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Gnesinka [82]
3 years ago
13

Flying against the wind, an airplane travels 2670 km in 3 hours. Flying with the wind, the same plane travels 11,070 km in 9 hou

rs. What is the rate of the plane in still air and what is the rate of the wind?
Physics
1 answer:
xxMikexx [17]3 years ago
5 0

Answer:

speed of plane in still air = 1060 km/h

speed of wind = 170 km/h

Explanation:

Let teh speed of plane in still air is vp and the speed of air is va.

Irt travels 2670 km in 3 hours against the wind

So,

vp - va = 2670 / 3 = 890 km/h ..... (1)

It travels 11070 km in 9 hours along the wind.

vp + va = 11070 / 9 = 1230 km/h .... (2)

Adding both the equations

2 vp = 2120

vp = 1060 km/h

and va = 1230 - vp = 1230 - 1060 = 170 km/h

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A charge of 12 c passes through an electroplating apparatus in 2. 0 min. what is the average current?
Studentka2010 [4]

A charge of 12 c passes through an electroplating apparatus in 2.0 min, then the average current is 0.1 ampere.

<h3>What is an electric charge?</h3>

Charged material experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. You can have a positive or negative electric charge.

Electric current is defined as the charge per unit of time.

The mathematical relation between current and the electric charge

I =Q/T

where I is the current flowing

Q is the total electric charge

T is the time period for which the current is flowing

As given in the problem A charge of 12 c passes through an electroplating apparatus in 2.0 min

Let us first convert the time period of minutes into seconds

1 min = 60 seconds

2 min = 2*60 seconds

         =120 seconds

By using the above relation between electric current and electric charge

and by substituting the respective values of the charge and the time period

I =Q/T

I = 12c/120 seconds

I = 0.1 Ampere

Thus, the average current flowing through the apparatus would be 0.1 Ampere.

Learn more about an electric charge from here

brainly.com/question/8163163

#SPJ4

5 0
1 year ago
If the load distance of a lever is 30 cm and the effort distance is 60 cm, calculate the amount of effort required to lift a loa
vladimir2022 [97]

Here,

Load distance (Ld) = 30 cm

Effort distance (Ed) = 60 cm

Load (L) = 200N

Effort (E) = ?

Now, By using formula,

or, E * Ed = L * Ld

or, E * 60 = 200 * 30

or, E = 6000/60

◆ E = 100N

This is a Right answer...

I hope you understand...

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3 years ago
Los huracanes y tornados se producen por las corrientes de convencion
Montano1993 [528]

Answer:

correct!

Explanation:

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Answer:

30.0625 W

Explanation:

325 g/h   x    (1h x 1kg)/(3600s x 1000g)   x   3,33 x 10^5 J/Kg = 30.0625 J/Kg = 30.0625 W

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