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Natasha2012 [34]
2 years ago
6

Why can you not put diesel in a regular car?

Physics
2 answers:
goblinko [34]2 years ago
6 0
It's not compatible. It's like putting metal in a microwave; it's not going to go well (sorry I don't have a scientific answer, I'm just thinking out loud). 
Ilia_Sergeevich [38]2 years ago
4 0
It will mess up the car's engine, and get stuck in the carburetor.
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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
castortr0y [4]

The vertical component of force exerted by the hi.nge on the beam will be,142.10N.

To find the answer, we need to know more about the tension.

<h3>How to find the vertical component of the force exerted by the hi.nge on the beam?</h3>
  • Let's draw the free body diagram of the system.
  • To find the vertical component of the force exerted by the hi.nge on the beam, we have to balance the total vertical force to zero.

                      F_V+T sin\alpha -mg=0\\F_V=mg-Tsin\alpha \\

  • To find the answer, we have to find the tension,

                     Tlsin\alpha - mg\frac{l}{2}sin\beta =0\\ \\Tlsin\alpha = mg\frac{l}{2}sin\beta\\\\Tsin57=\frac{mg}{2}sin90\\\\T=\frac{mg}{2sin57} =169.43N

  • Thus, the vertical component of the force exerted by the hi.nge on the beam will be,

                F_V=(29*9.8)-(169.43*sin57)=142.10N

Thus, we can conclude that, the vertical component of force exerted by the hi.nge on the beam will be,142.10N.

Learn more about the tension here:

brainly.com/question/28106868

#SPJ1

5 0
2 years ago
Read 2 more answers
When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin
mario62 [17]

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}

where n=1,2,3 or ...n

L=Length of string

T=Tension

\mu =Mass per unit length

When string is clamped at mid-point

Effecting length becomes L'=0.5 L

Thus new Frequency becomes

f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz      

4 0
3 years ago
The technology in which parabolic mirror arrays are placed in a desert with a water-filled pipe traversing the array is used to
andriy [413]

Answer:

The sun

Explanation:

In this system the energy of the sun heats the water in the pipe, producing a high pressured steam, which is used for moving a turbine and producing electricity, is a transformation of energy from solar to thermal, then to mechanical to electrical.

4 0
3 years ago
A bug splats against the windshield of a car traveling at high speeds down a backcountry road. Which statement correctly compare
zvonat [6]

Answer:

C. The bug's change in momentum is equal to the car's change in momentum.

Explanation:

As we know by Newton's 2nd law

F = \frac{\Delta P}{\Delta t}

here we have also know that when car hits the bug then force applied by wind shield on the bug is same as the force applied by the bug on the car's wind shield as per Newton's III law

F_{12} = F_{21}

so we know that

\frac{\Delta P_{12}}{\Delta t} = \frac{\Delta P_{21}}{\Delta t}

so we have

\Delta P_{12} = \Delta P_{21}

so correct answer will be

C. The bug's change in momentum is equal to the car's change in momentum.

6 0
3 years ago
Which statement correctly defines power?
Anna007 [38]

Answer:

Power=V\, I which corresponds to the second option shown: "voltage times amperage"

Explanation:

The electric power is the work done to move a charge Q across a given difference of potential V per unit of time.

Since such electrical work is the product of the potential difference V times the charge that moves through that potential, and this work is to be calculated by the unit of time, we need to divide the product by time (t) which leads to the following final simple equation:

Power=\frac{V\,Q}{t} =V\,\frac{Q}{t} = V\, I

Notice that we replaced the quotient representing charge per unit of time (Q/t) by the actual current running through the circuit.

This corresponds to the second option shown in the question: "Voltage times amperage".

6 0
3 years ago
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