Help quick p;leaseeeeeeeeeeee its about direction and magnitude

Physics

1 answer:

KonstantinChe [14]3 years ago

7 0

Explanation:

The net force of each square is the combination of the forces in each direction. The direction is the... direction the square would go in due to the net force. The magnitude of the net force is how large it is. So if you had a force pushing 2N to the left and 1N to the right, then the net force would be 1N to the left; because the two oppose eachother. If they were going in the same direction, then they'd add to each other. And perpendicular net forces (like one pushing up and another pushing left) can create net forces in diagonal directions.

I'm not going to do all of these for you because they're basically all the same thing and it's good practice for you anyway. But I'll do the first three just so you can get the idea:

1. The net force's magnitude is 4N and it's direction is to the right.

2. The net force's magnitude is 4N and it's direction is to the left.

3. The net force's magnitude is 0N and it has no direction because they are equal forces acting in opposite directions.

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Answer:

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Taya2010 [7]

Answer:

(a). The gauge pressure at the bottom of tube 1 is $P_{1}=\rho g h_{1}$

(b). The speed of the fluid in the left end of the main pipe $\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

Explanation:

Given that,

Gauge pressure at bottom = p₁

Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.

Find the speed of the fluid in the left end of the main pipe.

(a). We need to calculate the gauge pressure at the bottom of tube 1

Using bernoulli equation

$P_{1}=\rho g h_{1}$

(b). We need to calculate the speed of the fluid in the left end of the main pipe

Using bernoulli equation

Pressure for first pipe,

$P_{1}=\rho gh_{1}$ .....(I)