \frac{V^2}{k} \\\\U_f = \frac{1}{k} (\frac{1}{2} CV^2)\\\\U_f = \frac{U_i}{k} \\\\U_f = \frac{54.7 \ nJ}{80} \\\\U_f = 0.68 \ nJ[/tex]

Change in energy is calculated as follows;

[tex]\Delta U = U_i - U_f \\\\\Delta U = 54.7 \ nJ \ - \ 0.68 \ nJ\\\\\Delta U = 54.02 \ nJ[/tex]

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Answer:

Heat transferred = 22.9 watt

Explanation:

Given that:

[tex]T_1[/tex] = 27°C = (273 + 27) K = 300 K

[tex]T_2[/tex]= 600°C = (600 +273) K = 873 K

speed v = 2 m/s

length x = 40 cm = 0.4 cm

width = 1 cm = 0.001 m

The heat transferred from the plate can be calculate by using the formula:

Heat transferred = h×A ×ΔT

From the tables of properties of air, the following values where obtained.

[tex]k = 0.02476 \ W/m.k \\ \\ \rho = 1.225 \ kg/m^3 \\ \\ \mu = 18.6 \times 10^{-6} \ Pa.s \\ \\ c_p = 1.005 \ kJ/kg[/tex]

To start with the reynolds number; the formula for calculating the reynolds number can be expressed as:

reynolds number = [tex]\dfrac{\rho \times v \times x }{\mu}[/tex]

reynolds number = [tex]\dfrac{1.225 \times 2 \times 0.4}{18.6 \times 10^{-6}}[/tex]

reynolds number = [tex]\dfrac{0.98}{18.6 \times 10^{-6}}[/tex]

reynolds number = 52688.11204

Prandtl number = [tex]\dfrac{c_p \mu}{k}[/tex]

Prandtl number = [tex]\dfrac{1.005 \times 18.6 \times 10^{-6} \times 10^3}{0.02476}[/tex]

Prandtl number = [tex]\dfrac{0.018693}{0.02476}[/tex]

Prandtl number = 0.754963

The nusselt number for this turbulent flow over the flat plate  can be computed as follows:

Nusselt no = [tex]\dfrac{hx}{k} = 0.0296 (Re) ^{0.8} \times (Pr)^{1/3}[/tex]

[tex]\dfrac{h \times 0.4}{0.02476} = 0.0296 (52688.11204) ^{0.8} \times (0.754968)^{1/3}[/tex]

[tex]\dfrac{h \times 0.4}{0.02476} =161.4252008}[/tex]

[tex]h =\dfrac{161.4252008 \times 0.02476}{ 0.4}[/tex]

h = 9.992 W/m.k

Recall that:

The heat transferred from the plate can be calculate by using the formula:

Heat transferred = h×A ×ΔT

Heat transferred = [tex]h\times A \times (T_2-T_1)[/tex]

Heat transferred = 9.992 × (0.4 × 0.01) ×(873-300)

Heat transferred = 22.9 watt

Answer:

μ = 1

F = P√2

Explanation:

The parabola equation is: y = ½ x².

The slope of the tangent is dy/dx = x.

The angle between the tangent and the x-axis is θ = tan⁻¹(x).

At x = 1, θ = 45°.

Draw a free body diagram of the block.  There are three forces:

Weight force P pulling down,

Normal force N pushing perpendicular to the surface,

and friction force Nμ pushing up tangential to the surface.

Sum of forces in the perpendicular direction:

∑F = ma

N − P cos 45° = 0

N = P cos 45°

Sum of forces in the tangential direction:

∑F = ma

Nμ − P sin 45° = 0

Nμ = P sin 45°

μ = P sin 45° / N

μ = tan 45°

μ = 1

Draw a new free body diagram.  This time, friction force points down tangential to the surface, and applied force F pushes up tangential to the surface.

Sum of forces in the tangential direction:

∑F = ma

F − Nμ − P sin 45° = 0

F = Nμ + P sin 45°

F = (P cos 45°) μ + P sin 45°

F = P√2

Answer:

A:  magnitude and direction

B: Force that the field exerts on a test charge

C: its magnitude and direction is the same.

D: electrostatic machine

two rollers that are driven by a motor that generates a rotation

Explanation:

Answer:

Resistance of each bulb = 360 ohms

Explanation:

Let each bulb have a resistance r .

Since, even after removing one of the bulbs, the circuit is closed and the other bulbs glow. Therfore, the bulbs are connected in Parallel connection.

