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earnstyle [38]
3 years ago
13

Q. At what point in a waterfall do the drops of water contain the most kinetic energy ?

Physics
1 answer:
antoniya [11.8K]3 years ago
7 0

Answer:

2.When they reach the bottom of the fall

Explanation:

The potential energy of the waterfall is maximum at the maximum height and decreases with decrease in height. Based on the law of conservation of mechanical energy, as the potential energy of the water fall is decreasing with  decrease in height of the fall, its kinetic energy will be increasing and the kinetic energy will be maximum at zero height (bottom of the fall).

Thus, the correct option is "2" When they reach the bottom of the fall

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7. Water flows trough a horizontal tube of diameter 2.5 cm that is joined to a second horizontal tube of diameter 1.2 cm. The pr
Usimov [2.4K]

To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.

Our values are given as

d_1 = 2.5cm \rightarrow r_1=1.25cm=1.25*10^{-2}m

d_2 = 1.2cm \rightarrow r_2 = 0.6cm = 0.6*10^{-2}m

From the continuity equations in pipes we have to

A_1V_1 = A_2 V_2

Where,

A_{1,2} = Cross sectional Area at each section

V_{1,2} = Flow Velocity at each section

Then replacing we have,

(\pi r_1^2) v_1 = (\pi r_2^2) v_2

(1.25*10^{-2})^2 v_1 = ( 0.06*10^{-2})^2 v_2

v_2 = \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1

From Bernoulli equation we have that the change in the pressure is

\Delta P = \frac{1}{2} \rho (v_2^2-v_1^2)

7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)

7300= 8919.01 v_1^2

v_1 = 0.9m/s

Therefore the speed of flow in the first tube is 0.9m/s

6 0
4 years ago
When a scientific calculator shows the quantity below, what does it mean?<br> 1.5E8
jenyasd209 [6]
1.5 with 8 exponent of 10.
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150000000
6 0
2 years ago
Three resistors, 6.0-W, 9.0-W, 15-W, are connected in parallel in a circuit. What is the equivalent resistance of this combinati
ycow [4]

Answer:

2.9Ω

Explanation:

Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;

1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn

Where;

Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.

Note that;

R1= 6.0Ω

R2 = 9.0Ω

R3= 15.0 Ω

Therefore;

1/Req = 1/6 + 1/9 + 1/15

1/Req= 0.167 + 0.11 + 0.067

1/Req= 0.344

Req= (0.344)^-1

Req= 2.9Ω

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