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N the diagram, the arrow shows the movement of electric charges through a wire connected to a battery.

dolphi86 [110]

Answer:

D. A difference in resistance

5 0

2 years ago

Assuming that Bernoulli's equation applies, compute the volume of water ΔV that flows across the exit of the pipe in 1.00 s . In

OLEGan [10]

Answer:

discharge rate (Q) = 0.2005 m^{3} / s

Explanation:

if you read the question you would see that some requirements are missing, by using search engines, you can get the complete question as stated below:

Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0m , and the elevation of points 2 and 3 is 2.00 m . The cross-sectional area at point 2 is 4.80x10-2m ; at point 3, where the water is discharged, it is 1.60x10-2m. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe. Part A Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in 1.00 s . In other words, find the discharge rate \Delta V/Delta t. Express your answer numerically in cubic meters per second.

solution:

time = 1 s

elevation of point 1 (z1) = 10 m

elevation of point 2 (z2) = 2 m

elevation of point 3 (z3) = 2 m

cross section area of point 2 = 4.8 x 10^{2} m

cross section area of point 3 = 1.6 x 10^{2} m

g

acceleration due to gravity (g) = 9.8 m/s^{2}

find the discharge rate at point 3 which is the exit pipe.

discharge rate (Q) = A3 x V3

where A3 is the cross sectional area at point 3 and V3 is the velocity of the fluid and can be gotten by applying Bernoulli's equation below

$\frac{P1}{ρg} + \frac{V1^{2} }{2g} + Z1 = \frac{P3}{ρg} + \frac{V3^{2} }{2g} + Z3$

pressure at point 1 (P1) is the same as pressure at point 3 (P3), and at point 1, the velocity (V1) = 0. therefore the equation now becomes

$\frac{P1}{ρg} + Z1 = \frac{P1}{ρg} + \frac{V3^{2} }{2g} + Z3$

$Z1 = \frac{V3^{2} }{2g} + Z3$

V3 = $\sqrt{2g(Z1-Z3)}$

V3 = $\sqrt{2 x 9.8 x (10 - 3)}$

V3 = 12.53 m/s

discharge rate (Q) = A3 x V3 = 1.6 x 10^{-2} x 12.53

discharge rate (Q) = 0.2005 m^{3} / s

8 0

3 years ago

Question 2 (ID=81813)

zaharov [31]

Answer:

The answer is B

Explanation:

5/2=2.5

2.5x2=5

Hope this helps ik its kinda confusing lol

7 0

2 years ago

Read 2 more answers

A woman falls to the ground while wearing a parachute. The air resistance on the parachute of the parachute is 500N. If the woma

scoundrel [369]

The gravitational force on the woman is A) 500 N

Explanation:

There are two forces acting on the woman during her fall:

The force of gravity, $F_G$ , acting downward
The air resistance, $F_D$ , acting upward

According to Newton's second law, the net force acting on the woman is equal to the product between the woman's mass and her acceleration:

$\sum F=ma$

where m is the mass of the woman and a her acceleration.

The net force can be written as

$\sum F = F_G - F_D$

Also, we know that the woman falls at a constant velocity (5 m/s), this means that her acceleration is zero:

$a=0$

Combining the equations together, we get:

$F_G-F_D = 0$

which means that the magnitude of the gravitational force is equal to the magnitude of the air resistance:

$F_G=F_D=500 N$

Learn more about forces and Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0

2 years ago

Read 2 more answers

A cord of mass 0.65 kg is stretched between two supports 28 m apart. if the tension in the cord is 150 n, 2 how long will it tak

MaRussiya [10]

Ans: Time <span>taken by a pulse to travel from one support to the other = 0.348s
</span>
Explanation:
First you need to find out the speed of the wave.

Since
Speed = v = $\sqrt{ \frac{T}{\mu} }$

Where
T = Tension in the cord = 150N
μ = Mass per unit length = mass/Length = 0.65/28 = 0.0232 kg/m

So
v = $\sqrt{ \frac{150}{0.0232} }$ = 80.41 m/s

Now the time-taken by the wave = t = Length/speed = 28/80.41=0.348s

5 0

2 years ago