Specimen of steel has a rectangular cross section 20 mm wide and 40 mm thick, an elastic modulus of 207 GPa, and a Poisson’s ratio of 0.30. If this specimen is pulled in tension with a force of 60,000 N, what is the change in width if deformation is totally elastic? (A) Increase in width of 3.62  10−6 m (B) Decrease in width of 7.24  10−6 m (C) Increase in width of 7.24  10−6 m (D) Decrease in width of 2.18  10−6 m

Answer:

W_apparent = 93.1 kg

Explanation:

The apparent weight of a body is the weight due to the gravitational attraction minus the thrust due to the fluid where it will be found.

            W_apparent = W - B

The push is given by the expression of Archimeas

            B = ρ_fluide g V

            ρ_al = m / V

            m = ρ_al V

we substitute

            W_apparent = ρ_al V g - ρ_fluide g V

            W_apparent = g V (ρ_al - ρ_fluide)

we calculate

           W_apparent = 980 50 (2.7 - 0.8)

           W_apparent = 93100 g

            W_apparent = 93.1 kg

Answer:

A) 21.1 m

B) 17.3 m

C) 3.267x10^7 m/s

Explanation:

This is a case of special relativity.

Let the relative speed of astronauts ship to my ship be v.

According to my observation,

My craft is 21.1 m long, according to my observation, astronauts craft is 17.3 m long.

If we fix the reference frame as my ship, then the rest lenght of our identical crafts is 21.1 m and the relativistic lenght is 17.3 m

l' = 21.1 m

l = 17.3 m

From l = l'(1 - p^2)^0.5

Where p is c/v, and c is the speed of light

17.3 = 21.1 x (1 - p^2)^0.5

0.82 = (1 - p^2)^0.5

Square both sides

0.67 = 1 - p^2

P^2 = 0.33

P = 0.1089

Revall p = v/c

v/c = 0.1089

But c = speed of light = 3x10^8 m/s

Therefore,

v = 3x10^8 x 0.1089 = 3.267x10^7 m/s

Following are the response to the given points:

a) Its ship travels to the distance of [tex]21.1\ m[/tex]

b) The astronaut's craft would be at a range of  [tex]17.3\ m[/tex]

c) Relativity's use of length contraction:

[tex]\to L=L_0(\sqrt{1-\frac{v^2}{c^2}}) \\\\\to \frac{L}{L_0}=(\sqrt{1-\frac{v^2}{c^2}})[/tex]

Here,

 [tex]\to \frac{L}{L_0}=\frac{17.3}{21.1}=0.81[/tex]

Hence

[tex]\to 0.81=(\sqrt{1-\frac{v^2}{c^2}}) \\\\\to 0.6561=1-\frac{v^2}{c^2}\\\\\to \frac{v^2}{c^2} =1- 0.6561\\\\\to \frac{v^2}{c^2} =0.3439\\\\\to \frac{v}{c} =0.58\\\\\to v= 0.58 c[/tex]

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The mutual inductance of a Tesla coil system with two solenoids will double if the length and number of turns of the primary solenoid are both doubled, assuming all other factors remain constant.

When we are given a Tesla coil configuration with a solenoid of length l and cross-sectional area A, closely wound with N1 turns of wire, and another coil with N2 turns surrounding it at its center, we can calculate the mutual inductance based on the properties of the solenoids. If a new solenoid is introduced that has twice as many turns and is twice as long, the mutual inductance M of the system will be affected.

To understand how the mutual inductance changes, let's remember that for a closely wound solenoid, the mutual inductance can be calculated by a formula incorporating the number of turns, permeability of the core material, cross-sectional area, and the length of the solenoid. The mutual inductance is directly proportional to the product of the number of turns of each coil, the magnetic permeability of the core, and the area of the cross-section, and inversely proportional to the length of the solenoid. Therefore, if we double the length l and the number of turns N1 while keeping all other factors constant, the mutual inductance will also double.

