Technician A says that a heavy engine load results in high intake manifold vacuum and a high MAP sensor signal voltage. Technician B says that a manifold absolute pressure (MAP) sensor uses a perfect vacuum (zero absolute pressure) in the sensor to determine the pressure. Who is right?

Answers

Answer 1

Answer:

Technician B

Explanation:


Related Questions

Technician A says that the slip rings allow current to flow from the stationary end frame to the moving rotor. Technician B says that the stator is a rotating component in the alternator. Who is correct

Answers

Answer:

Technician A

Explanation:

The stator is a STATIONARY component of a rotary system. The stator is a stationary component in the alternator, therefore, technician B saying that it is a rotating component is WRONG.

Also, slip rings allows the flow of electrical signals and electrical power from a stationary structure to a rrotating structure. Therefore, technician A is correct in his take that the slip rings allow current to flow from the stationary end frame to the moving rotor.

Final answer:

Technician A is correct about slip rings allowing current flow to the rotor, while Technician B is incorrect as the stator is the stationary part of an alternator.

Explanation:

Technician A is correct in saying that slip rings allow current to flow from the stationary part to the moving rotor. Slip rings are made of a ring and brushes; the ring is attached to the rotating part of a machine (like an alternator's rotor), and the brushes are attached to the stationary part (such as the end frame), facilitating electrical contact as the rotor spins.

Technician B is incorrect; the stator is the stationary part of an alternator or electric motor that contains conductors where the electric current is induced by the motion of the rotor. In an alternator, the rotor is the rotating component, not the stator.

______ Was once widely used in the united states as a gasoline additive

Answers

Answer:

lead

Explanation:

Lead was once widely used in the United States as a gasoline additive.

Addition of lead is in the form of tetra ethyl lead(II).

It helps to improve the octane rating of gasoline and to produce more useful energy via each combustion step.

The compound containing lead was banned due to the huge health risk it poses to people. Lead poisoning is a known defect that affects people. When the gasoline is combusted, it releases lead as a by-product. Exposure to a high level of lead can cause brain damage and kidney failure.

A 2.3-kg toy locomotive is pulling a 1.0-kg caboose. The frictional force of the track on the caboose is 0.48 N backward along the track. If the train is accelerating forward is 2.6 m/s2, what is the magnitude of the force exerted by the locomotive on the caboose

Answers

Final answer:

The force exerted by the toy locomotive on the caboose is 3.08 N, which is calculated by adding the force required to overcome friction and the force necessary for acceleration according to Newton's second law.

Explanation:

To determine the magnitude of the force exerted by the toy locomotive on the caboose, we can use Newton's second law, which states that the force is equal to the mass times the acceleration (F = ma).

However, we also need to account for the frictional force acting on the caboose. The total force required to accelerate the caboose includes the force to overcome friction and the force needed for acceleration:

Step 1: Calculate the force needed to overcome friction, which is already given as 0.48 N.

Step 2: Determine the force needed to accelerate the caboose using the formula F = ma, where m is the mass of the caboose (1.0 kg) and a is the acceleration (2.6 m/s2).

Step 3: Add the force to overcome friction to the force required for acceleration to get the total force exerted by the locomotive on the caboose.

Step 2 in detail: F = ma = 1.0 kg × 2.6 m/s2 = 2.6 N.

Step 3 in detail: The total force exerted by the locomotive on the caboose is 2.6 N (to accelerate) + 0.48 N (to overcome friction) = 3.08 N.

Therefore, the force exerted by the toy locomotive on the caboose is 3.08 N.

What is the lewis structure of the covalent compound that contains one nitrogen atom

Answers

Full Question:

What is the Lewis structure of the covalent compound that contains one nitrogen atom, one hydrogen atom, and one carbon atom?

Explanation:

This covalent compound is the hydrogen cyanide, HCN.

The following steps are used to obtain it;

Step 1. Draw a skeleton structure of the compound based on the elements

Put the least electronegative atom C in the middle with H and Cl on either side.

H-C-N

Step 2. Count the valence (outermost) electrons you can use

H + C + N = 1 + 4 + 5 = 10

Step 3. Add these electrons to give every atom an octet

You have to put a triple bond between C and N.

Note: Each bond is made up of two electrons.

