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2 years ago

Why are water storage tanks usually placed on high towers

2 answers:

7 0

Thay are on high towers because if it was below how would the water flow. Putting it on high towers gives you an advantage of the gravity with means you got free pressure without having to use a pump.

<span>Water pressure = Height * density * gravity</span>

Ratling [72]2 years ago

4 0

<span>To increase the water pressure that flows from it. Allowing the water to go to the maximum amount of houses.</span>

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A way to prevent injuries in a collision is to

ratelena [41]

Always wear your proper safety restraints (or seat-belts)

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2 years ago

Which term best describes a fad diet

stira [4]

Answer:

a fad diet is basically a diet that was a trend or that was popular for a short amount of time and makes unreasonable claims for fast and very much wanted results

Explanation:

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2 years ago

4. What's the voltage drop if V, = 56 V, V2 = 35 V, and V3 = 3 V

Murrr4er [49]

Answer:

The correct option is A. 18 V

Therefore,

$V_{1}=18\ V$

Explanation:

Given:

Total Voltage,

V = 56 V,

Voltage drop

$V_{2}=35\ V$

$V_{3}=3\ V$

To Find:

$V_{1}=?$ (Assumed ,as it is not given clearly in the question)

Solution:

So in series combination total Voltage is given as

$V=V_{1}+V_{2}+V_{3}$

Substituting the values we get

$56=V_{1}+35+3\\\\V_{1}=56-38=18\ V$

Therefore,

$V_{1}=18\ V$

5 0

3 years ago

A wave of amplitude 20mm has intensity Ix. Another wave of the same frequency but of amplitude 5mm has an intensity Iy.

Alexeev081 [22]

Answer:

(C) 16

Explanation:

Given:

The amplitude of first wave (s₁) = 20 mm

The amplitude of second wave (s₂) = 5 mm

Intensity of first wave = Iₓ

Intensity of second wave = $I_y$

The intensity associated with a wave depends on the amplitude of the wave.

The intensity (I) is directly proportional to the square of the amplitude (s) of the wave and is expressed as:

$I=ks^2\\Where\ k\to constant\ of\ proportionality$

Now, the intensities of the two waves are given as:

$I_x=ks_1^2=k(20)^2\\\\I_y=ks_2^2=k(5)^2$

Dividing both the intensities, we get:

$\frac{I_x}{I_y}=\frac{k(20)^2}{k(5)^2}\\\\\frac{I_x}{I_y}=\frac{400}{25}\\\\\frac{I_x}{I_y}=16$

Therefore, the option (C) is correct.

5 0

2 years ago

A spring of negligible mass has force constant of 1600 Newtons per meter. (a) How far must the spring be compressed for 3.20 Jou

frez [133]

Answer:

a) x=63.0 $x10^{-3} m$ or 6.3cm

b) x=116.0 $x10^{-3} m$ or 11.6cm

Explanation:

a).

The elastic potential energy is modeling by equation :

$U1=\frac{1}{2}*k*x^{2} \\K=1600 \frac{N}{m}\\ U=3.2J\\m=1.2kg\\x^{2}=\frac{2*U}{k}\\x=\sqrt{\frac{2*3.2J}{1600 \frac{N}{m}}} \\x=\sqrt{4x10^{-3} m^{2}}\\ x=0.06324m$

b).

The work energy theorem explain which work is done in this case. the motion began from the rest so K1=K2 equal zero, Ug1 is no yet done and U2is also zero because is the potential energy

$Ug=K2\\mgy=\frac{1}{2}*k*x^{2} \\but \\y=h+x=0.80m+x\\m*g*(0.80m+x)=\frac{1}{2}*k*x^{2}$

Solving for x

$2*(m*g(h+x))=k*x^{2} \\k*x^{2}-(2*m*g*x)-(2*m*g*h)=0\\1600x^{2} -2*1.2kg*9.8\frac{m}{s^{2}}*x-2*1.2kg*0.80m=0\\1600x^{2} -23.52x-18.816m=0$

$x=-\frac{b+/-\sqrt{b^{2}-4*a*c } }{2*a} \\x=-\frac{-23.52+/-\sqrt{-23.52^{2}-4*1600*-18.816 } }{2*a} \\x=11.79 +/- 173.9\\x1=-0.10 m\\x2=0.116$

The negative is discard so

x=0.116m

7 0

3 years ago