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satela [25.4K]
3 years ago
5

Water flows at 10 m/s through a pipe with radius 0.025 m. The pipe goes up to the second floor of the building, 2.5 m higher, an

d the pressure remains unchanged. Which of the following statements is true concerning the velocity of water in the pipe and the radius of the pipe on the second floor?
O A. Since this is a closed system, the velocity of water and area of the pipe are the same on both floors.
OB. On the second floor, the velocity of the water increased and the area of the pipe is larger.
Oc. On the second floor, the velocity of the water increased and the area of the pipe smaller.
OD. On the second floor, the velocity of the water decreased and the area of the pipe is larger.
OE. On the second floor, the velocity of the water decreased and the area of the pipe is smaller.
OF. There is not enough information to determine what happens to the water and pipe size.

Physics
1 answer:
Lera25 [3.4K]3 years ago
5 0

Answer: from the information given, the velocity of the water will decrease but the pipe size will remain the same.

This can be proved with bernoulli's equation.

Explanation: careful analysis of the system using bernoulli's equation of flow is shown in the image attached

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Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
An experimental tungsten light bulb filament has a length of 5 cm and a diameter of 0.074 cm. The filament is basically just a w
adell [148]

Answer:

power emitted is 1.75 W

Explanation:

given data

length l = 5 cm = 5 ×10^{-2} m

diameter d = 0.074 cm = 74 ×10^{-5} m

total filament emissivity = 0.300

temperature = 3068 K

to find out

power emitted

solution

we find first area that is π×d×L

area = π×d×L

area = π×74 ×10^{-5}×5 ×10^{-2}

area = 1162.3892  ×10^{-5} m²

so here power emitted  is express as

power emitted  = E × σ × area × (temperature)^4

put here all value

power emitted  = 0.300× 5.67 × 1162.3892  ×10^{-5}  × (3068)^4

power emitted = 1.75 W

5 0
3 years ago
A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope in 10 s. The average power expended by the
Goshia [24]

To solve this problem we will use the concepts related to power, defined as the amount of energy applied over a period of time.

The energy in this case is the accumulated in the form of potential energy, over a period of time. Thus we will have that the mathematical expression of the power can be expressed as

P = \frac{E}{t}

Here,

E = Energy

t = time

As the energy is equal to the potential Energy we have tat

P = \frac{mgh}{t}

The weight (mg) of the man is 700N, the height (h) is 8m and the time is 10s, then:

P = \frac{700*8}{10}

P = 560W

Therefore the correct answer is A.

8 0
3 years ago
A body moving with an acceleration 2 m/s?then what is the change in velocity in 4sec.​
enot [183]

Answer:

As Per Provided Information

Moving body has 2m/s² acceleration

Time taken by body is 4 second

We are asked to find the 'change in velocity' ( ∆V) by the body.

<u>Formula Used here</u>

\boxed{\bf{\Delta \: V \:  =  acceleration \:  \times time \:}}

<u>Substituting </u><u>the </u><u>given </u><u>value</u>

<u>\sf\longrightarrow\Delta\:V \:  = 2 \times 4 \\  \\  \\ \sf\longrightarrow\Delta\:V \:  =8m {s}^{ - 1}</u>

<u>Therefore</u><u>,</u>

  • <u>Change </u><u>in </u><u>velocity </u><u>is </u><u>8</u><u> </u><u>m/</u><u>s</u>
7 0
2 years ago
In a LRC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect o
likoan [24]

Answer:

Impedance increases for frequencies below resonance and decreases for the frequencies above resonance

Explanation:

See attached file

Explanation:

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