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zalisa [80]
3 years ago
7

Two source charges are placed close together. The electric field vectors close to the charges all point toward the charge on the

right and away from the charge on the left. A positive test charge is released in the middle. Which way will the test charge move?
A.
to the right
B.
to the left
C.
up
D.
down
Physics
1 answer:
Leto [7]3 years ago
5 0
B is the correct answer i believe
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Water is flowing through a 45° reducing pipe bend at a rate of 200 gpm and exits into the atmosphere (P2 = 0 psig). The inlet to
Nataliya [291]

Answer:

F1=177.88 Newtons

Explanation:

Let's start with the Bernoulli's equation:

P_{1} + \frac{1}{2}\beta V_{1} ^{2} + \beta gh_{1}  =P_{2} + \frac{1}{2}\beta V_{2} ^{2} + \beta gh_{2}

Where:

P is pressure, V is Velocity, g is gravity, h is height and β is density (for water β=1000 kg/m3); at the points 1 and 2 respectively.

From the Bernoulli's equation and assuming that h is constant and P2 is zero (from the data), we have:

P_{1} + \frac{1}{2}\beta V_{1} ^{2} = \frac{1}{2}\beta V_{2} ^{2}

As we know, P1 must be equal to \frac{F_{1} }{A_{1}}, so, replacing P1 in the equation, we have:

P_{1} = \frac{F_{1}}{A_{1}} = \frac{1}{2}\beta(V_{2} ^{2} - V_{1} ^{2})

And

F_{1} = {A_{1}} ( \frac{1}{2}\beta(V_{2} ^{2} - V_{1} ^{2}))

Now, let's find the velocity to replace the values on the expression:

We can express the flow in function of velocity and area as Q = V A, where Q is flow, V is velocity and A is area. As the same, we can write this: Q_{1} = V_{1} A_{1}\\Q_{2} = V_{2} A_{2}. In the last two equations, let's clear Velocities.

V_{1} = \frac{Q_{1}}{A_{1}}\\V_{2} = \frac{Q_{2}}{A_{2}}

and replacing V1 and V2 on the last equation resulting from Bernoulli's (the one that has the force on it):

F_{1} = {A_{1}} ( \frac{1}{2}\beta((\frac{Q_{2}}{A_{2}})^{2} - (\frac{Q_{1}}{A_{1}})^{2}))

First, we have to consider that from a mass balance, the flow is the same, so Q1=Q2, what changes, is the velocity. Knowing this, let's write the areas, diameters, density and flow on International Units System (S.I.), because the exercise is asking us the answer in Newtons.

D_{1}=1.5 inches=0.0381 mts\\D_{2}=1 inches=0.0254 mts\\A_{1}=\frac{\pi D_{1}^{2} }{4}=0.00114mts^{2}\\A_{2}=\frac{\pi D_{2}^{2} }{4}=0.000507mts^{2}\\\beta=1000 kgs/m^{3}\\Q=200gpm=0.01mts^{3}/seg

Replacing the respective values in this last expression, we obtain:

F1 = 177.88 N

3 0
3 years ago
Which of the following is not a name of a scale for measuring temperature? A. Kelvin B. Celsius C. Absolute D. Fahrenheit
Lena [83]
Hello,

The answer is option C "absolute".

Reason:

The answer is option C because the three names for temperatures are: Kelvin, Celsius, and Fahrenheit therefore your answer is option C.

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit
5 0
3 years ago
How much mass of ice at 0℃ must melt when ∆Q=948000 J of heat energy is added to it?
butalik [34]
<h2>The ice melted is nearly 2.8 kg </h2>

Explanation:

The quantity of heat required to melt the ice can be found by the relation .

ΔQ = m x L

here L is used for the latent heat of fusion .

Its value is 334 J per gram for ice .

Thus m = \frac{\Delta Q}{L} = \frac{948000}{334}  = 2836 gram

or = 2.8 kg approx

6 0
3 years ago
. There are fewer organisms at the highest trophic levels, because
Amiraneli [1.4K]

Answer:

Because there is less energy at higher trophic levels

Explanation:

  • In a food chain that shows the feeding levels of organisms in an ecosystem, energy decreases as you move from the lowest trophic level to the highest trophic level.
  • The lowest trophic level consists of producers which have the highest energy as they trap energy directly from the sunlight.
  • The energy is lost from one trophic level to the next up to the highest trophic level which has the lowest energy.
  • Therefore, organisms at the highest trophic level are fewer in number but large in size.
7 0
3 years ago
While John is traveling along a straight interstate
lora16 [44]

Answi dont know

Explanation:

i dont know

7 0
3 years ago
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