The lewis structure is given in the image below;

A satellite of mass 230 kg is placed in Earth orbit at a height of 500 km above the surface. (a) Assuming a circular orbit, how long does the satellite take to complete one orbit

Answers

Answer:

Orbital period of satellite is 5.83 x 10³ s

Explanation:

The orbital period of satellite revolving around Earth is given by the equation :

[tex]T=\sqrt{\frac{4\pi ^{2} (R+h)^{3} }{GM} }[/tex]      .....(1)

Here R is radius of Earth, h is height of satellite from the Earth's surface, M is mass of Earth and G is gravitational constant.

In this problem,

Height of satellite, h = 500 km = 500 x 10³ m

Substitute 6378.1 x 10³ m for R, 500 x 10³ m for h, 5.972 x 10²⁴ kg for M and 6.67 x 10⁻¹¹ m³ kg⁻¹ s⁻² for G in equation (1).

 [tex]T=\sqrt{\frac{4\pi ^{2} [(6378.1+500)\times10^{3} ]^{3} }{6.67\times10^{-11} \times5.972\times10^{24} } }[/tex]

T = 5.83 x 10³ s

In general it is expected that ________. A) osmotic pressure will remain relatively consistent throughout the capillary bed B) osmotic pressure will be lower in the arteriole end of the capillary bed compared to the venous end C) osmotic pressure will be higher in the arteriole end of the capillary bed compared to the venous end D) hydrostatic pressure will remain constant throughout the capillary bed

Answers

Answer:

B. OSMOTIC PRESSURE WILL BE LOWER IN THE ARTERIOLE END OF THE CAPPILLARY BED COMPARED TO THE VENOUS END.

Explanation:

This is true for filtration to take place in the cappillary bed. Osmotic pressure is the net pressure that drives movement of fluid from the interstitial fluid back into the capillaries. Osmotic pressure increase favors reabsorption as water moves from region of higher water concentration in the interstitial fluid to the lower region of water concentration in the capillaries.

At the ends of a capillary bed, the difference in the hydrostatic and osmotic pressures provides a net filtration or reabsorption ratio. At the arteriole end of the capillary bed, hydrostatic pressure is greater than the osmotic pressure allowing movements of fluid to the interstitial fluid (filtration) while as the blood moves to the venous end, the osmotic pressure becomes greater than than hydrostatic pressure.

Final answer:

Osmotic pressure is usually higher at the arteriole end of the capillary bed than at the venous end (Option C). This happens because plasma proteins remain in the capillary causing water to move back into the capillary.

Explanation:

In general, it is expected that osmotic pressure will be higher in the arteriole end of the capillary bed compared to the venous end (option C). This is because during capillary exchange, fluids and solutes are filtered out at the arteriole end of capillaries due to higher blood pressure, and then reabsorbed at the venous end due to higher osmotic pressure. This helps maintain fluid balance and prevent edema.

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Radioactivity is a tendency for an element or a material to

Answers

Answer: Emit radiation


-radiate heat
-Emit radiation
-emit light
-emote electrons

At one instant of time, a car and a truck are traveling side by side in adjacent lanes of a highway. The car has a greater velocity than the truck has. Does the car necessarily have the greater acceleration?Explain

Answers

Answer:

NO

Explanation:

Acceleration is change in velocityΔv in respect to timeΔt

so if the velocity of the car is greater than the truck it does not mean that the car acceleration is greater than the truck.

Sometimes with constant velocity it means no accelaration ,but the truck may have accelaration

so, higher velocity of the car does not mean higher acceleration

nderstanding the high-temperature formation and breakdown of the nitrogen oxides is essential for controlling the pollutants generated by car engines. The second-order reaction for the breakdown of nitric oxide to its elements has rate constants of 0.0796 L/mol-s at 737°C and 0.0815 L/mol-s at 947°C. What is the activation energy of this reaction? Give your answer in scientific notation.