Answer:

The length of the side of the hole in the second cardboard sheet is [tex]L_2 = 0.01m[/tex]

Explanation:

From the question we are told that

     The distance of the point source  from the screen is  [tex]d = 1.0 m[/tex]

      The length of a side of the  first square hole is  [tex]L_1 = 0.020 \ m[/tex]

      The distance of the cardboard from the point source is [tex]D_1 = 0.50\ m[/tex]

   The distance of the second cardboard from the point source is [tex]D_2 = 0.25 \ m[/tex]

Let take the  [tex]\alpha_{max }[/tex] as  the angle at which the light is passing through the edges of the cardboards square hole

     Since the bright square casted on the screen by both  square holes on the   individual cardboards are then it means that

              [tex]\alpha_{max} __{1}} = \alpha_{max} __{2}}[/tex]

This implies that

             [tex]tan (\alpha_{max} __{1}}) = tan (\alpha_{max} __{2}})[/tex]      

Looking at this from the SOHCAHTOA concept

               [tex]tan (\alpha_{max} __{1}}) = \frac{opposite}{Adjacent}[/tex]

     Here opposite is  the length of the side of the  first cardboard square hole

     and    

      Adjacent is  the  distance of the from the  first cardboard square hole to the point source

And for  

            [tex]tan (\alpha_{max} __{2}}) = \frac{opposite}{Adjacent}[/tex]

    Here opposite is  the length of the side of the  second  cardboards square hole (let denote it with [tex]L_2[/tex])

and

Adjacent is the distance of the from the  second  cardboards square hole to the point source

         So

                 [tex]tan (\alpha_{max} __{1}}) = \frac{0.020}{0.50}[/tex]

         And  

                [tex]tan (\alpha_{max} __{2}}) = \frac{L_2}{0.25}[/tex]

Substituting this into the above equation

                 [tex]\frac{0.020}{0.50} = \frac{L_2}{0.25}[/tex]

Making [tex]L_2[/tex] the subject

                   [tex]L_2 = \frac{0.25 *0.020}{0.50}[/tex]

                 [tex]L_2 = 0.01m[/tex]

Since it is a square hole the sides are the same hence

The length of the side of the hole in the second cardboard sheet is [tex]L_2 = 0.01m[/tex]

Answer:20 cm

Explanation:

Given

Refractive index of material [tex]n=1.55[/tex]

Outer radius [tex]R_1=1.98\ cm[/tex]

Inner radius [tex]R_2=2.42\ cm[/tex]

using lens maker formula

[tex]\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})[/tex]

[tex]\frac{1}{f}=(1.55-1)(\frac{1}{1.98}-\frac{1}{2.42})[/tex]

[tex]f=\frac{10.895}{0.55}[/tex]

[tex]f=19.81\approx 20\ cm[/tex]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The minimum mass of [tex]M_1 = 90\ kg[/tex] correct option is  E

Explanation:

 Free body diagram of the set up  in the question is shown on the third uploaded image

  The mass of board is  [tex]M = 60kg[/tex]

   The length of the board is [tex]L = 6 \ m[/tex]

    The length extending over the edge is [tex]L_e = 4 \ m[/tex]

    The second mass is  [tex]M_2 = 30kg[/tex]

Now to obtain [tex]M_1[/tex] we take moment about the edge of the platform

               [tex]M_1 g L_1 = Mg \frac{L}{2} + M_2 g L_2[/tex]

              [tex]M_1 L_1 = M \frac{L}{2} + M_2 L_2[/tex]

  Substituting value  

               [tex]M_1 (2) = (60)(1) + (30)(4)[/tex]

               [tex]M_1 = 90 \ kg[/tex]

The minimum value of M1 needed to keep the board from falling off the platform is 90 kg.

From the information given, we are to find:

Given that:

Consider the center of gravity in the board that lies at the length of the board midpoint.

Then, the distance (D) of the gravity center from the platform end = 3 - 2

= 1 m

∴

Considering the moment about the platform end, the mass (M1) placed on the left side edge of the board can be computed as:

[tex]\mathbf{M_1gL_1 = MgD + M_2gL_2} \\ \\ \mathbf{M_1L_1 = MD + M_2gL_2} \\ \\ \mathbf{M_1(2) = 60 \ kg \times 1 + 30 \ kg \times (4)} \\ \\ \mathbf{ M_1 =\dfrac{60 \ kg + 120 kg }{2} } \\ \\ \mathbf{ M_1 =\dfrac{180 \ kg}{2} } \\ \\ \mathbf{ M_1 =90 \ kg }[/tex]

Therefore, we can conclude that the minimum value M1 needed to keep the board from falling off the platform is 90 kg.

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Answer:

9.6 ft

Explanation:

Distance is inversely proportional to weight

distance = k / (weight), where

k is a constant

or you could say,

distance * weight = k

In this scenario,

120 * 16 = 200 * distance

On rearranging, making, distance the subject of formula, we have

Distance = 120 * 16 / 200

Distance = 1920 / 200

Distance = 9.6 ft

So the 200 pounds person should sit 9.6 feet away from the centre to balance the see saw

Answer:

4.8 ft

Explanation:

torque = wt × distance

t1 = 120lb x 8 ft =960

t2 = 200lb x X ft

set them equal to each other.

