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timurjin [86]
3 years ago
12

A backyard swimming pool holds 150 cubic yards (yd3) of water. what is the weight of the water in pounds?

Physics
1 answer:
Crank3 years ago
5 0

solution:

1yd^3=(0.9144m)^3\\
therefore,\\
150yd^3=150yd^3\times\frac{(0.9144m)^3}{1yd^3}\\
=114.68m^3
convert m^3 to cm^3by the conversion factor as follow\\
1m^3=(100)cm^3\\
114.68m^3=114.68\times\frac{(100cm)3}{1m^3}\\
convert cm^3 to mL by the conversion factor as follow\\
1cm^3=1mL\\
114.68\times10^6cm^3=114.68\times10^6cm^3\times\frac{1mL}{1cm^3}\\
=114.68\times10^6mL\\
convert mL to g by the conversion factor as follow\\
1mL=1g\\therefore,\\
114.68\times10^6mL=114.68\times10^6mL\times\frac{1g}{1mL}\\
114.68\times10^6g\\
convert g to kg the conversion factor as follow\\
1g=10^-3kg\\
therefore,\\
114.68\times10^6=114.68\times10^6g\times\frac{10^-3kg}{1g}\\
=114.68\times10^3kg\\
convert kg to Ib by the conversion factor as follow\\
1kg=2.2Ib\\
therefore,\\
114.68\times10^3kg=114.68\times10^3kg\times\frac{2.2Ib}{1kg}\\
=2.52\times10^5Ib

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anygoal [31]

Answer:

a) 23.2 e V

b) energy of the original photon is 36.8 eV

Explanation:

given,

energy at ground level = -13.6 e V

energy at first exited state = - 3.4 e V

A photon of energy ionized from ground state and electron of energy K is released.

h ν₁ - 13.6 = K

K combine with photon in first exited state giving out photon of energy

h\nu_2 =\dfrac{hc}{\lambda}=\dfrac{12400}{466}

            = 26.6 e V

h c = 6.626 ×  10⁻³⁴ ×  3  × 10⁸  = 12400 e V A°

K + ( 3.4 ) = 26.6 e V

a) energy of free electron

K = 26.6 - 3.4 = 23.2 e V

b) energy of the original photon

h ν₁ - 13.6 = K

h ν₁  = 23.2 + 13.6

       = 36.8 e V

energy of the original photon is 36.8 eV

3 0
3 years ago
If a woman weighs 125 lb, her mass expressed in kilograms is x kg, where x is
adelina 88 [10]
The first thing you should know to solve this problem is the conversion of pounds to kilograms:
 1lb = 0.45 Kg
 We can solve this problem by a simple rule of three
 1lb ---> 0.45Kg
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 Clearing x we have:
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3 0
2 years ago
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Will give correct answer brainliest
levacccp [35]
The potential energy= mass times gravity times height. However, we are missing height. Gravity is a constant that is 9.8 on Earth. We then solve for height by dividing 350 by 10 and then 9.8 to get about 3.5.
TLDR: 3.5
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Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant
irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

a_g = \frac{GM}{R^2}

Here

M = \text{Mass inside the Orbit of the star}

R = \text{Orbital radius}

G = \text{Universal Gravitational Constant}

Mass inside the orbit in terms of Volume and Density is

M =V \rho

Where,

V = Volume

\rho =Density

Now considering the volume of the star as a Sphere we have

V = \frac{4}{3} \pi R^3

Replacing at the previous equation we have,

M = (\frac{4}{3}\pi R^3)\rho

Now replacing the mass at the gravitational acceleration formula we have that

a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho

a_g = \frac{4}{3} G\pi R\rho

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration.  So centripetal acceleration of the star is

a_c = \frac{4}{3} G\pi R\rho

At the same time the general expression for the centripetal acceleration is

a_c = \frac{\Theta^2}{R}

Where \Theta is the orbital velocity

Using this expression in the left hand side of the equation we have that

\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2

\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}

\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R

Considering the constant values we have that

\Theta = \text{Constant} \times R

\Theta \propto R

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0
3 years ago
A 1280 kg car is moving 4.92 m/s. a 509 N force then pushes it forward for 28.7 m. what is its final KE?(unit=J)PLEASE HELP
laila [671]

Answer:

The answer to your question is: total energy = 30100.4 J

Explanation:

Kinetic energy (KE) is the energy due to the movement of and object, its units are joules (J)

Data

mass = 1280 kg

speed = 4.92 m/s

Force = 509 N

distance = 28.7 m

Formula

KE = \frac{1}{2} mv^{2}

Work = Fd

Process

- Calculate Kinetic energy

- Calculate work

- Add both results

KE = \frac{1}{2} (1280)(4.92)^{2}

KE = 15492.1 J

Work = (509)(28.7)

Work = 14608.3 J

Total = 15492.1 + 14608.3

Total energy = 30100.4 J        

8 0
2 years ago
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