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Julli [10]
3 years ago
7

What would happen if one of the four-stroke combustion cycle parts never worked?​

Physics
2 answers:
-BARSIC- [3]3 years ago
5 0

Answer:

The engine will not work causing no movement in the wheels .

Explanation:

Remember the four stroke combustion cycle powers the wheels and if one of these parts do not work, then the car will not move .

Sergeu [11.5K]3 years ago
3 0

Answer:

An internal-combustion engine goes through four strokes: intake, compression, combustion (power), and exhaust. As the piston moves during each stroke, it turns the crankshaft.

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You are pushing on a heavy desk with a force of 65 N the desk does not slide the force of friction between the desk and the floo
astra-53 [7]

When an object does not move even on pushing , static frictional force acts on in opposite direction of the applied force to stop the object from moving. static frictional force is a self adjusting force and it adjust its value according to the applied force if the applied force is smaller than the maximum value of static frictional force. The object starts moving once the applied force on it becomes greater than the maximum static frictional force.  hence the statement is true.

7 0
3 years ago
Which of the following is NOT one of the four elements of emotion?
Nezavi [6.7K]

Answer:

Attention

Explanation:

Hello there, fellow peer! The answer to question is attention. Let's say someone is the control. The behavioral expression is an element of expression, so the control will feel emotions. Subjective Experience is when someone felt the way you feel and they are trying to help you. That is a type of emotion which can lead to empathy for you. This is also not the answer. Physiological Arousal is also not the answer because this is when you can feel what someone else is feeling and you try to give them therapy.

Using the process of elimination, our answer is therefore attention.

4 0
3 years ago
If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?
sdas [7]
Data:
f_{2} = 42 Hz
n (Wave node)
V (Wave belly) 
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
f_{n} =  \frac{nV}{2L}

Wire 2 → 2º Harmonic → n = 2

f_{n} = \frac{nV}{2L}
f_{2} = \frac{2V}{2L} &#10;
2V =  f_{2} *2L
V =  \frac{ f_{2}*2L }{2}
V =  \frac{42*2L}{2}
V =  \frac{84L}{2}
V = 42L

Wire 1 → 1º Harmonic or Fundamental rope → n = 1


f_{n} = \frac{nV}{2L}
f_{1} = \frac{1V}{2L}
f_{1} =  \frac{V}{2L}

If, We have:
V = 42L
Soon:
f_{1} = \frac{V}{2L}
f_{1} = \frac{42L}{2L}
\boxed{f_{1} = 21 Hz}

Answer:

<span>The fundamental frequency of the string:
</span>21 Hz

7 0
3 years ago
Read 2 more answers
Can someone help me?​
Gwar [14]
24- series
25- parallel
26- no, because they’re connected in series
27- yes, because they’re connected in parallel
7 0
3 years ago
John pushes a box with a constant force as shown in the graph below.
andrew11 [14]
From the graph, it can be seen that the constant force that John exerted in order to move the object is 14N. Work is calculated by multiplying the force with the distance to which the object moves in parallel with the direction of the force. 
                                      Work = Force x displacement
                                      Work = (14 N) x (8 m)
                                        Work = 112 J
The closest value is 110J. Thus, the answer to this item is the second choice. 
4 0
3 years ago
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