Answers

Answer:

[tex]E_a=1124.83 J/mol[/tex]

Explanation:

Given that second order equation

K₁ = 0.0796 L/mol-s , T₁= 737⁰C

T₁ = 737 + 273 K = 1010 K

K₂ = 0.0815 L/mol-s , T₂=947°C

T₂=947+273 K= 1220 K

The activation energy given as follows

[tex]\ln\dfrac{K_2}{K_1}=\dfrac{E_a}{R}\left ( \dfrac{1}{T_1}-\dfrac{1}{T_2} \right )[/tex]

Now by putting the values we can get

[tex]\ln\dfrac{0.0815}{0.0796}=\dfrac{E_a}{8.314}\left ( \dfrac{1}{1010}-\dfrac{1}{1220} \right )[/tex]

[tex]0.023=0.00017\times \dfrac{E_a}{8.314}[/tex]

[tex]E_a=0.023\times \dfrac{8.314}{0.00017}[/tex]

[tex]E_a=1124.83 J/mol[/tex]

Therefore the activation energy will be 1124.83 J/mol

A particular heat engine has a mechanical power output of 5.00 kW and an efficiency of 25.0%. The engine expels 8.00 3 103 J of exhaust energy in each cycle. Find (a) the energy taken in during each cycle and (b) the time interval for each cycle.

Answers

Answer:

Explanation:

Given

Power output [tex]P=5\ kW[/tex]

efficiency [tex]\eta =25\ \%[/tex]

Engine expels [tex]Q_r=8\times 10^3\ J[/tex]

Efficiency is given by

[tex]\eta =1-\dfrac{Q_r}{Q_s}[/tex]

where [tex]Q_s[/tex]=Heat supplied

[tex]0.25=1-\dfrac{8\times 10^3}{Q_s}[/tex]

[tex]0.75=\dfrac{8\times 10^3}{Q_s}[/tex]

[tex]Q_s=\dfrac{8\times 10^3}{0.75}[/tex]

[tex]Q_s=10.667\ kJ[/tex]

Work Produced by  cycle

[tex]W=Q_s-Q_r[/tex]

[tex]W=10.667-8[/tex]

[tex]W=2.667\ kJ[/tex]

Time interval for which power is supplied

[tex]P\times t=W[/tex]

[tex]t=\dfrac{W}{P}[/tex]

[tex]t=\dfrac{2.667}{5}[/tex]

[tex]t=0.5334\ s[/tex]  

Final answer:

The energy taken in each cycle of this particular heat engine is 2.0 x 10^4 J, and the time interval for each cycle is four seconds.

Explanation:

The efficiency of a heat engine (e) is derived from work output (W) divided by the energy input (Qin). Given the mechanical power output of the engine and its efficiency, we can use this formula to determine the energy input and the time interval for each cycle.

(a) Energy taken in during each cycle: Since efficiency e = Wout/Qin, then Qin = Wout / e = 5,000W (or 5 x 103 J/s) / 0.25 = 2.0 x 104 J. Where Wout is 5000W converted to Joules per second.

(b) Time interval for each cycle: The energy balance for one cycle is given by Qin = Wout + Qexhaust, where Wout is work output and Qexhaust is exhaust energy. The time for one cycle t = Qin / W = 2.0 x 104 J / 5,000 J/s = 4 seconds. The time interval for each cycle is four seconds.

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Water waves in a shallow dish are 7.0 cm long. At one point, the water moves up and down at a rate of 4.0 oscillations per second. (a) What is the speed of the water waves?

Answers

Answer:

0.28 m/s

Explanation:

From the equation of a moving wave,

V = λf.................. Equation 1

Where V = speed of the water wave, λ = wave length of the water wave, f = frequency of the water wave.

Given: λ = 7.0 cm = 0.07 m, f = 4 Hz.

Substitute into equation 1

V = 0.07(4)

V = 0.28 m/s.

Hence the speed of the water wave = 0.28 m/s

For the environment, why is the characteristic of regularity important?

Please use your own words.

Answers

Answer:

so we have a good place to live at.

Explanation:

Environment is understood to be the set of natural and human factors that surround man in his daily life. Thus, landforms, natural resources, buildings, etc., are part of it.

Taking these definitions into account, a regular environment is one in which all its conditions and components are found with the fewest possible alterations, or with human alterations that do not negatively affect its natural conditions.

In this way, an environment that has not definitively consumed its resources or that has not significantly affected the natural status of the region is considered regular. This characteristic is important because it allows the environment to not be negatively affected, allowing a normal development of human life.

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1) Calculate the torque required to accelerate the Earth in 5 days from rest to its present angular speed about its axis. 2) Calculate the energy required. 3) Calculate the average power required.

