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sweet-ann [11.9K]
3 years ago
6

At what rate is soda being sucked out of a cylindrical glass that is 6 in tall and has radius of 2 in? The depth of the soda dec

reases at a constant rate of 0.25 in/sec.
Physics
1 answer:
evablogger [386]3 years ago
4 0

Answer:

The soda is being sucket out at a rate of 3.14 cubic inches/second.

Explanation:

R= 2in

S= π*R²= 12.56 inch²

rate= 0.25 in/sec

rate of soda sucked out= rate* S

rate of soda sucked out=  3.14 inch³/sec

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A train moving at a constant speed on a surface inclined upward at 10.0° with the horizontal travels a distance of 400 meters in
Amiraneli [1.4K]
If the velocity of the train is v=s/t, where s is the distance and t is time, then v=400/5=80m/s. To get the vertical component of the velocity we need to multiply the velocity v with a sin(α): Vv=v*sin(α), where Vv is the vertical component of the velocity and α is the angle with the horizontal. So:

Vv=80*sin(10)=80*0.1736=13.888 m/s. 

So the vertical component of the velocity of the train is Vv=13.888 m/s.
7 0
3 years ago
0.0884 moles of a diatomic gas
Sloan [31]

Answer:

W = - 118.24 J (negative sign shows that work is done on piston)

Explanation:

First, we find the change in internal energy of the diatomic gas by using the following formula:

\Delta\ U = nC_{v}\Delta\ T

where,

ΔU = Change in internal energy of gas = ?

n = no. of moles of gas = 0.0884 mole

Cv = Molar Specific Heat at constant volume = 5R/2 (for diatomic gases)

Cv = 5(8.314 J/mol.K)/2 = 20.785 J/mol.K

ΔT = Rise in Temperature = 18.8 K

Therefore,

\Delta\ U = (0.0884\ moles)(20.785\ J/mol.K)(18.8\ K)\\\Delta\ U = 34.54\ J

Now, we can apply First Law of Thermodynamics as follows:

\Delta\ Q = \Delta\ U + W

where,

ΔQ = Heat flow = - 83.7 J (negative sign due to outflow)

W = Work done = ?

Therefore,

-83.7\ J = 34.54\ J + W\\W = -83.7\ J - 34.54\ J\\

<u>W = - 118.24 J (negative sign shows that work is done on piston)</u>

7 0
3 years ago
Read 2 more answers
An object is placed 5.00 cm beyond the focal point of a convex lens whose focal length is 10.0 cm. If the object height is 3.0 c
Aleks04 [339]

Answer:

The height of the image is, h' = 6.0 cm

The image is erect.

Explanation:

Given data,

The object distance, u = -5 cm

The focal length of convex lens, f = 10 cm

The object height, h = 3 cm

The lens formula,

                      \frac{1}{f}=\frac{1}{v}-\frac{1}{u}

                      \frac{1}{10}=\frac{1}{v}-\frac{1}{-5}

                      \frac{1}{v}=\frac{1}{10}-\frac{1}{5}

                      v = -10 cm

The magnification factor of lens

                     m=\frac{-10}{-5}

                     m = 2

                     m=\frac{h'}{h}

                     h'=h\times m

                     h'=3\times 2

                     h' = 6 cm

The height of the image is, h' = 6 cm

The image is erect.

4 0
3 years ago
In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables
Juli2301 [7.4K]
They made me do it I don’t even know what to say I’m so sorry
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3 years ago
what happens to the volume when temperature is held constant and the pressure is increased to 125 kPa
Semenov [28]
The volume decreases, by a factor of

(the original pressure/(125 kPa).
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3 years ago
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