120(8) = 200x

960 = 200x

x = 4.8 ft

Answer:2250J

Explanation:

mass(m)=20kg

velocity(v)=15m/s

Kinetic energy=(m x v^2)/2

Kinetic energy =(20 x 15^2)/2

Kinetic energy =(20x15x15)/2

Kinetic energy=4500/2

Kinetic energy=2250J

The kinetic energy of the ball will be 2250 joules.

We have a 20 kilogram ball rolls down a 10 meter ramp at the rate of 15 meters per second.

We have to determine its kinetic energy is joules.

The kinetic energy of the body is as follows -

K.E. = [tex]\frac{1}{2} mv^{2}[/tex]

According to the question -

Mass of ball = 20 kg

Velocity of ball = 15 m/s

Substituting the values in the above formula, we get -

K.E. = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2} \times 20\times 15\times 15[/tex] = 225 x 10 = 2250 joules.

Hence, the kinetic energy of the ball will be 2250 joules.

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Answer:

The longest wavelength in vacuum for which there is constructive interference for the reflected light, λ   = 3472.

Explanation:

Refractive index of Glass (given) = 1.5

For the case of a constructive interference,

2nt = (m + 1/2) λ

For case 1,

2nt = (m + 1/2) 496 nm

For case 2,

2nt = (m +1+ 1/2) 386 nm

2nt = (m+3/2) * 386 nm

(m + 1/2) 496 nm = (m+3/2) * 386 nm

m = 3

Inserting the value of m in 1.

2nt = (m + 1/2) 496 nm

2*1.5t = (3 + 1/2) * 496 nm

t = ((3 + 1/2) * 496 nm)/ 3

t = 578.6 nm

The thickness of the glass, t = 578.6 nm

b)

It is generally known that for constructive interference,

2nt = (m + 1/2) λ

λ = 2nt / ((m + 1/2))

For Longest Wavelength, m = 0

λ = 2*1.5*578.6/ (1/2)

λ = 3472 nm

Answer:

Explanation:

The point at which magnetic field is to be found lies outside wire so while applying Ampere's law we shall take the whole of current . If B be magnetic field which is circular around conductor.

Applying Ampere's law :-

∫ B dl = μ₀ I      ; I is current passing through ampere's loop

B x 2π x 2.00 = 4 x π x 10⁻⁷ x 2

B = 2 x 10⁻⁷ T.

The true statement is A. A cylindrical capacitor is a parallel-plate capacitor rolled into a tube, C. The charge on a capacitor increases quickly at first, then much more slowly as the capacitor charges E. One of the principal purposes of a capacitor is to store electric potential energy. F. A capacitor charges rapidly when connected to an RC circuit with a battery and the false statement are B. The dielectric constant indicates that the distance by which the two plates of a capacitor are separated and D. The voltage across a capacitor in an RC circuit increases linearly during charging.

Let's look at each statement one by one to categorize them as true or false.
a. True. A cylindrical capacitor can be thought of as a parallel-plate capacitor with the plates rolled into cylindrical shapes.

b. False. The dielectric constant is a measure of a material's ability to increase the capacitance of a capacitor, not a measure of the distance between the plates.

c. True. The charge on a capacitor follows an exponential curve, increasing rapidly at first and then more slowly as it approaches its maximum charge.

d. False. The voltage increases exponentially, not linearly, when charging a capacitor in an RC circuit.

e. True. Capacitors store electric potential energy in the electric field between their plates.

f. True. Initially, the capacitor in an RC circuit charges quickly, but the rate of charging decreases over time as it gets closer to full charge.

Answer:

A)

The emf is zero because the velocity of the rod is parallel to the direction of the magnetic field, so the charges experience no force.

B)

The emf is vBL

= (2.6 m/s)(4.50 T)(1.3 m)

= 15.21 V.

The positive end is end 2.

C)

The emf is zero because the magnetic force on each charge is directed perpendicular to the length of the rod.

Answer:

towards the object

Explanation:

Final answer:

To balance the forces on the 1 kg object, a force should be applied in the downward direction.