Answers

Answer:

a) τ = 4.47746 * 10^25 N-m

b) E = 2.06301 * 10^13 J

c) P = 3.25511*10^21 W

Explanation:

Given that,

The radius of earth r = 6.3781×10^6 m

The angular speed of earth w = 7.27*10^-5 rad/s

The time taken to reach above speed t = 5 yrs = 1.57784760 * 10^8 s

The mass of earth m = 5.972 × 10^24 kg

The inertia of sphere I = 2/5 * m* r^2

Solution:

angular acceleration of the earth from rest to w is given by α:

                               α = w / t

                               α = (7.27*10^-5) / (1.57784760 * 10^8)

                               α = 4.60754*10^-13 rad/s^2

The required torque τ is given by:

                               τ = I*α

                               τ = 2/5 * m* r^2 * α

 τ = 2/5 *(5.972 × 10^24) * (6.3781×10^6)^2 * (4.60754*10^-13)

 τ = 4.47746 * 10^25 N-m

Power required P to turn the earth to the speed w is:

                          P = τ*w

                          P = (4.47746 * 10^25)*(7.27*10^-5)

                          P = 3.25511*10^21 W

Energy E required is :

                          E = P / t

                          E = (3.25511*10^21) / (1.57784760 * 10^8)

                          E = 2.06301 * 10^13 J

Stars of spectral class m do not show strong lines of hydrogen in their spectra because

Answers

Answer:Hydrogen lines will be weak if the star is too hot or too cold.

Explanation:

At higher temperatures, the hydrogen atom ionizes due to the atomic collisions. M spectral class stars are mainly the main sequence and red stars. They are in a temperature range of 3000 K which means that these stars have maximum ionized calcium lines. In this star, hydrogen atoms have electrons in the lower energy state and it is difficult to absorb photons. These stars do not have enough temperature for absorption and undergo fusion.

The heat of fusion for water is 80. cal/g. How many calories of heat are released when 20.0 g of water at 0°C is frozen to ice?

Answers

Answer:

1600 cal

Explanation:

The formula

q=mΔ[tex]_{fus}[/tex][tex]H^{0}[/tex]

is used to calculate the heat required to melt a solid where q=amount of heat, m=mass and Δ[tex]_{fus}[/tex][tex]H^{0}[/tex] is the enthalphy of fusion

now we substitute, m=20g, Δ[tex]_{fus}[/tex][tex]H^{0}[/tex]=80cal/g

Therefore, q=20g x 80cal/g =1600 cal

I hope you find this information useful and interesting! Good luck!

Final answer:

The heat released when 20.0 g of water at 0°C is frozen to ice is 1600 calories.

Explanation:

When water freezes, it releases heat energy. The quantity of heat energy needed to convert a substance from a solid to a liquid or vice versa without affecting its temperature is known as the heat of fusion.

In this case, we are asked to find the number of calories of heat released when moment 20 g of water freezes to ice at zero degrees.

To calculate this, we can use the formula:

Heat released = mass of water * heat of fusion

Substituting the given values:

Heat released = 20.0 g * 80. cal/g = 1600 cal

Therefore, 1600 calories of heat are released when 20.0 g of water freezes to ice at 0 °C.

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Consider a current carrying a wire coming out of your computer screen towards you. Which statement below correctly describes the magnetic field created by the current in the wire?
1. The magnetic field encircles the wire in a counterclockwise direction
2. The magnetic field encircles the wire in a clockwise direction

Answers

Answer:

1. The magnetic field encircles the wire in a counterclockwise direction

Explanation:

When we have a current carrying wire perpendicular to the screen in which the current flows out of the screen then by the Maxwell's right-hand thumb rule we place the thumb of our right hand in the direction of the current and curl the remaining fingers around the wire, these curled fingers denote the direction of the magnetic field which is in the counter-clock wise direction.

Ever current carrying conductor produces a magnetic field around it.

If we use the Doppler method to measure the period with which a star alternately moves toward and away from us due to an orbiting planet, then we also know the __________

Answers

Answer:

Center of mass

Explanation:

The Doppler technique is a good method for discovering exoplanets. It uses the Doppler effect to analyze the motion and properties of the star and planet. Both the planet and the star are orbiting a common center of mass. This means that the star and the planet gravitationally attract one another, causing them to orbit around a point of mass central to both bodies.

If we use the Doppler method to measure the period with which a star alternately moves toward and away from us due to an orbiting planet, then we also know the planet's orbital period

When utilizing the Doppler method to measure the periodic shifts in a star's spectral lines caused by an orbiting planet, not only do we discern the star's radial velocity variations, but crucially, we also ascertain the orbital period of the planet. This period signifies how long it takes for the planet to complete one orbit around its host star.

