Which of the following describes the way a population can be dispersed?

Something has a mass of 40 kg and a velocity of 1 m/s

lesantik [10]

m=40kg
v=1m/s

$\\ \sf\Rrightarrow K.E=\dfrac{1}{2}mv^2$

$\\ \sf\Rrightarrow K.E=\dfrac{1}{2}(40)(1)^2$

$\\ \sf\Rrightarrow K.E=20J$

m=80kg

$\\ \sf\Rrightarrow K.E=\dfrac{1}{2}(80)(1)^2$

$\\ \sf\Rrightarrow K.E=40J$

Yes

6 0

3 years ago

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

Veronica claims that she can throw a dart at a dartboard from a distance of 2 m and hit the 5 cm wide bullseye if she throws the

wariber [46]

First, find the amount of time for the dart to hit the board using this equation: t = d/v

t = 2 m/ 15 m/s = 0.133 s

Then, find the height the dart has fallen from its initial point using this equation: h = 0.5gt²

h = 0.5(9.81 m/s²)(0.133 s)² = 0.0872 m or 8.72 cm

Since the diameter of the bull's eye is only 5 cm, and you started at the same level of the top of the bull's eye, that means the maximum allowance would only be 5 cm. Since it exceeded to 8.72 cm, it means that <em>Veronica will not hit the bull's eye.</em>

3 0

4 years ago

The quantity of heat Q that changes the temperature L1Tof a mass mof a substance isgiven by Q = cmt:T, where c is the specific h

White raven [17]

Answer:

a) Q = 80,000 cal

b) Q = 100,000 cal

c) Q = 540,000 cal

d) Q = 720,000 cal

Explanation:

a)1 kg from 0⁰ Ice to 0⁰ water, the heat produced is latent heat of fusion

$Q_{l} = ML_{f}$ = 1 * 80

$Q_{l}$ = 80 kCal = 80,000 cal

b) 1 kg of O°C ice water to 1 kg of 100°C boiling water

Specific heat capacity, c = 1000cal/kg.C

$Q_{c} = mc \delta T\\Q_{c} = 1 * 1000 * (100 - 0)\\Q_{c} =100000 cal$

c) 1 kg of 100°C boiling water to 1 kg of 100°C steam

Latent heat of vaporization is needed for this conversion

$Q_{v} = ML_{v} \\L_{v} = 540 kCal/kg\\Q_{v} =1* 540 \\Q_{v} = 540 kCal = 540000 cal$

d) 1 kg of O°C ice to 1 kg of 100°C steam.

Q = $Q_{L} + Q_{c} + Q_{v}$

Q = 80,000 + 100,000 + 540,000

Q = 720,000 cal

7 0

4 years ago

At an age of 380,000 years, the temperature of the universe had fallen to 3000 k, and electrons could then combine with protons

Andru [333]

The major transition occurred is that the universe became transparent to light for the first time. This is the consequence that might happen because of the changes in the universe at this time. The reaction between the elements that are present in the space can affect the present universe.

7 0

3 years ago