[tex] \frac{1}{r(equivalent)} = \frac{1}{r1} + \frac{1}{r2} + + + + \frac{1}{r15} [/tex]

[tex] \frac{1}{r(equivalent)} = \frac{15}{r} [/tex]

R(equivalent) = r/15

Now, As per Ohms Law :

V = I * R(equivalent)

120 V = 5 A * r/15

r = 360 ohms

Answer:

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Explanation:

The gravitational force of the earth keeps us bound to the earth. Gravitational force between earth and sun makes the earth move around the sun. Gravitational force between moon and earth makes the moon go around the earth.

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

We have that for the Question, it can be said that the average induced emf in the coil is

E=0.028565V

From the question we are told

A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to 0.24 T pointing down. What is the average induced emf in the coil?

Generally the equation for the Average emf induced   is mathematically given as

[tex]Emf_a=-NA\frac{dB}{dt}\\\\Where\\\\Area\\\\a=\pir^2\\\\a=\pi(0.056)^2\\\\a=0.00985\\\\[/tex]

Hence

[tex]dB=0.24-0.53\\\\dB=-0.29T[/tex]

Therefore

[tex]E=-\frac{1*0.00985*-0.29 }{0.10}[/tex]

E=0.028565V

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Answer:

3.57 MJ

Explanation:

ASSUMING it's fresh water with density of 1000 kg/m³

W = ΔPE = mgΔh = 14.0(1000)(9.81)(26.0) = 3,570,840 J

Salt water would require more.

Answer:

y is doubled

Explanation:

If x is halved, that means the value is doubled. Here is an exmaple:

y=1/2. If the denominater is doubled, y would equal y=1/1. So, the value of y has doubled from 0.5 to 1. Therefore, if the denominator is halved, the solution will be doubled.

Answer:

Electric field is zero at point 4.73 m

Explanation:

Given:

Charge place = 50 cm  = 0.50 m

change q1 = 5 µC

change q2 = 4 µC

Computation:

electric field zero calculated by:

[tex]E1 =k\frac{q1}{r^2} \\\\E2 =k\frac{q2}{R^2} \\\\[/tex]

Where electric field is zero,

First distance = x

Second distance = (x-0.50)

So,

E1 = E2

[tex]k\frac{q1}{r^2}=k\frac{q2}{R^2} \\\\[/tex]

[tex]\frac{5}{x^2}=\frac{4}{(x-50)^2} \\\\[/tex]

x = 0.263 or x = 4.73

So,

Electric field is zero at point 4.73 m

Answer:

d. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.

Explanation:

Let us take the momentum of a photon unit as u

we know that the rate of change of momentum is proportional to the force exerted.

For a absorbing surface, the photon is absorbed, therefore the final momentum is zero. From this we can say that

F = (u - 0)/t = u/t

for a unit time, the force is proportional to the momentum of the wave due to its energy density. Therefore,

F = u

For a reflecting surface, the momentum of the wave strikes the sail and changes direction. Since we know that the speed of light does not change, then the force is proportional to

F = (u - (-u))/t = 2u/t

just as the we did above, it becomes

F = 2u.

From this we can see that the force for a reflective sail is twice of that for an absorbing sail, and we know that the pressure is proportional to the force for a given area. From these, we conclude that the sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.

Answer: R=24.2Ω

Explanation: Power is rate of work being done in an electric circuit. It relates to voltage, current and resistance through the following formulas:

P=V.i

P=R.i²

[tex]P=\frac{V^{2}}{R}[/tex]

The resistance of the system is:

[tex]P=\frac{V^{2}}{R}[/tex]

[tex]R=\frac{V^{2}}{P}[/tex]

[tex]R=\frac{110^{2}}{500}[/tex]

R = 24.2Ω

For the device, resistance is 24.2Ω.

Answer:

A freely falling object has weight W=mg, where W-weight, m-mass of the object and g-acceleration produced due to the earth's gravity. ... This happens because the normal reaction force exerted on the object in the lift is equal to zero, and normal force equals to mg, which in turn equals the weight of the object

Explanation:

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Answer:

Gradually increases until the maximum weight reaches the surface of the earth

Explanation:

Answer:

a

    [tex]n = 1.119 *10^{18} \ photons[/tex]

b

  [tex]P = 1.6 \ W[/tex]

Explanation:

From the question we are told that

    The wavelength is  [tex]\lambda = 2780 nm = 2780 *10^{-9} \ m[/tex]

     The  energy  is  [tex]E = 80 mJ = 80 *10^{-3} \ J[/tex]

This energy is mathematically represented as

     [tex]E = \frac{n * h * c }{\lambda }[/tex]

Where  c is the speed of light with a value  [tex]c = 3.0 *10^{8} \ m/s[/tex]

             h is the Planck's  constant with the value  [tex]h = 6.626 *10^{-34} \ J \cdot s[/tex]

             n is the number of pulses

So

      [tex]n = \frac{E * \lambda }{h * c }[/tex]

substituting values

       [tex]n = \frac{80 *10^{-3} * 2780 *10^{-9}}{6.626 *10^{-34} * 3.0 *10^{8} }[/tex]