The Doppler effect causes a star's light to shift towards the blue end of the spectrum as it moves closer to Earth and towards the red end as it moves away. By analyzing these shifts, astronomers can deduce the star's motion induced by the gravitational pull of its companion planet. The periodicity of these shifts corresponds directly to the planet's orbital period. This information is fundamental in understanding the planet's characteristics, including its distance from the star and its mass, contributing significantly to our knowledge of exoplanetary systems in the vast cosmos.

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A 265 g mass attached to a horizontal spring oscillates at a frequency of 3.40 Hz . At t =0s, the mass is at x= 6.20 cm and has vx =− 35.0 cm/s . Determine The phase constant.

Answers

Answer:

The phase constant is 7.25 degree  

Explanation:

given data

mass = 265 g

frequency = 3.40 Hz

time t = 0 s

x = 6.20 cm

vx = - 35.0 cm/s

solution

as phase constant is express as

y = A cosФ ..............1

here A is amplitude that is = [tex]\sqrt{(\frac{v_x}{\omega })^2+y^2 }[/tex]  = [tex]\sqrt{(\frac{35}{2\times \pi \times y})^2+6.2^2 }[/tex]  =  6.25 cm

put value in equation 1

6.20 = 6.25 cosФ

cosФ  = 0.992

Ф = 7.25 degree  

so the phase constant is 7.25 degree  

A rope with a mass density of 1 kg/m has one end tied to a vertical support. You hold the other end so that the rope is horizontal and has a tension of 4 N. If you move the end of the rope back and forth, you produce a transverse wave in the rope with a wave speed of 2 m/s. If you double the amount of tension you exert on the rope, what is the wave speed?a. 2.8 m/sb. 1.0 m/sc. 2.0 m/sd. 0.25 m/se. 4.0 m/s

Answers

Answer: Option (a) is the correct answer.

Explanation:

The given data is as follows.

   mass per unit length ([tex]\mu[/tex]) = [tex]\frac{M}{l}[/tex] = 1 kg/m

      Tension = 4 N

    Speed (v) = [tex]\sqrt{\frac{F}{\mu}}[/tex]

So, when F is doubled then change in value of F will be as follows.

               F = 4 + 4 = 8 N

Therefore, speed will be calculated as follows.

           v = [tex]\sqrt{\frac{8 N}{1 kg/m}}[/tex]

              = 2.8 m/s

Thus, we can conclude that the wave speed is 2.8 m/s.

As a simple pendulum swings back and forth, the forces acting on the suspended object are the force of gravity, the tension in the supporting cord, and air resistance. Determine which forces will do work, and explain why or why not.

Answers

Final answer:

The force of gravity and air resistance do work on a simple pendulum, with gravity doing positive and negative work as the pendulum moves, and air resistance dissipating energy from the system. Tension in the string does not do work since it is always perpendicular to the pendulum's motion.

Explanation:

In the context of a simple pendulum, the forces capable of doing work are the force of gravity and air resistance. The force of gravity does work on the pendulum as it swings back and forth because it has a component along the direction of the pendulum's movement. Specifically, as the pendulum swings, gravity pulls it downward, causing it to accelerate towards the lowest point of its path. Upon reaching this lowest point, the pendulum has maximum kinetic energy because the potential energy due to its elevated position has been converted into kinetic energy. The pendulum then slows down as it climbs against gravity until it stops momentarily at the highest point of its swing, and gravity starts doing negative work, converting kinetic energy back into potential energy. The tension in the string, although a force present in the pendulum system, does no work because it acts perpendicular to the direction of the pendulum's motion at all times.

Air resistance does work, although typically very small, by opposing the motion of the pendulum and thus removing energy from the system, mostly in the form of thermal energy due to the friction between the air and the pendulum bob. In simple pendulum motion, air resistance is usually considered negligible, but it does contribute to the eventual stopping of the pendulum due to energy dissipation.

The force of gravity does work in both directions, the tension in the string does no work, and air resistance does work by dissipating the pendulum's energy.

Force of Gravity (Weight): The force of gravity does work on the pendulum bob. Gravity acts downward and provides the restoring force that accelerates the pendulum back toward its equilibrium position. As the pendulum swings, gravity causes the bob to move, thereby doing positive work as the bob descends and negative work (taking energy away from the pendulum's motion) as the bob ascends.

