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Sholpan [36]
3 years ago
6

A Toyota Camry of mass 1650 kg turns from Chaplin Road to Route 79, thereby accelerating from 35 MPH in the city till 70 MPH on

the freeway in 4.6 seconds. Neglecting the weight of the driver, friction and aerodynamic drag, (3a) (4p) Please calculate the work done by the engine for this acceleration, in kJ. (3b) (4p) Please calculate the average power delivered by an engine under these conditions, in kW. Hint: Make sure you convert MPH (miles per hour) into an appropriate SI unit.
Engineering
1 answer:
Gekata [30.6K]3 years ago
7 0

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cThe Mars Rover Spirit got stuck in the Martian sand. The wheels kept slipping. Attempts to free it were futile. Discuss the typ
IgorC [24]

Answer:

Improved/ advanced types of Actuators include servo systems, create a large range of actuator motion in response to the changing needs of the operational environment or process.

Actuators are local or automated suppliers of working motion.

Hydraulic and air cylinders can be classified as single-acting cylinders, meaning that the energy source result in movement in one direction and a spring is used for the other direction.

Explanation:

An actuator control system is referred to as any electronic, electrical, or electromechanical system often used to activate an actuator, control the direction as well as extent and duration of its output. Actuator control systems could take the form of extremely simple, manually-operated, start-and-stop stations, either sophisticated or programmable computer systems. The more improved/ advanced types include servo systems that produce a large range of actuator motion in response to the changing needs of the operational environment or process. This type of actuator control system uses an interface arrangement that assimilates feedback from the process or mechanism and adjusts the actuator in the right way. Most actuator systems will include at least a set of travel limits that prevent the actuator destroying itself or the secondary mechanism.

Actuators are local or automated suppliers of working motion. They are used to changes, adjust, or move a secondary mechanism, where a physical operator cannot intervene directly. They are denoted by a large range of varying types using electrical and electromagnetic, hydraulic, or pneumatic power sources to create linear or rotary outputs. One element they all have in common is the actuator control system used to start, stop, and adjust the range, speed, and duration of the working motion.

Actuators can produce a linear motion, rotary motion or oscillatory motion which means they can create motion in one direction, in a circular motion or in opposite directions at regular intervals. Hydraulic and air cylinders can be classified as single-acting cylinders, meaning that the energy source result in movement in one direction and a spring is used for the other direction.

7 0
3 years ago
The Energy Losses Associated with Valves and Fittings: a)- are generally associated with a K factor b)- are generally associated
madam [21]

Answer:

a)Are generally associated with factor.

Explanation:

We know that losses are two types

1.Major loss  :Due to friction of pipe surface

2.Minor loss  :Due to change in the direction of flow

As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and  it produce losses in the energy.

Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as

Losses=K\dfrac{V^2}{2g}

8 0
3 years ago
Steam enters a steady-flow adiabatic nozzle with a low inlet velocity (assume ~0 m/s) as a saturated vapor at 6 MPa and expands
Sergio [31]
Yea bro I don’t really know
7 0
2 years ago
To find the reactance XLXLX_L of an inductor, imagine that a current I(t)=I0sin(ωt)I(t)=I0sin⁡(ωt) , is flowing through the indu
Sophie [7]

Answer:

V(t) = XLI₀sin(π/2 - ωt)

Explanation:

According to Maxwell's equation which is expressed as;

V(t) = dФ/dt ........(1)

Magnetic flux Ф can also be expressed as;

Ф = LI(t)

Where

L = inductance of the inductor

I = current in Ampere

We can therefore Express Maxwell equation as:

V(t) = dLI(t)/dt ....... (2)

Since the inductance is constant then voltage remains

V(t) = LdI(t)/dt

In an AC circuit, the current is time varying and it is given in the form of

I(t) = I₀sin(ωt)

Substitutes the current I(t) into equation (2)

Then the voltage across inductor will be expressed as

V(t) = Ld(I₀sin(ωt))/dt

V(t) = LI₀ωcos(ωt)

Where cos(ωt) = sin(π/2 - ωt)

Then

V(t) = ωLI₀sin(π/2 - ωt) .....(3)

Because the voltage and current are out of phase with the phase difference of π/2 or 90°

The inductive reactance XL = ωL

Substitute ωL for XL in equation (3)

Therefore, the voltage across inductor is can be expressed as;

V(t) = XLI₀sin(π/2 - ωt)

3 0
3 years ago
What’s the number of gold atoms in a nanogram? a picogram?
zvonat [6]

Answer :

The number of gold atoms in nanogram is, 3.057\times 10^{12}

The number of gold atoms in picogram is, 3.057\times 10^{9}

Explanation :

As we know that the molar mass of gold is, 196.97 g/mole. That means, 1 mole of gold has 196.97 grams of mass of gold.

As we know that,

1 mole contains 6.022\times 10^{23} number of atoms.

First we have to determine the number of gold atoms in a nanogram.

As, 196.97 grams of gold contains 6.022\times 10^{23} number of gold atoms

And, 1 grams of gold contains \frac{1g}{196.97g}\times (6.022\times 10^{23}) number of gold atoms

So, 10^{-9} nanograms of gold contains \frac{1g}{196.97g}\times (10^{-9})\times (6.022\times 10^{23})=3.057\times 10^{12} number of gold atoms

The number of gold atoms in nanogram is, 3.057\times 10^{12}

Now we have to determine the number of gold atoms in a picogram.

As, 196.97 grams of gold contains 6.022\times 10^{23} number of gold atoms

And, 1 grams of gold contains \frac{1g}{196.97g}\times (6.022\times 10^{23}) number of gold atoms

So, 10^{-12} picograms of gold contains \frac{1g}{196.97g}\times (10^{-12})\times (6.022\times 10^{23})=3.057\times 10^{9} number of gold atoms

The number of gold atoms in picogram is, 3.057\times 10^{9}

8 0
3 years ago
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