Explanation:

The forces on the 1 kg object can be balanced by applying a force in the opposite direction to the net force acting on the object. In this case, the net force is the sum of the weight of the object and the tension in the string. Since the weight acts downward and the tension in the string acts upward, the force should be applied in the downward direction to balance the forces on the 1 kg object.

Complete Question

The complete question is shown on the first uploaded image

Answer:

The gauge pressure that water has at the House A  [tex]P_A = 257020.68 Pa[/tex]

The gauge pressure that water has at the House B  [tex]P_B = 188454 \ Pa[/tex]

Explanation:

From the question we are told that

    The mass of water when full is  [tex]m_f = 7.41* 10^{5} kg[/tex]

Generally the volume of water in this tank is mathematically represented as

              [tex]V = \frac{m }{\rho}[/tex]

Where  [tex]\rho[/tex] is the density of water with a value of with a value of [tex]\rho = 1000 kg /m^3[/tex]

   substituting values

                  [tex]V = \frac{7.41 *10^5}{10^3}[/tex]

                  [tex]V = 741 m^3[/tex]

This volume is the volume of a sphere since the tank is spherical so

            [tex]V = \frac{4 \pi ^3}{3}[/tex]

  making r the subject of the formula

           [tex]r =\sqrt[3]{ \frac{741 *3 }{4\pi} }[/tex]              

        [tex]r = 5.6134 m[/tex]

Now we can use this parameter to obtain the diameter

  So

        [tex]d = 2 * r[/tex]

substituting values

        [tex]d = 2 * 5.6134[/tex]

        [tex]d = 11.23m[/tex]

The pressure  the water has at  faucet in House A is mathematically evaluated as

        [tex]P_A = \rho g h_A[/tex]

This height is obtained as follows

                       [tex]h_A = d+ 15[/tex]

The value 15 is gotten from the diagram

  so

          [tex]h_A = 15 + 11.23[/tex]

          [tex]h_A = 26.22 m[/tex]

Now substituting values

         [tex]P_A = 26.23 * 9.8 * 1000[/tex]

         [tex]P_A = 257020.68 Pa[/tex]

      The pressure  the water has at  faucet in House B is mathematically evaluated as

        [tex]P_B = \rho g h_B[/tex]

This height is obtained as follows

                       [tex]h_B = d+ 15[/tex]

The value 15 is gotten from the diagram

  so

          [tex]h_B = d + 15 -h[/tex]

substituting values

         [tex]h_B =11.23 + 15 -7[/tex]

          [tex]h_A = 19.23 m[/tex]

Now substituting values

         [tex]P_B = 19.23 * 9.8 * 1000[/tex]

         [tex]P_B = 188454 \ Pa[/tex]

Answer:

Apparent

The key discovery about Cepheid variable stars that led in the 1920s to the resolution of the question of whether spiral "nebulae" were separate and distant galaxies or part of the Milky Way Galaxy was the direct relationship between the pulsation period and the absolute brightness or luminosity of the Cepheid variables. A measurement of apparent brightness of a variable star could then be used to determine the distance to the "nebula" containing it.

Explanation:

A variable star is a star with changing apparent brightness. The changes can occur over years or in a fraction of seconds. For example the sun whose energy output varies by approximately 0.1 percent of its magnitude, over an 11-year solar cycle. This variable(apparent brightness) can be used to determine how far a variable star is (distance). Therefore, a measurement of apparent brightness of a variable star could then be used to determine the distance to the "nebula" containing it.

Answer:

The total angular momentum is 292.59 kg.m/s

Explanation:

Given that :

Rotation of the horizontal circular platform [tex]\omega[/tex] = 0.919 rad/s

mass of the platform (m) = 90.7 kg

radius (R) = 1.91 m

mass of the poodle [tex]m_p[/tex] = 20.5 kg

Your mass  [tex]m'[/tex] = 73.5 kg

speed v = 1.05 m/s with respect to the platform

[tex]V_p = \frac{1.05}{2} \ m /s \\ \\ = 0.525 \ m /s \\ \\r = \frac{R}{2}[/tex]

r = 0.955

Mass of the mutt [tex]m_m[/tex] = 18.5 kg

[tex]r' = \frac{3}{4} \ R[/tex]

Your angular momentum is calculated as:

Your angular velocity relative to the platform is [tex]\omega' = \frac{v}{R} = \frac{1.05}{1.91 } = 0.5497 \ rad/s[/tex]

[tex]Actual \ \omega_y = \omega - \omega ' = (0.919 - 0.5497) \ rad/s = 0.3693 \ rad/s[/tex]

