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sineoko [7]
3 years ago
12

Sand falls from an overhead bin and accumulates in a conical pile with a radius that is always threethree times its height. Supp

ose the height of the pile increases at a rate of 22 cm divided by scm/s when the pile is 1616 cm high. At what rate is the sand leaving the bin at that​ instant?
Physics
1 answer:
hammer [34]3 years ago
5 0

Answer:

159241.048 cm³/s

Explanation:

r = Radius = 3×height = 3h

h = height = 16 cm

Height of the pile increases at a rate = \frac{dh}{dt}=22\ cm/s

\text{Volume of cone}=\frac{1}{3}\pi r^2h\\\Rightarrow V=\frac{1}{3}\pi (3h)^2h\\\Rightarrow V=3\pi h^3

Differentiating with respect to time

\frac{dv}{dt}=9\pi h^2\frac{dh}{dt}\\\Rightarrow \frac{dv}{dt}=9\pi 16^2\times 22\\\Rightarrow \frac{dv}{dt}=159241.048\ cm^3/s

∴ Rate is the sand leaving the bin at that​ instant is 159241.048 cm³/s

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sveta [45]

Answer:

Ionization potential of C⁺⁵ is 489.6 eV.

Wavelength of the transition from n=3 to n=2 is 1.83 x 10⁻⁸ m.

Explanation:

The ionization potential of hydrogen like atoms is given by the relation :

E = \frac{13.6Z^{2} }{n^{2} } eV     .....(1)

Here <em>E</em> is ionization potential, <em>Z</em> is atomic number and <em>n</em> is the principal quantum number which represents the state of the atom.

In this problem, the ionization potential of Carbon atom is to determine.

So, substitute 6 for <em>Z</em> and 1 for <em>n</em> in the equation (1).

E = \frac{13.6\times(6)^{2} }{1^{2} }

<em> E = </em>489.6 eV

The wavelength (λ)  of the photon due to the transition of electrons in Hydrogen like atom is given by the relation :

\frac{1}{\lambda} =RZ^{2}[\frac{1}{n_{1} ^{2}}-\frac{1}{n_{2} ^{2} }]     ......(2)

R is Rydberg constant, n₁ and n₂ are the transition states of the atom.

Substitute 6 for Z, 2 for n₁, 3 for n₂ and 1.09 x 10⁷ m⁻¹ for R in equation (2).

\frac{1}{\lambda} =1.09\times10^{7} \times6^{2}[\frac{1}{2 ^{2}}-\frac{1}{3 ^{2} }]

\frac{1}{\lambda}  = 5.45 x 10⁷

λ = 1.83 x 10⁻⁸ m

7 0
4 years ago
If the mass of a car is 1,100 kg and its momentum is 22,000 kg<br> m/s, then what is its velocity?
egoroff_w [7]
Momentum = mass*velocity

In variable form, this is the same as p=mv.

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22,000=1,100v
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3 years ago
A plane is flying horizontally with speed 237 m/s at a height 1620m above the ground when a package is dropped for the plane. Th
Thepotemich [5.8K]

1. 0.5g*t^2 = 2010 m.

4.9t^2 = 2010.

t = 20.3 s. = Fall time.

D = Xo*t. = 193m/s * 20.3s = 3909 m.

2. V=sqrt(Xo^2+Yo^2)=sqrt(193^2+58^2) = 202 m/s.

3. Vo*t + 0.5g*t^2 = 2010 m.

58*t + 4.9*t^2 = 2010.

4.9t^2 + 58t - 2010 = 0.

Use Quadratic Formula.

t = 15.2 s. = Fall time.

D = 193m/s * 15.2s = 2934 m.

4 0
3 years ago
A dragster completed a 402.3-(0.2500-mi) run in 5.023s. If the car had a constant acceleration, what was its acceleration and fi
Ray Of Light [21]

Answer:

v = 80.092 m/s

a = 15.945 m/s²

Explanation:

Given,

The distance completed by the dragster, d = 402.3 m

The time taken by the dragster to complete that distance is, t = 5.023 s

The initial velocity of the dragster, u = 0

The final velocity of the dragster, v = ?

The acceleration of the dragster, a = ?

The velocity of the body is given by the formula

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Substituting the above values inn the equation

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So, the velocity of the dragster is, v = 80.092 m/s

The acceleration of the body is given by the formula

                                   a = (v - u)/t  m/s²

Substituting the above values in the equation

                                      = (80.092 - 0) / 5.023

                                      = 15.945 m/s²

Hence, the acceleration of the dragster is, a = 15.945 m/s²

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