Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 300 kN/m, fc = 100 kN/m, Dy = 300 kN, spanAB = 6m, span BC = 6m, spanCD = 6m

Experts, ACE, Genius... Can Anybody Calculate For The Reactions At Supports A And B Please? Will Give

Answers

Answer 1

Answer:

Support at Cy = 1.3 x 10³ k-N

Support at Ay = 200 k-N

Explanation:

given:

fb = 300 k-N/m

fc = 100 k-N/m

D = 300 k-N

L ab = 6 m

L bc = 6 m

L cd = 6 m

To get the reaction A or C.

take summation of moment either A or C.

Support Cy:

∑ M at Ay = 0

      (( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )

Cy = -------------------------------------------------------------------

                                      ( L ab + L bc )

Cy = 1.3 x 10³ k-N

Support Ay:

Since ∑ F = 0,           A + C - F - D = 0

                                   A = F  + D - C

                                  Ay = 200 k-N

Answer 2

Answer:

i was going to but its to late

Explanation:


Related Questions


a cylindrical jar is 10cm long and has a cross sectional area of 36cm. if it is completely filled with a fluid of relative density 0.2, calculate the mass of the fluid in the jar​

Answers

Answer:

The mass of the fluid is 72 g.

Explanation:

The following data were obtained from the question:

Height (h) = 10 cm

Area of cross section (A) = 36cm²

Relative density = 0.2

Mass =..?

Next, we shall determine the volume of the cylinder. This can be achieved by doing the following:

Volume = Area x Height

Volume = 36 x 10

Volume = 360 cm³

Next, we shall determine the density of the liquid.

This can be obtained as follow:

Relative density = density of substance/density of water.

Relative density = 0.2

Density of water = 1 g/cm³

Density of fluid =...?

Relative density = density of substance/density of water.

0.2 = density of fluid / 1 g/cm³

Cross multiply

Density of fluid = 0.2 x 1 g/cm³

Density of fluid = 0.2 g/cm³

Finally, we shall determine the mass of fluid as follow:

Volume = 360 cm³

Density of fluid = 0.2 g/cm³

Mass of fluid =...?

Density = mass /volume.

0.2 g/cm³ = mass of fluid /360 cm³

Cross multiply

Mass of fluid = 0.2 g/cm³ x 360 cm³

Mass of fluid = 72 g

Therefore, the mass of the fluid in the jar is 72 g.

The principles of magnetism apply everywhere on earth. What does this tell us about God and His character?

Answers

Answer:

God is omnipresent.

Explanation:

This means God is everywhere and He works where ever we are in the world

A bullet with a mass of 20 g and a speed of 960 m/s strikes a block of wood of mass 4.5 kg resting on a horizontal surface. The bullet gets embedded in the block. The speed of the block immediately after the collision is:________.
A) cannot be found because we don't know whether the surface is frictionless.
B) is 0.21 km/s.
C) is 65 m/s.
D) is 9.3 m/s.
E) None of these is correct

Answers

Answer:

4.25m/sE. None of the option is correct

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

A cylindrical shell of radius 7.00 cm and length 2.59 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 20.1 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C.
A) Use approximate relationships to find the net charge on the shell.
B) Use approximate relationships to find the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell.

Answers

the answer is c yw :D

A platinum wire is used to determine the melting point of indium. The resistance of the platinum wire is 2.000 Ω at 20°C and increases to 3.072 Ω as indium starts to melt. What is the melting point of indium? (The temperature coefficient of resistivity for platinum is 3.9 ×

Answers

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:

When a nerve cell depolarizes, charge is transferred across the cell membrane, changing the potential difference. For a typical nerve cell, 9.0 pC of charge flows in a time of 0.50 ms. Part A What is the average current

Answers

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

7.00 kg piece of solid copper metal at an initial temperature T is placed with 2.00 kg of ice that is initially at -20.0°C. The ice is in an insulated container of negligible mass and no heat is exchanged with the surroundings. After thermal equilibrium is reached, there is 0.90 kg of ice and 1.10 kg of liquid water.

Required:
What was the initial temperature of the piece of copper?

Answers

Answer:

122°C

Explanation:

From the data Final temperature is 0 deg C since there is 0.9kg of ice and 1.10kg of liquid water.