Tension in the String: The tension in the string does no work on the pendulum bob. This is because the tension force always acts perpendicular to the direction of the bob's instantaneous motion. Work is defined as the force component in the direction of displacement times the displacement itself. Since there is no displacement in the direction of the tension force, it does no work.

Air Resistance: Air resistance does work on the pendulum bob, but in the opposite sense to gravity. Air resistance acts against the direction of motion of the bob, causing it to lose energy over time. This dissipation of energy due to air resistance results in damping, gradually reducing the amplitude of the pendulum's oscillation.

An rv travels 45 km east and stays the night at a KOA. The next day it travels for 3 hours to the north l, traveling 110 km. What is the displacement over the two days for the RV?

Answers

Answer:

The displacement of RV for the two days is 118.85 km at an angle of 67.75° with the east direction.

Explanation:

Given:

Distance moved in the East direction (d) = 45 km

Distance moved in the North direction (D) = 110 km

Displacement is defined as the difference of final position and initial position.

Let us draw a diagram representing the above situation.

Point A is the starting point and point C is the final position of RV.

So, the displacement of RV in two days is given as:

Displacement = Final position - Initial position = AC

Now, triangle ABC is a right angled triangle with AB = 45 km, BC = 110 km, and AC being the hypotenuse.

Using Pythagoras theorem, we have:

[tex]AC^2=AB^2+BC^2\\\\AC=\sqrt{AB^2+BC^2}[/tex]

Plug in the given values and solve for AC. This gives,

[tex]AC=\sqrt{(45\ km)^2+(110\ km)^2}\\\\AC=\sqrt{2025+12100}\ km\\\\AC=\sqrt{14125}\ km\\\\AC=118.85\ km[/tex]

Now, the direction of displacement with the east direction is given as:

[tex]\theta=\tan^{-1}(\frac{BC}{AB})\\\\\theta =\tan^{-1}(\frac{110}{45})=67.75^\circ[/tex]

Therefore, the displacement of RV for the two days is 118.85 km at an angle of 67.75° with the east direction.

The displacement of the rv over the two days is 118.85 km.

The given parameters;

initial displacement of the rv = 45 km eastfinal displacement of the rv, = 110 km north

The displacement of the rv over the two days is calculated by applying Pythagoras theorem as follows;

[tex]c^2 = a^2 + b^2\\\\c = \sqrt{a^2 + b^2} \\\\c = \sqrt{(45)^2 + (110)^2} \\\\c = 118.85 \ km[/tex]

Thus, the displacement of the rv over the two days is 118.85 km.

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A boy is pulling a wagon with a force of 70.0 N directed at an angle of 30 degrees above the horizontal. What are the x and y components of this force?

Answers

Answer:

x=61

y=35

Explanation:

vertical component= 70sin(30)

Horizontal component=70cos(30)

A 4.0-kg object is moving with speed 2.0 m/s. A 1.0-kg object is moving with speed 4.0 m/s. Both objects encounter the same constant braking force, and are brought to rest. Which object travels the greater distance before stopping?

Answers

Both the objects travel equal distance before stopping.

Explanation:

Given-

Mass of object 1, m₁ = 4kg

Speed of object 1, v₁ = 2m/s

Mass of object 2, m₂ = 1kg

Speed of object 2, v₂ = 4m/s

Force₁ = Force₂ = F

Distance, s = ?

We know,

[tex]v^2 - u^2 = 2as\\\\[/tex]

Where, v is the final velocity

            u is the initial velocity

            a is the acceleration

            s is the distance

When the brake is applied, the object comes to rest and the final velocity, v becomes 0. So,

[tex]s = \frac{u^2}{2a}[/tex]

We know,

[tex]a = \frac{F}{m}[/tex]

The stopping distance becomes,

[tex]s = \frac{u^2m}{2F}[/tex]

For object 1:

[tex]s = \frac{(2)^2 X 4 }{F}[/tex]

[tex]s = \frac{16}{F}[/tex]

For object 2:

[tex]s = \frac{(4)^2 X 1}{F}\\ \\s = \frac{16}{F}[/tex]

For both the objects the distance travelled is same.

Therefore, both the objects travel equal distance before stopping.

