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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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Answer:

proof in explanation

Explanation:

First, we will calculate the number of half-lives:

n = \frac{t}{t_{1/2}}

where,

n = no. of half-lives = ?

t = total time passed = 2100 million years

t_{1/2} = half-life = 700 million years

Therefore,

n = \frac{2100\ million\ years}{700\ million\ years}\\\\n = 3

Now, we will calculate the number of uranium nuclei left (n_u):

n_u = \frac{1}{2^{n} }(total\ nuclei)\\\\n_u = \frac{1}{2^{3} }(6400\ million)\\\\n_u = \frac{1}{8}(6400\ million)\\\\n_u =  800\ million

and the rest of the uranium nuclei will become thorium nuclei (u_{th})

n_{th} = total\ nuclei - n_u\\n_{th} = 6400\ million-800\ million\\n_{th} = 5600\ million

dividing both:

\frac{n_{th}}{n_u}=\frac{5600\ million}{800\ million} \\\\n_{th} = 7n_u

<u>Hence, it is proven that after 2100 million years there are seven times more thorium nuclei than uranium nuclei in the rock.</u>

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2 years ago
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Choose a substance you're familiar with. what are its physical and chemical properties? How would you measure its density? What
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A car is running at the speed of 45km/hr see a child 25 meter ahead and suddenly apllies a brakes. If the retradation of the car
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Answer:

The car stops in 7.78s and does not spare the child.

Explanation:

In order to know if the car stops before the distance to the child, you take into account the following equation:

x=x_o+v_ot-\frac{1}{2}at^2        (1)

vo: initial speed of the car = 45km/h

a: deceleration of the car = 2 m/s^2

t: time

xo: initial distance to the child = 25m

x: final distance to the child = 0m

It is necessary that the solution of the equation (1) for time t are real.

You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:

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You take the first value t1 because it has physical meaning.

The solution for t is real, then, the car stops in 7.78s and does not spare the child.

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<span>When the green arrow and solid red light is illuminated, </span>means you turn in the direction of the arrow.
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