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4 years ago

I kick ball has a 10 in diameter what is the volume of the ball

1 answer:

5 0

The answer is 523.60 rounded. To get this use the formula: V=4/3 • π • r^3 = 4/3·π·(53) ≈ 523.59878. Which is four over three, multiplied by “pie” multiplied by the radius cubed. (Which is 5 • 5 • 5). To get the radius, you split the diameter in half. For you equation, “1/2 of ten is five” so the radius is five. Hope this helps!

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Kitty [74]

Answer:you in connections too?

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3 years ago

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A 500.0-g chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating

mixas84 [53]

Answer:

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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

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3 years ago

Siska hit a golf ball into the air with an initial velocity of 56 feet per second. The height h in feet of the ball above the gr

olasank [31]

To find the time it takes for the ball to reach the given height, we need an equation that relates the height and time it travel which is given to be <span>h=-16t^{2}+56t. We just substitute the height of 49 feet then solve for t. We do as follows:

</span><span>h=-16t^{2}+56t
49 = </span>-16t^{2}+56t
t = 1.75 seconds

Hope this answers the question. Have a nice day.

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3 years ago

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In a physics experiment, two equal-mass carts roll towards each other on a level, low-friction track. One cart rolls rightward a

xz_007 [3.2K]

Answer:

The correct answer is option a.

Explanation:

Conservation of momentum :

$m_1u_1+m_2u_2=m_1v_1+m_1v_2$