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Ipatiy [6.2K]
3 years ago
10

Consider a horizontal layer of the dam wall of thickness dx located a distance x above the reservoir floor. What is the magnitud

e dF of the force on this layer that results from adding the water to the reservoir? Express your answer in terms of x, dx, the magnitude of the acceleration due to gravity g, and any quantities from the problem introduction.

Physics
1 answer:
adoni [48]3 years ago
6 0

Answer:

Explanation:

Attached is the solution

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stiks02 [169]

Answer:

Explanation:

It probably would

7 0
3 years ago
The poles of the Earth's magnetic field line up with the geographic poles of the Earth.
Varvara68 [4.7K]

Answer: False

Explanation: A magnetic compass does not point to the geographic north pole. A magnetic compass points to the earth's magnetic poles, which are not the same as earth's geographic poles. Furthermore, the magnetic pole near earth's geographic north pole is actually the south magnetic pole.

8 0
3 years ago
Read 2 more answers
The diagram below shows the velocity vectors for two cars that are moving relative to each other.
scoundrel [369]

Answer:

The answer is "5 \ \frac{m}{s} \ west"

Explanation:

\to \vec{V_1} = (25 \frac{m}{s}) (\hat{-i})\\\\\to  \vec{V_2} = (20 \frac{m}{s}) (\hat{-i})\\\\

velocity of car | respect to car :

\to \vec{V_{12}} = \vec{V_1} - \vec{V_2}\\\\

          =\vec{-25} \hat{i}+ \vec{20} \hat{i}\\\\= 5 \ \frac{m}{s} \ west

7 0
2 years ago
Two circular holes, one larger than the other, are cut in the side of a large water tank whose top is open to the atmosphere. Ho
lidiya [134]

Answer:\frac{r_1}{r_2}=1.565

Explanation:

Given

two holes are made with different sizes

Hole 1 is large in size and hole 2 is small

If the volume flow rate of water is same for both the hole then small hole must be below the large hole because for same flow rate, velocity of water is large while cross-sectional area is small so it compensate to give same flow for both the holes.

Now for radius apply Bernoulli's theorem at hole 1 and 2

P_1+\rho gh_1=P_{atm}+\frac{1}{2}\rho v_1^2

P_2+\rho g6h_2=P_{atm}+\frac{1}{2}\rho v_2^2

if hole 1 is h distance below water surface then h_2=6h

and P_1=P_2=P_{atm}

Also v_1=\sqrt{2gh}

v_2=\sqrt{2g(6h)}

and Q=A_1v_1=A_2v_2

A=\pi r^2

thus \dfrac{r_1}{r_2}=\sqrt{\dfrac{v_2}{v_1}}

\dfrac{r_1}{r_2}=\sqrt{\dfrac{\sqrt{6h}}{\sqrt{h}}}

\frac{r_1}{r_2}=1.565

5 0
3 years ago
When an atom loses an electron, it becomes a
mel-nik [20]
When an atom loses an electron, it becomes a positive ion.

a. positive ion.
4 0
3 years ago
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