That means that 1.10kg of the ice undergoes Heat of Fusion which is 3.34x10^5 J/kg...

Heat lost by copper = Heat gained by ice + Heat of fusion

-> (7.0kg)(390J/kg*C)(0-T) = (2.00kg)(2100J/kg*C)(0 - (-20) + (1.10kg)(3.34x10^5 J/kg)

-> T(2730) = 334001

-> T = 122°C

A person standing 180m from the foot of a high building claps hi
hand and hears the echo 0.03minutes later. What is the speed
sound in air at that temperature?
A) 331m/s
B) 240m/s C) 200m/s D) 300m/s

Answers

Answer:

C) 200 m/s

Explanation:

The sound travels a total distance of 360 m in 0.03 minutes.

v = (360 m) / (0.03 min × 60 s/min)

v = 200 m/s

A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string

Answers

Answer:

The time interval is [tex]t = 5.48 *10^{-3} \ s[/tex]

Explanation:

From the question we are told that

   The length of the string is  [tex]l = 3.00 \ m[/tex]

    The  mass of the string is [tex]m = 5.00 \ g = 5.0 *10^{-3}\ kg[/tex]

     The  tension on the string is  [tex]T = 500 \ N[/tex]

   

The  velocity of the pulse is mathematically represented as

      [tex]v = \sqrt{ \frac{T}{\mu } }[/tex]

Where [tex]\mu[/tex] is the linear density which is mathematically evaluated as

       [tex]\mu = \frac{m}{l}[/tex]

substituting values

     [tex]\mu = \frac{5.0 *10^{-3}}{3}[/tex]

     [tex]\mu = 1.67 *10^{-3} \ kg /m[/tex]

Thus  

     [tex]v = \sqrt{\frac{500}{1.67 *10^{-3}} }[/tex]

    [tex]v = 547.7 m/s[/tex]

The time taken is evaluated as

    [tex]t = \frac{d}{v}[/tex]

substituting values

      [tex]t = \frac{3}{547.7}[/tex]

      [tex]t = 5.48 *10^{-3} \ s[/tex]

At t=0 a 2150kg rocketship in outer space fires the engine which exerts a force=At2, and F(1.25s)=781.25N in the x direction. Find the impulse J during the interval t=2.00s and t=3.5s

Answers

Answer:

5.81 X 10^3 Ns

Explanation:

Given that

F = At² and F at t = 1.25 s is 781.25 N ?

A = F/t² at t = 1.25 s => F = 781.25/(1.25)² = 500 N/s²

d(Impulse) = Fdt

Impulse = ∫Fdt =∫At²dt evaluated in the interval 2.00 s ≤ t ≤ 3.50 s

Impulse = At³/3 = (500/3)(t³) = 166.7t³ between t = 2.00 s and t = 3.50 s

Impulse = 166.7[3.5³ - 2³] = 166.7[42.875 - 8] = 166.7[34.875] = 5813.7 Ns

5.81 X 10^3 N.s

what is electric field strength​

Answers

Answer:

Electric field strengh is a measure of the strength of an electric field at a given point in space, equal to the field would induce on a unit electric charge at that point.

Electric field strength is also known as Electric Field Intensity .

Explanation:

Electric Field is also defined as force per charge. The unit will be force unit divided by charge unit. In this case, it will be Newton/Coulomb or N/C.

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A 0.10 kg point mass moves in a circular path with a radius of 0.36 m with a net force of 10.0 N toward the center of the circle. Select all of the following that are true statements.

a. The velocity of the object is 6 m/s toward the center of the circle.
b. The speed of the object is 6 m/s and decreasing.
c. The speed of the object is 6 m/s and increasing.
d. The velocity of the object is a constant 6 m/s.
e. The speed of the object is a constant 6 m/s.

