An auto repair technician who specializes in the installation, troubleshooting, and repair of heating and air conditioning systems is called a/an A. passenger comfort specialist. B. maintenance and light repair (MLR) specialist. C. service manager. D. electrical system specialist.

Answers

Answer 1

Answer: c is the correct answer

Answer 2

Answer:

B is correct because I know it is :)

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A solid cylinder of cortical bone has a length of 500mm, diameter of 2cm and a Young’s Modulus of 17.4GPa. Determine the spring constant ‘k’. Please explain.

Answers

Answer:

The spring constant k is $1.115* 10^(9) N/m$

Solution:

As per the question:

Length of the solid cylinder, L = 500 mm = $500* 10^(- 3) = 0.5 m$

Diameter pf the cylinder, D = 2 cm = 0.02 m

As the radius is half the diameter,

Radius, R = 1 cm = 0.01 m

Young's Modulus, E = 17.4 GPa = $17.4* 10^(9) Pa$

Now,

The relation between spring constant, k and Young's modulus:

$kL = EA$

where

A = Area

Area of solid cylinder, A = $2\piR(L + R)$

$0.5k = 17.4* 10^(9)* 2\piR(L + R)$

$k = (17.4* 10^(9)* 2\pi* 0.01(0.01 + 0.5))/(0.5)$

k = $1.115* 10^(9) N/m$

Young's modulus, E is the ratio of stress and strain

And

Stress = $(Force or thrust)/(Area)$

Strain = $(length, L)/(elongated or change in length, \Delta L)$

Also

Force on a spring is - kL

Therefore, we utilized these relations in calculating the spring constant.

A neutral metal ball is suspended by a string. A positively charged insulating rod is placed near the ball, which is observed to be attracted to the rod. This is because:____________. a. the ball becomes negatively charged by induction
b. the ball becomes positively charged by induction
c. the string is not a perfect insulator
d. there is a rearrangement of the electrons in the ball
e. the number of electrons in the ball is more than the number in the rod

Answers

Answer:

d. there is a rearrangement of the electrons in the ball

Explanation:

Inside the neutral metal ball, there are equal no. of positive charges (protons) and negative charges (electrons). Normally, the charges are distributed evenly throughout the ball.

However, when the positively charged insulating rod is brought near, since positive charges and negative charges attract each other, the electrons (-ve charges) in the metal ball moves towards the side nearest to the rod. The metal ball gets attracted to the rod.

a and b are not correct because the rod is insulating, so electrons cannot be transferred between them to induce a net charge in the metal ball. the no. of electrons is unrelated to the attraction between opposite charges , so e is incorrect as well.

A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10.0 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30.0 m/s. If the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10.0 m/s, what is the magnitude of the velocity of the stone just before it hits the ground? Ignore any effects of air resistance.

Answers

Answer:

vf = 30 m/s : (the magnitude of the velocity of the stone just before it hits the ground)

Explanation:

Because the stone moves with uniformly accelerated movement we apply the following formulas:

vf²=v₀²+2*g*h Formula (1)

Where:

h: displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

g: acceleration due to gravity in m/s²

Free fall of the stone

Data

v₀ = 10 m/s

vf = 30.0 m/s

g = 9,8 m/s²

We replace data in the formula (1) to calculate h:

vf²=v₀²+2*g*h

(30)² = (10)² + (2)(9.8)*h

(30)²- (10)²= (2)(9.8)*h

h =( (30)²- (10)²) /( 2)(9.8)

h = 40.816 m

Semiparabolic movement of the stone

Data

v₀x = 10 m/s

v₀y = 0 m/s

g = 9.8 m/s²

h= 40.816 m

We replace data in the formula (1) to calculate vfy :

vfy² = v₀y² + 2*g*h

vfy² = 0 + (2)(9.8)( 40.816)

$v_(fy)=√(2*9.8*40.816) = 28.284 (m)/(s)$

$v_(f)=\sqrt{v_(ox)^2+v_(fy)^2}=√((10)^2+(28.284)^2) = 30(m)/(s)$

The magnitude of the velocity of the stone just before it hits the ground is 30 m/s.