[tex]I_y = m'R^2 = 73.5 *1.91^2= 268.14 \ kgm^2[/tex]

[tex]L_Y = I_y*\omega_y= 268.14*0.3683= 98.76 \ kg.m/s[/tex]

For poodle :

[tex]Relative \ \ \omega' = \frac{V_p}{R/2} = 0.550 \ rad/s[/tex]

Actual [tex]\omega_p = \omega - \omega' = 0.919 -0.550 = 0.369 \ rad/s[/tex]

[tex]I_p = m_p(\frac{R}{2} )^2 = 20.5(\frac{1.91}{2} )^2 = 18.70 \ kgm^2[/tex]

[tex]L_p = I_p *\omega_p = 18.70*0.369 = 6.9003 \ kgm/s[/tex]

[tex]I_M = m_m(\frac{3}{y4} R)^2 = 18.5 (\frac{3}{4} 1.91)^2 = 37.96 \ kgm^2[/tex]

[tex]L_M = I_M \omega = 37.96 * 0.919= 34.89 \ kg.m/s[/tex]

Disk [tex]I = \frac{mr^2}{2} = \frac{90.7*1.91^2}{2}= 165.44 \ kgm^2[/tex]

[tex]L_D = I \omega = 165.44*0.919 =152.04 \ kg.m/s[/tex]

Total angular momentum of system is:

L = [tex]L_D +L_Y+L_P+L_M[/tex]

= (152.04 + 98.76 + 6.9003 + 34.89) kg.m/s

= 292.59 kg.m/s

Answer:

Dogs were probably domesticated by accident, when wolves began trailing ancient hunter-gatherers to snack on their garbage. Docile wolves may have been slipped extra food scraps, the theory goes, so they survived better, and passed on their genes. Eventually, these friendly wolves evolved into dogs

Selective breeding, also known as artificial selection, is a process used by humans to develop new organisms with desirable characteristics. Breeders select two parents that have beneficial phenotypic traits to reproduce, yielding offspring with those desired traits.

Hope it helps!

Answer:

Explanation:

In series connection of resistors, same current flows in the circuit.

Power dissipated by a resistor is

P = i²R

And since same current flows in them, it implies that the power is directly proportional to Resistance, so the higher the resistance of the resistor the higher the power dissipated in the series connection. Also, the lower the resistance, the lower the power dissipated.

2. In parallel connection, same voltage is applied across the resistor.

So, power dissipated by each resistor is

P = V² / R

So, since the same voltage is applied across parallel connection, then, power dissipated in each resistor is inversely proportional to the resistance.,

So, the higher the resistance, the lower the power and the lower the resistance, the higher the power.

Answer: 10.2 kg if g = 9.8, 10 if g = 10.

Explanation:

Weight or the "force of gravity" on a person is simply defined by the equation: F = ma. In this case, the acceleration is g, which is 9.8 but can be rounded up to 10. Based on this, we have:

F = mg

100 = m*9.8

m = 10.2(or 10 if we set g to 10).

The child's mass on the earth  if the force of gravity on the child’s body is 100 N will be equal to 10.2 kg.

The fundamental force of attraction operating on all matter is recognized as gravity, also spelled gravity, in mechanics.

It has no impact on identifying the interior properties of common matter because it is the weakest force known to exist in nature.

The formation and growth of planets, galaxies, and the universe are all under the influence of this long-range, cosmic force, which further determines the trajectories of objects throughout the universe and the entire universe.

As per the given information in the question,

Weight, w = 100 N

Use the formula,

W = m × g

100 N = m × 9.8

m = 100/9.8

m = 10.2 kg

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Answer: 20 grams

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=m\times c\times \Delta T[/tex]

Q = Heat absorbed by liquid = 5.23 kcal = 5230 cal    (1kcal=1000cal)

C = heat capacity of liquid = [tex]2.09cal/g^0C[/tex]

Initial temperature of the liquid  = [tex]T_i[/tex] = 101 K

Final temperature of the liquid  = [tex]T_f[/tex]  = 225 K

Change in temperature ,[tex]\Delta T=T_f-T_i=(225-101)K=124K[/tex]

Putting in the values, we get:

[tex]5230=m\times 2.09cal/g^0C\times 124K[/tex]

[tex]m=20g[/tex]

Thus the sample of liquid in grams is 20

Answer:

[tex]W=\frac{1}{2}kq_1q_2[\frac{1}{r_2}-\frac{1}{r_1}][/tex]