Answers

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

The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s2. At an instant when the magnitude of the total acceleration is 6.0 m/s2, what is the speed of the particle? Group of answer choices

Answers

Answer:

The speed of the particle is 2.86 m/s

Explanation:

Given;

radius of the circular path, r = 2.0 m

tangential acceleration,  [tex]a_t[/tex] = 4.4 m/s²

total magnitude of the acceleration, a = 6.0 m/s²

Total acceleration is the vector sum of  tangential acceleration and radial acceleration

[tex]a = \sqrt{a_c^2 + a_t^2}\\\\[/tex]

where;

[tex]a_c[/tex] is the radial acceleration

[tex]a = \sqrt{a_c^2 + a_t^2}\\\\a^2 = a_c^2 + a_t^2\\\\a_c^2 = a^2 -a_t^2\\\\a_c = \sqrt{a^2 -a_t^2}\\\\a_c = \sqrt{6.0^2 -4.4^2}\\\\a_c = \sqrt{16.64}\\\\a_c = 4.08 \ m/s^2[/tex]

The radial acceleration relates to speed of particle in the following equations;

[tex]a_c = \frac{v^2}{r}[/tex]

where;

v is the speed of the particle

[tex]v^2 = a_c r\\\\v= \sqrt{a_c r} \\\\v = \sqrt{4.08 *2}\\\\v = 2.86 \ m/s[/tex]

Therefore, the speed of the particle is 2.86 m/s

When the reflected path from one surface of a thin film is one full wavelength different in length from the reflected path from the other surface and no phase change occurs, will the result be destructive interference or constructive interference?

Answers

Answer:

destructive interference

Explanation:

As we know that , when the phase difference between the path of two wavelength is 180°, then its known as destructive interference . And when the phase difference between the path of two wavelength is 0°, then its known as constructive interference.

In the constructive interference , the resulting amplitude will be maximum while in the destructive interference , the resulting amplitude will be zero(minimum).

Therefore the answer will be destructive interference.

The velocity selector in in a mass spectrometer consists of a uniform magnetic field oriented at 90 degrees to a uniform electric field so that a charge particle entering the region perpendicular to both fields will experience an electric force and a magnetic force that are oppositely directed. If the uniform magnetic field has a magnitude of 37.8 ~\text{mT}37.8 mT, then calculate the magnitude of the electric field that will cause a proton entering the velocity selector at 40.640.6 km/s to be undeflected. Give your answer in units of kV/m.

Answers

Answer:

50k/h is the answer to iy

A long, current-carrying solenoid with an air core has 1800 turns per meter of length and a radius of 0.0165 m. A coil of 210 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system

Answers

Answer:

The mutual inductance is  [tex]M = 0.000406 \ H[/tex]

Explanation:

From the question we  are told that

    The  number of turns per unit length  is  [tex]N = 1800[/tex]

    The radius is  [tex]r = 0.0165 \ m[/tex]

     The  number of turns of the solenoid is  [tex]N_s = 210 \ turns[/tex]

   

Generally the mutual inductance of the  system is mathematically represented as

       [tex]M = \mu_o * N * N_s * A[/tex]

Where A is the cross-sectional area of the system which is mathematically represented as

       [tex]A = \pi * r^2[/tex]

substituting values

      [tex]A = 3.142 * (0.0165)^2[/tex]

       [tex]A = 0.0008554 \ m^2[/tex]

also   [tex]\mu_o[/tex] is the permeability of free space with the value  [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

So  

      [tex]M = 4\pi * 10^{-7} *1800 * 210 * 0.0008554[/tex]

      [tex]M = 0.000406 \ H[/tex]

The electric field of a sinusoidal electromagnetic wave obeys the equation E = (375V /m) cos[(1.99× 107rad/m)x + (5.97 × 1015rad/s)t].
(a) What is the speed of the wave?
(b) What are the amplitudes of the electric and magnetic fields of this wave?
(c) What are the frequency, wavelength, and period of the wave? Is this light visible to humans?

Answers

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

If your metal car moves over a wide, closed loop of wire embedded in a road surface, is the magnetic field of the Earth within the loop altered? Does this produce a current pulse?

Answers

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.