       [tex]n = 1.119 *10^{18} \ photons[/tex]

Given that the pulses where emitted 20 times in one second then the period of the pulse is

       [tex]T = \frac{1}{20}[/tex]

      [tex]T = 0.05 \ s[/tex]

Hence the average power of photons in one 80-mJ pulse during 1 s is mathematically represented as

       [tex]P = \frac{E}{T}[/tex]

substituting values

       [tex]P = \frac{ 80 *10^{-3}}{0.05}[/tex]

        [tex]P = 1.6 \ W[/tex]

Answer:

3.4093

Explanation:

NPSHa = hatm + hel + hf +hva

the elevation head is the hel

friction loss head is hf

NPSHa is the head of vapour pressure of fluid

atmospheric pressure head is hatm

log₁₀P* = [tex]A -\frac{B}{C+T}[/tex]

[tex]A, B, C are fixed[/tex]

log₁₀Pv = [tex]4.07827-\frac{1343.943}{387.15-53.773}[/tex]

= 4.07827 - 1343.943/333.377

=4.07827 - 4.0313009

= 0.0469691

we take the log

p* = 1.114218

we convert this value to get 111421.8

hvap = 111421.8 * 1/776.14 * 1/9.81

= 14.63

hatm = 1.1 *101325/1 * 1/9.81 *1/776.14

=14.64

hf = 7000/1 * 1/776.14 * 1/9.81

= 0.9193

NPSHa = 2.5

hel = 0.9193 + 2.5 + 14.63 - 14.64

hel = 3.4093

The NSPH values are used to calculate cavitation. The vapor pressure of the liquid is 1.114 atm.

The vapor pressure can be calculated by,

[tex]\mathrm {NPSH_A}= ( \frac {p_i}{\rho g} + \frac {V_i^2}{2g})- \frac {p_v}{\rho g}[/tex]

Where,

[tex]\mathrm {NPSH_A}[/tex] = available NPSH

[tex]p_i[/tex]     = absolute pressure at the inlet = 1.1 atm

[tex]V_i[/tex]     = average velocity at the inlet = 10, 000 kg/h

[tex]\rho[/tex] = fluid density = 886 kg/m3.  

g = acceleration of gravity = 9.8 m/s²

[tex]p_v[/tex] = vapor pressure of the fluid = ?

Put the values in the equation, we get

[tex]p_v = 1.114\ atm[/tex]

Therefore, the vapor pressure of the liquid is 1.114 atm.

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Answer:

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

2000 counts per second = 1 meter

x counts per second = 3 meter

Cross multiply to get x

2000 * 3 = 1* x

6000 = x

This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample

Answer:

0.05cos10t

Explanation:

X(t) = Acos(wt+φ)

The oscillation angular frequency can be calculated using below formula

w = √(k/M)

Where K is the spring constant

But we were given body mass of 5.0 kg

We know acceleration due to gravity as 9.8m)s^2

The lenghth of spring which stretches =10 cm

Then we can calculate the value of K

k = (5.0kg*9.8 m/s^2)/0.10 m

K= 490 N/m

Then if we substitute these values into the formula above we have

w = √(k/M)

w = √(490/5)

= 9.90 rad/s=10rads/s(approximately)

Its position as a function of time can be calculated using the below expresion

X(t) = Acos(wt+φ)

We were given amplitude of 5 cm , if we convert to metre = 0.05m

w=10rads/s

Then if we substitute we have

X(t)=0.05cos(10×t)

X(t)= 0.05cos10t

Therefore,Its position as a function of time=

X(t)= 0.05cos10t

Answer:

D Is 430m

Explanation:

See attached file

Answer:

halve the slit separation

Explanation:

As we know that

In YDS experiment, the equation of fringe width is as follows

[tex]\beta = \frac{\lambda D}{d}[/tex]

where,

D denotes the separation in the middle of screen and slits

d denotes the distance in the middle of two slits

And to increase the Δx we have to decrease the d i.e, the distance between the two slits

Hence, the first option is correct

Answer:

Not sure but I believe predictable.

Explanation:

Phenomena usually consists of :

- A history, a date in which the physical phenomenon has occurred.

- A source, a place or reason to why or where the physical phenomena has occured.

According to this, I want to say predictable.

It is not repeatable, there are one-time phenomenons that have occurred that scientists to this day still have not recorded again such as the Big Bang.

It is not provable. Most of the theories earlier scientists and historians have predicted have not yet been proved. Yet rather, somehow, they have been explored and investigated.

It is not readable. This is self explanatory, some things scientists investigate are not written down, nor read. It starts with a mental theory and then immediately goes to the next phase of investigation.