The given parameters;

initial vertical velocity of the stone, $v_y_0$ = 10 m/s

final vertical velocity of the stone, $v_y_f$ = 30 m/s

The height traveled by the stone before it hits the ground is calculated as;

$v_y_f^2 = v_y_0^2 + 2gh\n\nh = (v_y_f^2- v_y_0^2)/(2g) \n\nh = ((30)^2 - (10)^2)/(2* 9.8) \n\nh = 40.82 \ m$

If the the stone is projected horizontally with initial velocity of 10 m/s;

the initial vertical velocity = 0

Final vertical velocity of the stone is calculated as follow;

$v_y_f^2 = v_y_0^2 + 2gh\n\nv_y_f^2 = 0 + 2* 9.8* 40.82\n\nv_y_f^2 = 800.07\n\nv_y_f = √(800.07) \n\nv_y_f = 28.28 \ m/s$

The horizontal velocity doesn't change.

the final horizontal velocity, $v_x_f$ = initial horizontal velocity = 10 m/s

The resultant of the final velocity of the stone before it hits the ground;

$v _f= √(v_x_f^2 + v_y_f^2) \n\nv_f = √(10^2 + 28.28^2) \n\nv_f= 29.99 \ m/s \approx 30 \ m/s$

Thus, the magnitude of the velocity of the stone just before it hits the ground is 30 m/s.

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Suppose that we are designing a cardiac pacemaker circuit. The circuit is required to deliver pulses of 1ms duration to the heart, which can be modeled as a 500 ohm resistance. The peak amplitude of the pulses is required to be 5 V. However, the battery delivers only 2.5 V. Therefore, we decide to charge two equal value capacitors in parallel from the 2.5V battery and then switch the capacitors in series with the heart during the 1ms pulse. What is the minimum value of the capacitances required so the output pulse amplitude remains between 4.9 V and 5.0 V throughout its 1ms duration

Answers

Answer:

Minimum capacitance = 200 μF

Explanation:

From image B attached, we can calculate the current flowing through the capacitors.

Thus;

Since V=IR; I = V/R = 5/500 = 0.01 A

Maximum charge in voltage is from 5V to 4.9V. Thus, each capacitor will have 2.5V. Hence, change in voltage(Δv) for each capacitor will be ; Δv = 0.05 V

So minimum capacitance will be determined from;

i(t) = C(dv/dt)

So, C = i(t)(Δt/Δv) = 0.01[0.001/0.05]

C = 0.01 x 0.0002 = 200 x 10^(-6) F = 200 μF

If you have two substances, one with a density of 2.0 g/cm3 and one with a density of 1.3 g/cm3 and you combined them, which one would float on topother and why?

Answers

Explanation:

Assuming the substances are fluids that do not mix, the lighter substance (ρ = 1.3 g/cm³) will float on top of the heavier substance (ρ = 2.0 g/cm³). This is due to Archimedes' Principle, which explains buoyancy.

A solid 0.6950 kg ball rolls without slipping down a track toward a vertical loop of radius ????=0.8950 m . What minimum translational speed ????min must the ball have when it is a height H=1.377 m above the bottom of the loop in order to complete the loop without falling off the track? Assume that the radius of the ball itself is much smaller than the loop radius ???? . Use ????=9.810 m/s2 for the acceleration due to gravity.

Answers

Answer:

The minimum transnational speed is 4.10 m/s.

Explanation:

Given that,

Mass of solid ball = 0.6950 kg

Radius = 0.8950 m

Height = 1.377 m

We need to calculate the minimum velocity of the ball at bottom of the loop to complete the track

Using formula velocity at lower point

$v_(min)=√(5gR)$

Put the value into the formula

$v_(min)=√(5*9.8*0.8950)$

$v_(min)=6.62\ m/s$

We need to calculate the velocity

Using conservation of energy

P.E at height +K.E at height = K.E at the bottom

$mgH+(1)/(2)mv^2=(1)/(2)m(√(5gR))^2$

$v^2=(√(5gR))^2-2gH$

$v^2=(6.62)^2-2*9.8*1.377$

$v^2=16.8352$

$v=√(16.8352)$

$v=4.10\ m/s$

Hence, The minimum transnational speed is 4.10 m/s.

Final answer:

The minimum translational speed the solid ball must have when it is at a height H=1.377 m above the bottom of the loop to successfully complete the loop without falling off the track is approximately 7.672 m/s. This was derived using principles of energy conservation.

Explanation:

The minimum translational speed must be sufficient enough to maintain contact with the track even at the highest point of the loop. Using the principle of energy conservation, the total energy at the height H, assuming potential energy to be zero here, should be equal to the total energy at the highest point of the loop. Here, the total energy at height H will consist of both kinetic and potential energy while at the top of the loop it consists of potential energy only. Setting these equations equal to each other: 0.5 * m * v² + m * g * H = m * g * 2R Solving the above equation for v:v = √2g (2R-H). Substituting known values henceforth gives us √2*9.81*(2*0.895-1.377) = 7.672 m/s. Hence, the ball must have a minimum translational speed of approximately 7.672 m/s at height H to complete the loop without falling.

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