Explanation:

To find the work W to put the negative charge in the new orbit you can use the following formula:

[tex]W=\Delta K\\\\K=\frac{1}{2}mv^2[/tex]

That is, the total work is equal to the change in the kinetic energy of the negative charge. Then you calculate the speed of the electron, by using the second Newton Law and the expression for the electrostatic energy:

[tex]F=ma_c\\\\-k\frac{(q_1)(q_2)}{r_1^2}=m\frac{v^2}{r_1}\\\\v^2=k\frac{q_1q_2}{mr_1}[/tex]

r1: radius of the first orbit

m: mass of the negative charge

v: velocity of the charge

k: Coulomb's constant

q1: charge of the fixed particle at point P

q2: charge of the negative charge

Hence, the velocity of the charge in a new orbit with radius r2 is:

[tex]v'^2=k\frac{q_1q_2}{mr_2}[/tex]

Finally the work required to put the charge in the new orbit is:

[tex]W=\Delta K =\frac{1}{2}m[v'^2-v^2]\\\\W=\frac{1}{2}m[k\frac{q_1q_2}{mr_2}-k\frac{q_1q_2}{mr_1}]\\\\W=\frac{1}{2}kq_1q_2[\frac{1}{r_2}-\frac{1}{r_1}][/tex]

Answer:

v(t) = 21.3t

v(t) = 5.3t

[tex]v(t) = 48 -48 e ^{ \frac{t}{9}}[/tex]

Explanation:

When no sliding friction and no air resistance occurs:

[tex]m\frac{dv}{dt} = mgsin \theta[/tex]

where;

[tex]\frac{dv}{dt} = gsin \theta , 0 < \theta < \frac{ \pi}{2}[/tex]

Taking m = 3 ; the differential equation is:

[tex]3 \frac{dv}{dt}= 128*\frac{1}{2}[/tex]

[tex]3 \frac{dv}{dt}= 64[/tex]

[tex]\frac{dv}{dt}= 21.3[/tex]

By Integration;

[tex]v(t) = 21.3 t + C[/tex]

since v(0) = 0 ; Then C = 0

v(t) = 21.3t

ii)

When there is sliding friction but no air resistance ;

[tex]m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta[/tex]

Taking m =3 ; the differential equation is;

[tex]3 \frac{dv}{dt}=128*\frac{1}{2} -\frac{\sqrt{3} }{4}*128*\frac{\sqrt{3} }{4}[/tex]

[tex]\frac{dv}{dt}= 5.3[/tex]

By integration; we have ;

v(t) = 5.3t

iii)

To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :

[tex]m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta - kv[/tex]

The differential equation is :

= [tex]3 \frac{dv}{dt}=128*\frac{1}{2} - \frac{ \sqrt{ 3}}{4}*128 *\frac{ \sqrt{ 3}}{2}-\frac{1}{3}v[/tex]

= [tex]3 \frac{dv}{dt}=16 -\frac{1}{3}v[/tex]

By integration

[tex]v(t) = 48 + Ce ^{\frac{t}{9}[/tex]

Since; V(0) = 0 ; Then C = -48

[tex]v(t) = 48 -48 e ^{ \frac{t}{9}}[/tex]

Answer:

0.05m

Explanation:

Density of water = ρ(w) = 1000 kg/ m³ ;  

Density of Mercury = ρ(m) = 13628.95 kg/ m³  

Total pressure at bottom of cylinder=1.1atm

Therefore, pressure due to water and mercury =1.1-1 =atm

0.1atm=10130pa

The pressure at the bottom is given by,

ρ(w) x g[0.4 - d] + ρ(m) x g x d  = 10130

1000 x 9.8[0.4 - d] + 13628.95 x 9.8 d = 10130

3924 - 9810d + 133416d= 10130

123606d= 6206

d= 6202/123606

d= 0.05m

Depth of mercury alone =  d = 0.05m

We have that for the Question " determine the depth of the mercury."

