How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole

Answers

Answer 1

Answer:

Answer: hello your question is incomplete below is the complete question

Water stands at a depth H in a large open tank whose side walls are vertical . A hole is made in one of the walls at a depth h below the water surface. Part B How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole

answer :

At Height ( h ) from the bottom of Tank

Explanation:

Determine how far above the bottom of the tank a second hole be cut

For the second hole to have the same range as the first hole

Range of first hole = Velocity of efflux of water * time of fall of water

= √ (2gh) * √( 2g (H - h) / g)

= √ ( 4(H-h) h)

Hence the Height at which the second hole should be placed to exercise same range of stream emerging = h from the bottom of the Tank

Answer 2

Answer:

Final answer:

The second hole should be cut at the same height as the first hole to have the same range for the stream.

Explanation:

In order for the stream emerging from the second hole to have the same range as the first hole, the second hole should be cut at the same height as the first hole. This is because the range of the stream depends on the initial velocity and the vertical distance traveled. If the second hole is higher or lower than the first hole, the vertical distance traveled will be different and the range of the stream will be affected.

Learn more about Range of stream from a hole here:

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15.23.....................

I think it would be 15.23 not so sure but
hope this helps! (:

An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and a quality of 0.25. An electric heater inside the tank is turned on to heat this H2O until the pressure increases to 200 kPa. Please determine the change in total entropy of water during this process. Hint: See if you can find the electrical work consumed during this process.

Answers

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

$v_1 = vf + x(vg - vf)$

$= 0.001053 + 0.25(1.1594 - 0.001053)$

$v_1 = 0.20964 m^3/kg$

$s_1 = Sf + x(Sfg)$

$= 1.4337 + 0.25 * 5.7894 = 2.88 kJ/kg K$

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume so $v_1 - v_2 = 0.29064 m^3/kg$

$v_2 = v_1 = vf + x(vg - vf)$

$=0.29064 = x_2(0.88578 - 0.001061)$

$x_2 = 0.327$

$s_2 = 1.5302 + 0.32 * 5.5968 = 3.36035 kJ/kg K$

Change in entropy $\Delta s = m(s_2 - s_1)$

=3( 3.36035 - 2.88) = 1.44 kJ/kg

Two point charges are placed on the x axis.The firstcharge, q1= 8.00 nC, is placed a distance 16.0 mfromthe origin along the positive x axis; the second charge,q2= 6.00 nC, is placed a distance 9.00 mfrom the originalong the negative x axis.[Give the x and y components of the electric fieldas an ordered pair. Express your answer innewtons per coulomb to three significant figures.Keep in mind that an x component that points tothe right is positive and a y component thatpoints upward is positive.]

Answers

Answer:

E = (0, 0.299) N

Explanation:

Given,

Charge $q_1\ =\ 8.0\ nC$
Charge $q_2\ =\ 6.0\ nC$
Distance of the first charge from the origin = (16m, 0)
Distance of the second charge from the origin = (-9, 0)
Point where the electric field required = (0, 12m)

Let $\theta_1\ and\ theta_2$ be the angle of the electric fields by first and second charge at the point A.

$\therefore sin\theta_1\ =\ (12)/(20)\n\Rightarrow \theta_1\ =\ sin^(-1)\left ((12)/(20)\ \right )\n\Rightarrow \theta_1\ =\ 36.87^o\n\n\therefore sin\theta_1\ =\ (12)/(9)\n\Rightarrow \theta_1\ =\ sin^(-1)\left ((12)/(9)\ \right )\n\Rightarrow \theta_1\ =\ 53.13^o\n$

Electric field by charge $q_1$ at point A,

$F_1\ =\ (kq_1)/(r_1^2)\n\Rightarrow F_1\ =\ (9* 10^9* 8* 10^(-9))/(20^2)\n\Rightarrow F_1\ =\ 0.18\ N/C$

Electric field by the charge $q_2$ at point A,

$F_1\ =\ (kq_1)/(r_1^2)\n\Rightarrow F_1\ =\ (9* 10^9* 6.0* 10^(-9))/(16^2)\n\Rightarrow F_1\ =\ 0.24\ N/C$

Now,

Net electric field in horizontal direction at point A $F_x\ =\ F_(1x)\ +\ F_(2x)\n\Rightarrow F_x\ =\ F_1cos\theta_1\ +\ F_2cos\theta_2\n\Rightarrow F_x\ =\ 0.18*( -cos36.87^o)\ +\ 0.24* cos53.13^o\n\Rightarrow F_x\ =\ -0.144\ +\ 0.144\ N/C\n\Rightarrow F_x\ =\ 0\ N/C$

Net electric field in vertical direction at point A.

$F_y\ =\ F_(1y)\ +\ F_(2y)\n\Rightarrow F_y\ =\ F_1sin\theta_1\ +\ F_2sin\theta_2\n\Rightarrow F_y\ =\ 0.18* sin36.87^o\ +\ 0.24* sin53.13^o\n\Rightarrow F_y\ =\ 0.180\ +\ 0.192\n\Rightarrow F_y\ =\ 0.299\ N/C$

Hence, the net electric field at point A,

$F\ =\ ( 0, 0.299 )\ N/C.$

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