Select the correct answer.

A teenager was pulling a prank and placed a large stuffed penguin in the middle of a roadway. A driver is traveling on this leve

Anvisha [2.4K]

Whats the question????

3 0

4 years ago

Examples of reciprocating motion in daily life

bonufazy [111]

Answer:

Examples of reciprocating motion in daily life are;

1) The needles of a sewing machine

2) Electric powered reciprocating saw blade

3) The motion of a manual tire pump

Explanation:

A reciprocating motion is a motion that consists of motion of a part in an upward and downwards $(\updownarrow)$ or in a backward and forward (↔) direction repetitively

Examples of reciprocating motion in daily life includes the reciprocating motion of the needles of a sewing machine and the reciprocating motion of the reciprocating saw and the motion of a manual tire pump

In a sewing machine, a crank shaft in between a wheel and the needle transforms the rotary motion of the wheel into reciprocating motion of the needle.

8 0

3 years ago

technician a says that the higher the gear selected the more torque is available. technician b says that a transaxle contains ge

umka21 [38]

Technician A is correct. Technician B is wrong because a gear's transmission is used to increase or decrease torque.

The relation torque is relying on multiplying the circumferential detail with the resource of the usage of the radius; massive gears experience a greater amount of torque, at the same time as smaller gears experience a great deal much less torque. Similarly, the torque ratio is equal to the ratio of the gears' radii. A gear's transmission torque modifications as it will boom or decreases speed. Commonly, with the resource of the usage of lowering the speed, a small torque on the doorway issue is transferred as a massive torque at the output issue. The calculation of torque is quantified with the resource of the usage of an extensive form of teeth.

Learn more about the torque at brainly.com/question/28220969

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3 0

1 year ago

For an Otto cycle, plot the cycle efficiency as a function of compression ratio from 4 to 16.

Elza [17]

Assumptions:

Steady state.
Air as working fluid.
Ideal gas.
Reversible process.
Ideal Otto Cycle.

Explanation:

Otto cycle is a thermodynamic cycle widely used in automobile engines, in which an amount of gas (air) experiences changes of pressure, temperature, volume, addition of heat, and removal of heat. The cycle is composed by (following the P-V diagram):

Intake <em>0-1</em>: the mass of working fluid is drawn into the piston at a constant pressure.

Adiabatic compression <em>1-2</em>: the mass of working fluid is compressed isentropically from State 1 to State 2 through compression ratio (r).

$r =\frac{V_1}{V_2}$

Ignition 2-3: the volume remains constant while heat is added to the mass of gas.
Expansion 3-4: the working fluid does work on the piston due to the high pressure within it, thus the working fluid reaches the maximum volume through the compression ratio.

$r = \frac{V_4}{V_3} = \frac{V_1}{V_2}$

Heat Rejection 4-1: heat is removed from the working fluid as the pressure drops instantaneously.
Exhaust 1-0: the working fluid is vented to the atmosphere.

If the system produces enough work, the automobile and its occupants will propel. On the other hand, the efficiency of the Otto Cycle is defined as follows:

$\eta = 1-(\frac{1}{r^{\gamma - 1} } )$

where:

$\gamma = \frac{C_{p} }{C_{v}} : specific heat ratio$

Ideal air is the working fluid, as stated before, for which its specific heat ratio can be considered constant.

$\gamma = 1.4$

Answer:

See image attached.

5 0

4 years ago

A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calcula

makkiz [27]

Answer:

F_y = 151319.01N = 15.132 KN

Explanation:

From the linear momentum equation theory, since flow is steady, the y components would be;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

We are given;

Length; L = 5ft = 1.52.

Initial diameter;d1 = 12in = 0.3m

Exit diameter; d2 = 24 in = 0.6m

Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s

Initial pressure;p1 = 30 psi = 206843 pa

Thus,

initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²

Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²

Now, we know that volume flow rate of water is given by; Q = A•V

Thus,

At exit, Q2 = A2•V2

So, 0.28 = 0.28•V2

So,V2 = 1 m/s

When flow is incompressible, we often say that ;

Initial mass flow rate = exit mass flow rate.

Thus,

ρ1 = ρ2 = 1000 kg/m³

Density of water is 1000 kg/m³

And A1•V1 = A2•V2

So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

So, from initial equation of y components;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

6 0

3 years ago