Which does a reference point provide

1 answer:

5 0

Point of reference - an indicator that orients you generally; "it is used as a reference for comparing the heating and the electrical energy involved"

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A plane flies 1800 miles in 9 hours, with a tailwind all the way. the return trip on the same route, now with a headwind, tak

Fittoniya [83]

Initially its moving with tail wind so here the speed of wind will support the motion of the plane

so we can say

$V_{plane} + v_{wind} = \frac{distance}{time}$

$V_{plane} + v_{wind} = \frac{1800}{9}$

$V_{plane} + v_{wind} = 200 mph$

now when its moving with head wind we can say that wind is opposite to the motion of the plane

$V_{plane} - v_{wind} = \frac{distance}{time}$

$V_{plane} - v_{wind} = \frac{1800}{12}$

$V_{plane} - v_{wind} = 150mph$

now by using above two equations we can find speed of palne as well as speed of wind

$V_{plane} = 175 mph$

$v_{wind} = 25 mph$

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3 years ago

A 2.5m long steel piano wire has a diameter of 0.5cm how great is the tention in the wire if it stretches by 0.45cm when tighten

UkoKoshka [18]

Answer: Their u go i found it their was about 3 pages i did not no what pages u had to do.

Explanation:

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2 years ago

A person is initially driving a car east down a straight road. the magnitude of the instantaneous acceleration is decreasing wit

Alexxandr [17]

By definition, acceleration is the change in velocity per change of time. As time passes by, the time increases in value. So, when the acceleration is decreasing while the time is increasing, then that means that the change of velocity is also decreasing with time. So, optimally, the initial velocity and the velocity at any time are very relatively close to each other,

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2 years ago

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Your cell phone typically consumes about 390 mW of power when you text a friend. If the phone is operated using a lithium-ion ba

yulyashka [42]

Answer:

I = 0.11 A

Explanation:

In an electric circuit, the power delivered to a load, is just the product of the potential difference between the load terminals, times the current flowing through it, as follows:

$P = V*I$

In this case, the power is the one consumed by the cell phone = 390 mW, and the voltage the one produced by the internal energy of the battery, 3.5 V, neglecting the voltage loss at the internal resistance of the battery.
So, we can solve the above equation for the current I, as follows:

$I = \frac{P}{V} = \frac{0.39W}{3.5V} = 0.11 A$