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tamaranim1 [39]
3 years ago
8

A cylindrical water tank has a height of 20cm and a radius of 14cm. If it is filled to 2/5 of its capacity, calculate.

Physics
2 answers:
Makovka662 [10]3 years ago
8 0

Answer:

4.926 L Y 7.389 L

Explanation:

first you calculate the tank volume

V = π(14 cm)^{2}(10 cm = 12315 cm^{3}

then you convert to liters

12315 cm^{3} = 12.315 l

then you calculate the liters of water

2/5(12.35 l) = 4.926 l

finally we calculate the amount without water

12.315 l - 4.926 l = 7.389 l

HERE IS MORE INFORMATION ON THE SUBJECT. THEY REMOVED THE

ENGLISH SITE BUT YOU CAN USE TRANSLATOR

LINK: https://gscourses.thinkific.com/courses/fisicai

den301095 [7]3 years ago
5 0

Answer:

Explanation:

Total capacity is

V = πr²h = π(14²)(20) = 12,315 cm³

I. Quantity of water in the tank 12,315(2/5) = 4,926 cm³

II. Quantity of water left to fill the tank to its capacity.

12,315 - 4,926 = 7,389 cm³

or

12,315(3/5) = 7,389 cm³

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Answer:

w=694.575\ m

Explanation:

Given:

velocity of the sound, v=343\ m.s^{-1}

time lag in echo form one wall of the valley, t= 1.55\ s

time lag in echo form the other wall of the valley, t'=2.5\ s

<u>distance travelled by the sound in the first case:</u>

d=v.t

d=343\times 1.55

d=531.65\ m

Since this is the distance covered by the sound while going form the source to the walls and then coming back from the wall to the source so the distance of the wall form the source will be half of the distance obtained above.

s=\frac{d}{2}

s=\frac{531.65}{2}

s=265.825\ m

<u>distance travelled by the sound in the second case:</u>

d'=v.t'

d'=343\times 2.5

d'=857.5\ m

Since this is the distance covered by the sound while going form the source to the walls and then coming back from the wall to the source so the distance of the wall form the source will be half of the distance obtained above.

s'=\frac{d'}{2}

s'=\frac{857.5}{2}

s'=428.75\ m

<u>Now the width of valley:</u>

w=s+s'

w=265.825+428.75

w=694.575\ m

6 0
4 years ago
The reactants for a certain chemical reaction are shown. Cu(NO3)2 + NaOH → Which of the following shows the correct product or p
valentinak56 [21]

Answer : The products of this reaction are, Cu(OH)_2 and NaNO_3

Explanation :

First we have to write the balanced chemical reaction.

Cu(NO_3)_2+2NaOH\rightarrow Cu(OH)_2+2NaNO_3

In this reaction, 1 mole of copper nitrate react with the 2 moles of sodium hydroxide to give 1 mole of copper hydroxide and 2 moles of sodium nitrate.

This reaction is a double displacement reaction.

Double displacement reaction : A type of reaction in which a positive cation and a negative anion of two reactants react to exchange their places to form two new products.

The general representation of this reaction is,

XY+AB\rightarrow XB+AY

(X and A are the cations, Y and B are the anions)

Hence, the products of this reaction are, copper hydroxide,Cu(OH)_2 and sodium nitrate,NaNO_3

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3 years ago
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<span>b. the weather patterns outside

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3 years ago
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4 years ago
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beks73 [17]

Answer:

v_max = (1/6)e^-1 a

Explanation:

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v(t)=ate^{-6t}   (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]  (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

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For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a

hence, the maximum speed is v_max = ((1/6)e^-1)a

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