A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s at a distance of 120 m from its starting point. When the truck has travelled a distance of 60 m from its starting point, its speed is v1 m/s.
Which of the following statements concerning v1 is true?
a. v1< 12.5m/s
b. v1= 12.5m/s
c. v1 >12.5m/s

Answers

Answer:

the correct answer is c     v₁> 12.5 m / s

Explanation:

This is a one-dimensional kinematics exercise, let's start by finding the link to get up to speed.

            v² = v₀² + 2 a₁ x

as part of rest v₀ = 0

           a₁ = v² / 2x

           a₁ = 25² / (2 120)

           a₁ = 2.6 m / s²

now we can find the velocity for the distance x₂ = 60 m

           v₁² = 0 + 2 a1 x₂

           v₁ = Ra (2 2,6 60)

           v₁ = 17.7 m / s

these the speed at 60 m

we see that the correct answer is c     v₁> 12.5 m / s

one arm of a u shaped tube contains water and the other alcohol. if the two fluids meet exactly at the bottom of the U and the alcohol is at a height of 18 cm at what height will water be

Answers

Complete Question

One arm of a U-shaped tube (open at both ends) contains water, and the other alcohol.

If the two fluids meet at exactly the bottom of the U, and the alcohol is at a height of 18 cm, at what height will the water be?Assume the density of alcohol is [tex]\rho_a = 790\ kg/m^3[/tex]

Answer:

The  height of water is  [tex]h_w = 0.142 \ m[/tex]

Explanation:

From the question we are told that

    The height of the alcohol is  [tex]h_a =18 \ cm = 0.18 \ m[/tex]

     The  density of the alcohol is  [tex]\rho_a = 790\ kg/m^3[/tex]

Generally the pressure on both arm of the tube are equal given that they are both open

i,e    [tex]P_a = P_w[/tex]

Where  [tex]P_a[/tex] is pressure of alcohol and  [tex]P_w[/tex] is pressure of water

   So the pressure on the arm of the tube containing the alcohol is mathematically evaluated as

         [tex]P_a = g * h * \rho[/tex]

substituting values

         [tex]P_a =9.8 * 0.18 * 790[/tex]

         [tex]P_a = 1394 \ Pa[/tex]

Generally the pressure on the arm of the tube containing the water is mathematically evaluated as

       [tex]P_w = g * h_w * \rho_w[/tex]

where  [tex]\rho_w[/tex] is the density of water which has  a value [tex]\rho _w = 1000 \ kg/m^3[/tex]

So  

      [tex]1394 = 9.8 * h_w * 1000[/tex]

=>    [tex]h_w = \frac{1394}{9800}[/tex]

=>  [tex]h_w = 0.142 \ m[/tex]

A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hits the center in 0.455 s. (Neglect any effects due to air resistance.)At what angle relative to the floor was the dart thrown?

Answers

Answer:

The  angle is  [tex]\theta = 15.48^o[/tex]

Explanation:

From the question we are told that  

     The distance of the dartboard from the dart is  [tex]d = 3.66 \ m[/tex]

     The time taken is  [tex]t = 0.455 \ s[/tex]

   

The  horizontal component of the speed of the dart is mathematically represented as

      [tex]u_x = ucos \theta[/tex]

where u is the the velocity at dart is lunched

  so

      [tex]distance = velocity \ in \ the\ x-direction * time[/tex]

substituting values

      [tex]3.66 = ucos \theta * (0.455)[/tex]

 =>   [tex]ucos \theta = 8.04 \ m/s[/tex]

From projectile kinematics the time taken by the dart can be mathematically represented as

         [tex]t = \frac{2usin \theta }{g}[/tex]

=>    [tex]usin \theta = \frac{g * t}{2 }[/tex]

       [tex]usin \theta = \frac{9.8 * 0.455}{2 }[/tex]

      [tex]usin \theta = 2.23[/tex]

=>   [tex]tan \theta = \frac{usin\theta }{ucos \theta } = \frac{2.23}{8.04}[/tex]

       [tex]\theta = tan^{-1} [0.277][/tex]

      [tex]\theta = 15.48^o[/tex]

     

A circular loop in the plane of a paper lies in a 0.75 T magnetic field pointing into the paper. The loop's diameter changes from 18.0 cm to 6.8 cm in 0.46 s.
A) Determine the direction of the induced current and justify your answer.
B) Determine the magnitude of the average induced emf.
C) If the coil resistance is 2.5 Ω, what is the average induced current?

Answers

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

what is transmission of heat?​

Answers

Answer:

Heat transfer is the transmission of heat energy from a body at higher temperature to lower temperature. The three mechanisms of heat transfer are

Conduction ConvectionRadiation.