It can be said that

From the question we are told

the cylinder is 0.4 m tall and the absolute (or total) pressure at the bottom is 1.1 atmospheres, determine the depth of the mercury. (Assume the density of mercury to be 1.36 104 kg/m^3, and the ambient atmospheric pressure to be 1.013e5 Pa)

Therefore,

Absolute pressure at the bottom of the container =

[tex]P = 1.1 atm = 1.1 * (1.013*105) Pa\\\\= 1.1143 * 10^5 Pa[/tex]

Where,

Height of the cylinder = H = [tex]0.4 m[/tex]

Height of the water in the cylinder = [tex]H_1[/tex]

Height of the mercury in the cylinder = [tex]H_2[/tex]

Therefore,[tex]H = H_1 + H_2\\\\H_1 = H - H_2\\\\P = P_{atm} + \rho_1gH_1 + \rho_2gH_2\\\\P = P_{atm} + \rho_1g(H - H_2) + \rho_2gH_2\\\\1.1143*10^5 = 1.013*10^5 + (1000)(9.81)(0.4 - H_2) + (1.36*10^4)(9.81)H_2\\\\1.013*10^4 = 3924 - 9810H_2 + 133416H_2\\\\143226H_2 = 6206\\\\H_2 = 4.333*10^{-2} m[/tex]

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Answer:

Decrease the voltage,and do not change the resistance,the current will also decrease

Explanation:

Decrease the voltage,and do not change the resistance, the current will also decrease, because voltage is directly proportional to current

Answer:

B.If we decrease the voltage, and do not change the resistance, then the current will also decrease.

Explanation:

i just take the (pre)test

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Answer:0.15 V

Explanation:

Given

Dimension of coil [tex]4cm\times 5cm[/tex]

Area of coil [tex]A=4\times 5=20\ cm^2[/tex]

Magnetic field [tex]B=0.75\ T[/tex]

Time of rotation [tex]t=0.1\ s[/tex]

No of turns [tex]N=10[/tex]

Initial flux associated with the coil

[tex]\phi_i=N(B\cdot A)[/tex]

[tex]\phi_i=N(BA\cos \theta )[/tex]

where [tex]\theta [/tex]=angle between magnetic field and area vector of coil

[tex]\phi_i=N(BA\cos 90 )[/tex]

Finally when coil is perpendicular to the field

[tex]\phi_f=N(B\cdot A)[/tex]

[tex]\phi_i=N(BA\cos 0 )[/tex]

and induced emf is given by

[tex]e=-\frac{d\phi }{dt}[/tex]

[tex]e=-\frac{\phi_1-\phi_2}{t-0}[/tex]

[tex]e=-\frac{(0-10\times 0.75\times 20\times 10^{-4})}{0.1}[/tex]

[tex]e=0.15\ V[/tex]

Answer: Vertical displacement = -27.6m

And takes 2.375 s

A pumpkin thrown at a horizontal speed of 4.0 m/s and travels 9.5 m horizontally before hitting the ground will take 2.375 seconds to hit the ground.

Given the following information:

Horizontal speed of the pumpkin = 4.0 m/s

Horizontal distance traveled before hitting the ground = 9.5 m

Ignoring air resistance

Use the following formula to calculate the time it takes the pumpkin to hit the ground:

time = horizontal distance / horizontal speed

time = 9.5 m / 4.0 m/s = 2.375 seconds

Therefore, it takes the pumpkin 2.375 seconds to hit the ground.

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Answer:

Magnitude of induced emf is 0.00635 V

Explanation:

Radius of circular loop r = 45 mm = 0.045 m

Area of circular loop [tex]A=\pi r^2[/tex]

[tex]A=3.14\times 0.045^2=0.00635m^2[/tex]

Magnetic field is increases from 250 mT to 350 mT

Therefore change in magnetic field [tex]dB=250-350=100mT[/tex]

Emf induced is given by

[tex]e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}[/tex]

[tex]e=-0.00635\times \frac{100\times 10^{-3}}{0.10}=-0.00635V[/tex]

Magnitude of induced emf is equal to 0.00635 V

Answer:

0.52rad/s^2

Explanation:

To find the angular acceleration you use the following formula:

[tex]\omega^2=\omega_o^2+2\alpha\theta[/tex]   (1)

w: final angular velocity

wo: initial angular velocity

θ: revolutions

α: angular acceleration

you replace the values of the parameters in (1) and calculate α:

[tex]\alpha=\frac{\omega^2-\omega_o^2}{2\theta}[/tex]

you use that θ=22 rev = 22(2π) = 44π

[tex]\alpha=\frac{(12rad/s)^2-(0rad/s)^2}{2(44\pi)}=0.52\frac{rad}{s^2}[/tex]

hence, the angular acceñeration is 0.52rad/s^2

Answer:

If Patty remains on Earth then at the time Peggy will come back from her trip, Peggy will be much younger than her sister Patty because of time-dilation.

Explanation:

Peggy and Patty are sisters. Peggy is an astronaut and Patty is an astronomer.

