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tatiyna
3 years ago
13

Pa answer po plsssss​

Physics
1 answer:
tankabanditka [31]3 years ago
4 0

Refer to the attachment

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All you have to do is find the description for a b c and d words and find the meaning 8 9 10 11 and 12
insens350 [35]
A-11 polar easterlies
b-8 winds blowing between the equator and 30° N and south
c-10
d-9
4 0
4 years ago
Several short trips taken from a cold start can use ....... as much fuel as a longer multi-purpose trip covering the same distan
Lerok [7]

Several short trips taken from a cold start can use ...twice... as much fuel as a longer multi-purpose trip covering the same distance when the engine is warm.

In cold weather, properly designed gasoline aids in engine starting, while in hot weather, it helps prevent vapor lock. In order to meet the requirements of a modern engine, the fuel must have the volatility for which the engine's fuel system was built and an antiknock quality strong enough to prevent knock during routine operation.

During the intake phase, the air and fuel are combined before being introduced into the cylinder. The spark ignites the fuel-air mixture after the piston compresses it, resulting in combustion. During the power stroke, the piston is propelled by the expansion of the combustion gases.

To learn more about engine and fuel please visit -
brainly.com/question/5181209
#SPJ4

4 0
2 years ago
Proof that the kinetic energy of a moving body is half mv square​
Ivenika [448]

Answer:

1

Explanation:

4 0
3 years ago
A man pulled a 13.0 kg object 11.8 cm vertically with his teeth. (a) How much work (in J) was done on the object by the man in t
jonny [76]

Answer:

(a)The work done by the man is -15.03J.

(b)The force exerted on the object is 127.4N.

Explanation:

Mass of the object pulled by the man is -13kg

Object is lifted 11.8 cm vertical with his teeth it means (displacement = +11.8cm = +0.118m)

Acceleration due to gravity is 9.8 \mathrm{m} / \mathrm{s}^{2}

(a) <u>Calculating the work done</u>:

Work done = mgh

Where "m" is mass of an object, "g" is acceleration due to gravity and "h" is the displacement.

\text { Work }=-13 \times 9.8 \times(+0.118 \mathrm{m})

\text { Work }=-15.03 \mathrm{J}

The work done by the man is -15.03J.

(b) <u>Calculating the force</u>:

Probably the man and the object are close to the exterior of the earth. If the rigidity required to maintained the object of consistent velocity interior the gravitational field of the earth is \mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}

Thus the weight of the object is balanced by the force of the man's teeth on the object. That is

F = mg

\mathrm{F}=13 \times 9.8

F = 127.4N

The force exerted on the object is 127.4N.

4 0
3 years ago
Suppose the Sun appeared to you 900 times dimmer than it does now. How far away from the Sun would you be? (A) 1/9 AU (B) 3 AU
Harrizon [31]

Answer:

The correct answer is option 'c': 30 AUs

Explanation:

For a spherical wave front emitted by sun with total energy 'E' the energy density over the surface when it is at a distance 'r' from the sun is given by

e=\frac{E}{4\pi r^{2}}

This energy per unit area is sensed by observer as intensity of the sun.

Let the initial intensity of sun at a distance r_{1} be e_{1}

Thus if the sun becomes 900 times dimmer we have

e'=\frac{e_{1}}{900}\\\\\frac{E}{4\pi r_{2}^{2}}=\frac{1}{900}\times \frac{E}{4\pi r_{1}^{2}}\\\\\Rightarrow r_{2}^{2}={r_{1}^{2}}\times 900\\\\\therefore r_{2}={r_{1}}\times {30}

Thus the distance increases 30 times.

5 0
4 years ago
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