Example of Conduction:

Heating a metal

Example of Convection:

Sea Breeze

Example of Radiation:

Sun

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Answer:

Transmission of heat is the movement of thermal energy from one thing to another thing of different temperature.

There are three(3) different ways heat can transfer and they are:

a) Conduction (through direct contact).

b) Convection (through fluid movement).

c) Radiation (through electromagnetic waves).

Examples: 1.Heating a saucepan of water using a coalpot.(conduction&convection).

2. Baking a pie in an oven(radiation).

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A beach ball filled with air is pushed about 1 m below the surface of a swimming pool and released from rest. Which of the following statements are valid, assuming the size of the ball remains the same?
a) The buoyant force on the ball decreases as the ball approaches the surface of the pool.
b) As the ball rises in the pool, the buoyant force on it increases.
c) The buoyant force on the ball equals its weight and remains constant as the ball rises.
d) The buoyant force on the ball while it is submerged is approximately equal to the weight of the volume of water that could fill the ball.
e) When the ball is released, the buoyant force exceeds the gravitational force, and the ball accelerates upward.

Answers

Answer:

e is correct

Explanation:

When a ball is pushed below the surface of a pool, it is submerged when the buoyant force is approximately equal to the water's weight of the volume that could fill the ball. When the ball is released, the buoyant force becomes greater than the gravitational force so that the ball accelerates upward.

What is buoyant force?

The buoyant force can be described as the upward force exerted on an object wholly or partially immersed in a fluid and is also called Upthrust. A body submerged partially or fully in a fluid due to the buoyant force appears to lose its weight.

The following factors affect buoyant force the density of the fluid, the volume of the fluid displaced, and the local acceleration due to gravity.

When an object immerses in water, the object experiences a force from the downward direction opposite to the gravitational pull, which causes a decrease in its weight. The difference in this pressure gives the upward force on the object, as buoyancy.

Therefore, options (d), (e) are correct.

Learn more about buoyant force, here:

https://brainly.com/question/21990136

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Consider the uniform electric field E = (8.0ĵ + 2.0 ) ✕ 103 N/C. What is its electric flux (in N · m2/C) through a circular area of radius 9.0 m that lies in the xy-plane? (Enter the magnitude.)

Answers

Answer:

5.09 x 10⁵ Nm²/C

Explanation:

The electric flux φ through a planar area is defined as the electric field Ε times the component of the area Α perpendicular to the field. i.e

φ = E A

From the question;

E = (8.0j + 2.0k) ✕ 10³ N/C

r = radius of the circular area = 9.0m

A = area of a circle = π r²           [Take π = 3.142]

A = 3.142 x 9² = 254.502m²

Now, since the area lies in the x-y plane, only the z-component of the electric field is responsible for the electric flux through the circular area.

Therefore;

φ = (2.0) x 10³ x 254.502

φ = 5.09 x 10⁵ Nm²/C

The electric flux is 5.09 x 10⁵ Nm²/C

10. A manufacturer knows from experience that the resistance of resistors she produces is normal with mean µ = 150Ω and the standard deviation σ = 5Ω. What percentage of the resistors will have resistance between 148 Ω and 152 Ω? Between 140 Ω and 160 Ω?

Answers

Answer:

Explanation:

Using the formula for finding the Z score

Z = x-µ/σ

x is the sample size

µ is the sample mean

σ is the standard deviation

For percentage of the resistors will have resistance between 148 Ω and 152 Ω, or is calculated as shown

P(148≤x152) = Z(152-150/5) - Z(148-150/5)

P(148≤x152) = Z(0.4)-Z( - 0.4)

P(148≤x152) = 0.6554-0.3446

The Z values are from the normal distribution table.

P(148≤x152) = 0.3108

The percentage of resistor that will have between 148 and 152 ohms is 0.3108×100% = 31.08%

Similarly for resistances between 140 Ω and 160 Ω

P(140≤x160) = Z(160-150/5) - Z(140-150/5)

P(140≤x160) = Z(2.0)-Z( - 2.0)

P(140≤x160) = 0.9775-0.02275

The Z values are from the normal distribution table.