There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted so that it can be rotated about its axis by turning a hand crank. The outer wall of the cylinder is perforated with small holes. You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is 13.1 cm. When the cylinder is rotating at 2.99 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall

Answers

Answer:[tex]a_c=46.24\ m/s^2[/tex]

Explanation:

Given

radius of container [tex]r=13.1\ cm[/tex]

cylinder is rotating with [tex]N=2.99 rps[/tex]

Centripetal acceleration at the outer wall is given by

[tex]a_c=\omega ^2\times r[/tex]

where [tex]\omega [/tex]=Angular speed

[tex]\omega =2\pi N[/tex]

[tex]\omega =2\pi \times 2.99[/tex]

[tex]\omega =18.78\ rad/s[/tex]

[tex]a_c=(18.78)^2\times 0.131[/tex]

[tex]a_c=46.24\ m/s^2[/tex]

A battery has a terminal voltage of 12.0 V when no current flows. Its internal resistance is 2.0 Ω. If a 4.6 Ω resistor is connected across the battery terminals, what is the terminal voltage and what is the current through the 4.6 Ω resistor?

Answers

Answer:

Check attachment for solution

Explanation:

Given that 12V battery

Answer:

Terminal voltage = 8.36 V

Current = 1.82 A

Explanation:

E.M.F of battery = 12V

Internal resistance of battery (r) = 2Ω

Resistance of resistor (R) = 4.6Ω

Now the formula for terminal voltage across the battery is;

V = ε - Ir

Where ε is EMF and I is electric current

Using ohms law, we know that V = IR and I = V/R.

Thus, let's put V/R for current in the potential difference equation;

V = ε - r(V/R)

Thus, lets make V the subject of the formula ;

V + (rV/R) = ε

V(1 + r/R) = ε

So, V = ε/(1 + r/R)

V = 12/(1 + (2/4.6))

V = 12/(1 + 0.4348)

V = 12/1.4348 = 8.36 V

Thus from V=IR, we can find current. So 8.36 = I(4.6)

I = 8.36/4.6 = 1.82 A

Calculate the magnitude of the force exerted by each wire on a 1.20-m length of the other.

Answers

Incomplete question.The complete question is attached below as screenshot along with figure

Answer:

[tex]F=6.00*10^{-6}N[/tex]

Force is repulsive

Explanation:

Given data

Current I₁=5.00A

Current I₂=2.00A

Length L=1.20 m

Radius r=0.400m

To find

Force F

Solution

As the force is repulsive because currents are in opposite direction

From repulsive force we know that:

[tex]F=\frac{u_{o}I_{1}I_{2}L}{2\pi r}[/tex]

Substitute the given values

[tex]F=\frac{u_{o}(5.00A)(2.00A)(1.20m)}{2\pi (0.400m)}\\ F=6.00*10^{-6}N[/tex]

A child on a merry-go-round takes 3.9 s to go around once. What is his angular displacement during a 1.0 s time interval?

Answers

Answer:

Angular displacement=2π/3.9 rad

Explanation:

Given data

Time t=3.9s

Required

The angular displacement during a 1.0 s time interval

Solution

In 3.9 second the child covers a full circle=2π rad

Angular displacement after 1.0 second is given as:

[tex]=\frac{2\pi }{3.9} rad[/tex]

Angular displacement=2π/3.9 rad

Final answer:

The child's angular displacement on the merry-go-round during a 1.0s time interval would be approximately 1.609 radians.

Explanation:

To find the child's angular displacement on the merry-go-round, we first need to know the rate at which the merry-go-round is turning. This is called the angular velocity and is measured in radians per seconds (rad/s). If it takes 3.9 seconds for the merry-go-round to make one full revolution, this equals 2π radians. Therefore, the angular speed of the merry-go-round is 2π/3.9 rad/s.

Now, if we want to know how much the child displaces in 1.0 second, we simply multiply the angular speed by the time interval. So the angular displacement is (2π/3.9 rad/s)*1.0s = 1.609 rad.

Therefore, the child's angular displacement during a 1.0s time interval would be approximately 1.609 radians.

Learn more about Angular Displacement here:

https://brainly.com/question/14769426

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A negative test charge will accelerate toward regions of ________ electric potential and ________ electric potential energy.

