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stealth61 [152]
2 years ago
15

A piston in an internal combustion engine applies torque during 150° of each full rotation of the crankshaft to which it is conn

ected. Suppose an 8-cylinder engine (8 such pistons) is running with its crankshaft turning at 2500 rpm and produces 410J of work each second. What is the torque applied to the crank shaft by one of the engine’s eight pistons?
Physics
1 answer:
aalyn [17]2 years ago
4 0

Answer:

T=1.566 N.m

Explanation:

Given that:

rotational speed of the scrank shaft, N = 2500 rpm

power produced by one cylinder, P = 410 W

We know, in case of rotational power:

P=\frac{2\pi.N.T}{60}

where: T= torque

Substituting the respective values in the above eq.

410=\frac{2\pi\times2500\times T}{60}

T= \frac{410\times60}{2\pi\times2500}

T=1.566 N.m is the torque applied by the each piston of the engine.

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Jonah rode his skateboard 350 meters down a straight road at a constant velocity. He skated
Marysya12 [62]

Answer:

250 m/min down the road

Explanation:

Velocity is equivalent to speed but it considers the direction of the object. Velocity is also calculated by dividing the distance travelled by time. Therefore, v=\frac {d}{t} where d and t are distance and time respectively. Given that d is given as 350 m and t is 1.4 s then by substitution v=\frac {350}{1.4}=250 m/min and the direction is down the road.

Velocity is 250 m/min down the road

8 1
3 years ago
The tibia is a lower leg bone (shin bone) in a human. The maximum strain that the tibia can experience before fracturing corresp
IRISSAK [1]

Answer:

a

   k    =  11600000 N/m

b

   \Delta  L  =  3.2323 *10^{-5} \ m

c

  F =  3750.28 \  N  

Explanation:

From the question we are told that

    The Young modulus is  E =  1.4 *10^{10} \  N/m^2

     The length is  L  =  0.35 \ m

      The  area is  2.9 \ cm^2  =  2.9 *10^{-4} \ m ^2

   

Generally the force acting on the tibia is mathematically represented as

       F =  \frac{E *  A  *  \Delta  L }{L}    derived from young modulus equation

Now this force can also be mathematically represented as

      F =  k *  \Delta  L    

So

     k    =  \frac{E *  A  }{L}

substituting values

     k    =  \frac{1.4 *10^{10} *  2.9 *10^{-4}  }{ 0.35}

     k    =  11600000 N/m

    Since the tibia support half the weight then the force experienced by the tibia is  

        F_k  =  \frac{750 }{2}  =  375 \  N

 From the above equation the extension (compression) is mathematically represented as

          \Delta  L  =  \frac{ F_k  *  L  }{ A *  E }        

substituting values

           \Delta  L  =  \frac{  375   *  0.35  }{ (2.9 *10^{-4}) *   1.4*10^{10} }

           \Delta  L  =  3.2323 *10^{-5} \ m

From the above equation the maximum force is  

        F =  \frac{1.4*10^{10} *  (2.9*10^{-4})  *  3.233*10^{-5} }{ 0.35}  

         F =  3750.28 \  N  

4 0
3 years ago
A 160 g basketball has a 32.7 cm diameter and may be approximated as a thin spherical shell. Starting from rest, how long will i
mafiozo [28]

Answer:

   t = 0.24 s

Explanation:

As seen in the attached diagram, we are going to use dynamics to resolve the problem, so we will be using the equations for the translation and the rotation dyamics:

Translation:  ΣF = ma

Rotation:      ΣM = Iα ; where α = angular acceleration

Because the angular acceleration is equal to the linear acceleration divided by the radius, the rotation equation also can be represented like:

                    ΣM = I(a/R)

Now we are going to resolve and combine these equations.

For translation:     Fx - Ffr = ma

We know that Fx = mgSin27°, so we substitute:

         (1)                 mgSin27° - Ffr = ma  

For rotation:         (Ffr)(R) = (2/3mR²)(a/R)

The radius cancel each other:

        (2)                Ffr = 2/3 ma

We substitute equation (2) in equation (1):

                            mgSin27° - 2/3 ma = ma

                            mgSin27° = ma + 2/3 ma

The mass gets cancelled:

                            gSin27° = 5/3 a

                            a = (3/5)(gSin27°)

                            a = (3/5)(9.8 m/s²(Sin27°))

                            a = 2.67 m/s²

If we assume that the acceleration is a constant we can use the next equation to find the velocity:

                           V = √2ad; where  d = 0.327m

                           V = √2(2.67 m/s²)(0.327m)

                            V = 1.32 m/s

Because V = d/t

                             t = d/V

                             t = 0.327m/1.32 m/s

                             t = 0.24 s

7 0
3 years ago
A 32-cm-long solenoid, 1.8 cm in diameter, is to produce a 0.30-T magnetic field at its center. If the maximum current is 4.5 A,
daser333 [38]

Answer:

<h2>16,931 turns</h2>

Explanation:

The magnetic field produced is expressed using the formula

B = \frac{\mu_0NI}{L}

B is the magnetic field = 0.30T

I is the current produced in the coil = 4.5A

\mu_0 is the magnetic permittivity in vacuum = 1.26*10^-6Tm/A

L is the length of the solenoid = 32 cm = 0.32 m

N is the number of turns in the solenoid.

Making N the subject of the formula from the equation above;

B = \frac{\mu_0NI}{L}\\\\BL = \mu_0NI\\\\Dividing\ both\ sides \ by \ \mu_0I\\\\\frac{BL}{\mu_0I} =\frac{\mu_oNI}{\mu_0I} \\\\

N = \frac{BL}{\mu_0I}

Substituting the give values to get N;

N = \frac{0.3*0.32}{1.26*10^{-6} * 4.5}\\\\N = \frac{0.096}{0.00000567} \\\\N = 16,931.21

The number of turns the solenoid must have is approximately 16,931 turns

4 0
3 years ago
Please help me with a speech.The topic of my speech is Passion.​
Vikentia [17]

Answer:

Explanation:

Passion

For me, standing on the summit of Mt Everest was the result of following a process. The process of mountaineering. I love mountaineering. I am passionate about it. I love the months of planning for an expedition, the months of sweating and training to prepare my body physically. The meticulous preparation of my equipment. Most of all I love the huge mental challenge I have to overcome before each climb to confront my own fear. All these reasons are why I climb, they are why I climbed Mt Everest and that is why I continue to climb.

Passion is an enormously powerful force. It gives us the strength to get through hard times and setbacks. It gives us strength to overcome our fears, to ignore what other people think of us, to be disciplined and make sacrifices in pursuit of our dreams. Passionate people do not want to take shortcuts – they consider that ‘learning the process’ is an important part of the journey.

In mountaineering it’s easy to spot those who are not passionate about the process. They want to stand on top of the mountain but they are not really interested in the process of climbing the mountain. I feel for these people. Success without hard work is a hollow, empty feeling. They never last long in the sport.

Just as in life, successful mountaineers are the ones who are passionate. They are not there just to stand on the summit. Their passion gives them the energy to work the hardest, fight the longest, and in the words of Winston Churchill “never, never. never give-up”.

3 0
3 years ago
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