P(140≤x160) = 0.9547

The percentage of resistor that will have between 140 and 160ohms is 0.9547×100% = 95.47%

The percentage of the resistors will have resistance between 148 Ω and 152 Ω is 31.08%

The percentage of the resistors will have resistance between 140 Ω and 160 Ω is  95.47%

Percentage of resistors:

To solve for the probability we will use the standard score of the Z score, which is given by:

Z = (x - µ)/σ

where x is the sample size

µ is the sample mean = 150Ω

σ is the standard deviation =

The probability of the resistors with resistance between 148 Ω and 152 Ω, will be:

P(148 ≤ 152) = Z((152-150)/5) - Z((148-150)/5)

P(148 ≤ 152) = Z(0.4)-Z( - 0.4)

P(148 ≤ 152) = 0.6554-0.3446

P(148 ≤ 152) = 0.3108

So, the percentage will be:

0.3108×100% = 31.08%

Similarly for resistances between 140 Ω and 160 Ω

P(140 ≤ 160) = Z((160-150)/5) - Z((140-150)/5)

P(140 ≤ 160) = Z(2.0)-Z( - 2.0)

P(140 ≤ 160) = 0.9775-0.02275

P(140 ≤ 160) = 0.9547

The percentage of resistor that will have between 140 and 160ohms is 0.9547×100% = 95.47%

Learn more about standard score:

https://brainly.com/question/25875323?referrer=searchResults

Do the math: How many seconds would it take an echo sounder’s ping to make the trip from a ship to the Challenger Deep (10,994 meters) and back? Recall that depth in meters = ½ (1500 m/sec × Echo travel time in seconds). Round your answer to two decimal places.

Answers

Answer:

14.66secs

Explanation:

Given the formula for calculating the depth in metres expressed as

depth in meters = ½ (1500 m/sec × Echo travel time in seconds)

Given depth of the challenger = 10, 994 meters, we will substitute this given value into the formula given to calculate the time take for the echo to travel.

10, 994 = depth in meters = ½ * 1500 m/sec × Echo travel time in seconds

10,994 = 750 * Echo travel time in seconds

Dividing both sides by 750;

Echo travel time in seconds = 10,994 /750

Echo travel time in seconds ≈ 14.66secs (to two decimal places)

Therefore, it would take an echo sounder’s ping 14.66secs to make the trip from a ship to the Challenger Deep and back

At which temperature do the lattice and conduction electron contributions to the specific heat of Copper become equal.

Answers

Answer:

At 3.86K

Explanation:

The following data are obtained from a straight line graph of C/T plotted against T2, where C is the measured heat capacity and T is the temperature:

gradient = 0.0469 mJ mol−1 K−4 vertical intercept = 0.7 mJ mol−1 K−2

Since the graph of C/T against T2 is a straight line, the are related by the straight line equation: C /T =γ+AT². Multiplying by T, we get C =γT +AT³ The electronic contribution is linear in T, so it would be given by the first term: Ce =γT. The lattice (phonon) contribution is proportional to T³, so it would be the second term: Cph =AT³. When they become equal, we can solve these 2 equations for T. This gives: T = √γ A .

We can find γ and A from the graph. Returning to the straight line equation C /T =γ+AT². we can see that γ would be the vertical intercept, and A would be the gradient. These 2 values are given. Substituting, we f ind: T =

√0.7/ 0.0469 = 3.86K.

Three identical capacitors are connected in series to a battery. If a total charge of Q flows from the battery, how much charge does each capacitor carry?

Answers

Answer:

Each of the capacitor carries the same charge, Q

Explanation:

When capacitors are connected in series, the battery voltage is divided equally across the capacitors. The total voltage across the three identical capacitors is calculated as follows;

[tex]V_T = V_1 + V_2 + V_3[/tex]

We can also calculate this voltage in terms of capacitance and charge;

[tex]V = \frac{Q}{C} \\\\V_T = V_1 + V_2 +V_3 \\\\(given \ total \ charge \ as \ Q, then \ the \ total \ voltage \ V_T \ can \ be \ written \ as)\\\\V_T = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} \\\\V_T = Q(\frac{1}{C_1 } +\frac{1}{C_2} + \frac{1}{C_3 })\\\\[/tex]

Therefore, each of the capacitor carries the same charge, Q

Why does front side of spoon forms inverted image but the back side form opposite of inverted image?

Answers

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.

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