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2 years ago

7

calculate the pressure exerted by a mercury column of 76 cm high at its bottom.Given that density of mercury is 13600kg/m^3 and

g=9.8

1 answer:

Amanda [17]2 years ago

5 0

Answer:

Explanation:I don't say you have to mark my ans as brainliest but if you think it has really helped you plz don't forget to thank me ...

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Pls answerrrrr thisssss

Jobisdone [24]

The amplitude of wave-c is 1 meter.

The speed of all of the waves is (12meters/2sec)= 6 m/s.

The period of wave-a is 1/2 second.

4 0

3 years ago

Which of the following terms corresponds to #2 on the image?

Mila [183]

Answer:

Trough

Explanation:

cuz physics you see

7 0

3 years ago

tensa zangetsu [6.8K]

Answer:

C. amount of charge on the source charge.

Explanation:

Electric field lines can be defined as a graphical representation of the vector field or electric field.

Basically, it was first introduced by Michael Faraday and it is typically a curve drawn to the tangent of a point is in the direction of the net field acting on each point.

The number, or density, of field lines on a source charge indicate the amount of charge on the source charge. Therefore, the density of field lines on a source charge is directly proportional to quantity of charge on the source.

8 0

2 years ago

1. Add 17.35 g, 25.6 g and 8.498 g. chaper 1 physical quantity 11class .physic

Korvikt [17]

51.448 g is the required answer!

8 0

3 years ago

Your low-flow showerhead is delivering water at 1.2×10−4m3/s, about 2.0 gallons per minute.

Oksanka [162]

To solve this problem it is necessary to apply the fluid mechanics equations related to continuity, for which the proportion of the input flow is equal to the output flow, in other words:

$Q_1 = Q_2$

We know that the flow rate is equivalent to the velocity of the fluid in its area, that is,

$Q = VA$

Where

V = Velocity

A = Cross-sectional Area

Our values are given as

$Q_2 = 1.2*10^{-4}m^3/s$

$d = 0.021m$

$r = \frac{0.021}{2} = 0.0105m$

Since there is continuity we have now that,

$V_1A_1 = Q_2$

$V_1A_1 = 1.2*10^{-4}$

$V_1 = \frac{1.2*10^{-4}}{A_2}$

$V_1 = \frac{1.2*10^{-4}}{\pi r^2}$

$V_1 = \frac{1.2*10^{-4}}{\pi (0.0105)^2}$

$V_1 =0.347m/s$

Therefore the speed of the water's house supply line is 0.347m/s

7 0

3 years ago