Choose the appropriate answer combination to fill in the blanks correctly.

higher; higher
higher; lower
lower; higher
lower; lower

Answers

Final answer:

A negative test charge moves toward regions of higher electric potential and lower electric potential energy, which is the opposite direction of the electric field defined by positive charges.

Explanation:

A negative test charge will accelerate toward regions of higher electric potential and lower electric potential energy. This happens because a negative charge moves oppositely to the electric field direction, which is defined from high to low potential. When a negative charge, such as an electron, moves toward a higher potential, it is moving towards a region where it would have a lower potential energy if it were positive; however, since it is negative, its potential energy actually decreases as it moves in this direction.

Understanding electric fields and potentials, consider that a positive test charge is repelled by positive charges and attracted to negative charges. Since the field lines point away from positive charges and toward negative charges, a negative test charge would move in the direction opposite to the field lines, meaning it moves from lower to higher potential but in doing so, it lowers its electric potential energy.

Assume that charge −q−q-q is placed on the top plate, and +q+q+q is placed on the bottom plate. What is the magnitude of the electric field EEE between the plates?

Answers

Answer:

Magnitude of electric field = E = q/Aε0

Explanation:

Consider plates are placed at a distance of d. As given in the question the charge stored on the plates have magnitude q and given by:

                                          q = CV

And  

                                          V = q/C    ……. (i)

The capacitance is given by the following equation:  

                                         C = Aε0/d ……. (ii)

Put equation (ii) in (i) ,

                                          V = qd/ Aε0 …..(iii)    

The electric field is defined as:  

                                            E = V/d   …… (iv)

Put equation (iii) in (iv),

                                            E = qd/ Aε0d

                                            E = q/Aε0

Hence, the magnitude of electric field will be q/Aε0 .

                                         

Final answer:

The electric field between two plates with charges −q and +q is directed from the positive to the negative plate and can be calculated using the charge density and vacuum permittivity. For systems of parallel conducting plates, the electric field can be found using potential differences and distances, while the interaction of a charge with a dipole requires considering forces from each dipole component.

Explanation:

The magnitude of the electric field E between two plates with charges −q and +q can be found using the formula E = σ/ε₀, where σ is the surface charge density (σ = q/A, A is the area of the plate) and ε₀ is the vacuum permittivity. The direction of the electric field is from the positive to the negative plate. Considering a scenario with two parallel conducting plates 15 cm apart and each with a charge of −0.225 C and +0.225 C, you would first calculate the electric field strength using the aforementioned formula and then state that its direction is from the positive to the negative plate.

For the case involving three parallel conducting plates with specified potentials, you calculate the average electric field experienced by an electron by finding the potential difference between the plates and dividing it by the distance between the plates. The electric field between the plates is uniform if we assume the plates are infinitely large and parallel.

The net force on a charge −q placed equidistant from a dipole with charges +2q and −2q can be found using Coulomb's law. Since the forces due to the +2q and −2q charges will be equal in magnitude but opposite in direction, the result will be a null net force in the direction along the line connecting the two dipole charges. However, there will be a resultant force perpendicular to this line, attracting the −q charge towards the dipole's center.

Technician A says that most TSBs involve a specific stored DTC. Technician B says that the ECT and IAT reading should be close to the same temperature after the vehicle sits for several hours. Which technician is correct?

A) Technician A only

B) Technician B only

C) Both technicians

D) Neither technician

Answers

Answer:

C

Explanation:

Both technicians are correct.

Cheers

Technician B is correct because the ECT and IAT should read similarly after a car has been sitting for a while, reflecting the same ambient temperature. TSBs can address a variety of issues, not necessarily involving a specific DTC as Technician A suggests.

Technician B is correct. Technical Service Bulletins (TSBs) typically address widespread problems or issues found in a particular model of a vehicle. While these bulletins may include information about specific Diagnostic Trouble Codes (DTCs), they do not always involve a specific stored DTC, as Technician A suggests. TSBs can cover various issues, including non-DTC-related performance, noise, and other iterative improvements that are not related to any stored codes.

Technician B is accurate in saying that the Engine Coolant Temperature (ECT) sensor and the Intake Air Temperature (IAT) sensor readings should be close to the same after the vehicle has been sitting for several hours, typically overnight. This is because both the coolant in the engine and the air in the intake manifold would have reached ambient temperature, reflecting similar temperatures if the vehicle's sensors are functioning correctly.

Answer choice B, which credits Technician B, is the correct option.

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