Peggy goes for mission to Alpha Centauri, the nearest star system to the Solar System at 80% of the speed of light, and will come back after 1010 years.

If Patty remains on Earth then at the time Peggy will come back from her trip, Peggy will be much younger than her sister Patty because of time dilation.

This is due to the fact that time moves slower in Alpha Centauri because of its massive gravitational force which bends space time. Moreover, It is known that Peggy's spacecraft moves at 80% of the speed of light, it will result in velocity time dilation since time moves slow if you travel at a speed near to the speed of light.

Final answer:

Peggy, the astronaut twin traveling at 80% of the speed of light, will experience less time due to time dilation, and upon her return will be younger than her Earth-bound twin sister, Patty.

Explanation:

The question is about the relativistic effects that occur when one twin travels at significant fraction of the speed of light while the other remains on Earth.

According to the theory of relativity, time dilation will cause the traveling twin, Peggy, to age more slowly compared to her twin sister, Patty, who remains on Earth.

If Peggy travels to Alpha Centauri, which is 4.3 light years away, at 80% of the speed of light, and assuming the round trip takes 10 years for the Earth-bound twin, we can calculate that the moving twin will experience less than 10 years of elapsed time due to the effects of time dilation.

This happens because the faster Peggy travels, the more pronounced the effect of time dilation will be. This is a well-known result predicted by Einstein's special theory of relativity and has been confirmed through experiments involving high-speed particles and precise clocks.

Thus, when Peggy returns, she will be younger than her twin sister Patty, who has experienced the full 10 years on Earth.

Answer:

L = 182.4 m      

Explanation:

Given:-

- The number of turns of the coil, N = 50

- The shape of the coil = square

- The angle between the coil and magnetic field, θ = 30°

- The change in magnetic field, ΔB = ( 700 - 250 ) μT

- The time duration in which magnetic field changes, Δt = 0.3 s

- The induced emf, E = 60.0 mV

Solution:-

- The problem at hand is an application of Faraday's law. The law states that the induced emf ( E ) is proportional to the negative rate of change of magnetic flux ( ΔФ / Δt ) and number of turns of the coil ( N ).

- The Faraday's law is mathematically expressed as:

                    E =  - N* ( ΔФ / Δt )

Where,

- The flux ( Ф ) through a current carrying with an cross-sectional area ( A ) at a normal angle ( θ ) to the direction of magnetic field ( B ) is given by the following relationship.

                    Ф = B*A*cos ( θ )

- We need the rate of change of magnetic flux ( ΔФ / Δt ) for the Faraday's law. I.e the induced emf ( E ) is proportional to rate of change in magnetic field ( ΔB / Δt ), rate of change of angle between the coil and magnetic field ( Δθ / Δt ) or rate of change of cross-sectional area of the coil under the influence of magnetic field.

- To determine the exact relationship. We will derive the multi-variable function of flux ( Ф ) with respect to time "t":

                     Ф ( B , A , θ ) = B*A*cos ( θ )

- The first derivative would be ( Use chain and product rules )

    ( ΔФ / Δt ) = ΔB / Δt*A*cos ( θ ) + B*ΔA/Δt*cos ( θ ) - B*A*sin ( θ )*Δθ/Δt

- For the given problem the only dependent parameter that is changing is magnetic field ( B ) with respect to time "t". Hence, ( ΔA/Δt = Δθ/Δt = 0 ):

                        ΔФ / Δt  = (ΔB/Δt)*A*cos ( θ )

- Substitute the rate of change of magnetic flux  ( ΔФ / Δt ) into the expression for Faraday's Law initially stated:

                        E =  - N*(ΔB/Δt)*A*cos ( θ )

- Plug in the values and evaluate the Area of the square coil:

                       A =  - E / ( N*(ΔB/Δt)*cos ( θ ) )

                       A = - 0.06 / ( 50*[ (250-700)*10^-6/0.3 ] *cos ( 30° ) )

                       A = - 0.06 / -0.07216

                       A = 0.8314 m^2

- The square coil has equal sides ( x ). The area of a square A is given by:

                      A = x^2

                      x = √0.8314

                      x = 0.912 m

- The perimeter length of a single coil in terms of side length "x" is given as:

                      P = 4x

Whereas for a coil of N turns the total length ( L ) would be:

                      L = N*P

                      L = 4Nx

                      L = 4 * 50 * 0.912

                      L = 182.4 m                 ... Answer