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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.

Water pressure = Height * density * gravity

Ratling [72]2 years ago

4 0

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

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An apple contains 165 Calories. How many actual calories does it contain? How many joules does it contain

jasenka [17]

Answer:

95 calories

Explanation:

A medium-sized apple has only 95 calories but plenty of water and fiber.

8 0

2 years ago

A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for 3000 m in 210 seconds. How fast is the car moving

dedylja [7]

Answer:

I dont know i just need the points

6 0

2 years ago

A chimpanzee sitting against his favorite tree gets up and walks 70.9 m due east and 31.9 m due south to reach a termite mound,

DIA [1.3K]

A right triangle is formed by the 70.9 m walked east and 31.9 m walked south.
The legs of this right triangle are 70.9 and 31.9.
The shortest distance between two points is a straaight line. Therefore the hypotenuse of this triangle is going to be that shortest distance.
We can use the Pythagorean Theorem to find the hypotenuse of this triangle.
$a^2+b^2=c^2\\70.9^2+31.9^2=c^2\\5026.81+1017.61=c^2\\6044.42=c^2\\c=\sqrt{6044.42}$

$c=\sqrt{6044.42}\approx\boxed{77.7458681}\ (decimal\ form)\\\\c=\sqrt{6044.42}=\sqrt{\frac{604442}{100}}=\boxed{\frac{\sqrt{604442}}{10}}\ (exact\ form)$

As for the second part of the question, we want to find the angle formed by the hypotenuse and the 31.9 walked east.
We could use any of the three trigonometric ratios here since we know all 3 sides.
sine = opposite / hypotenuse
cosine = adjacent / hypotenuse
tangent = opposite / adjacent
I am going to use tangent, because then I won't have to deal with the hypotenuse and so the answer will be more accurate.

If you haven't already drawn yourself a diagram, now is a good time to.
The side opposite our angle is the 31.9, and the adjacent is 70.9.
Therefore, $\tan(m\angle)=\frac{31.9}{70.9}$ .

We can use inverse trig ratios here to find the measure of our angle.
$\tan^{-1}(\tan(m\angle))=\tan^{-1}(\frac{31.9}{70.9})\\\\m\angle=\tan^{-1}(\frac{31.9}{70.9})\approx\boxed{24.2243851\°\ or\ 0.422795279\ rad}$

4 0

2 years ago

Future space rockets might propel themselves by firing laser beams, rather than exhaust gases, out the back. The acceleration wo

Nina [5.8K]

Answer:

Acceleration = 0.0282 m/s^2

Distance = 13.98 * 10^12 m

Explanation:

we will apply the energy theorem

work done = ΔK.E ( change in Kinetic energy ) ---- ( 1 )

where :

work done = p * t

= 15 * 10^6 watts * ( 1 year ) = 473040000 * 10^6 J

( note : convert 1 year to seconds )

and ΔK.E = 1/2 mVf^2 given ; m = 1200 kg and initial V = 0

back to equation 1

473040000 * 10^6 = 1/2 mv^2

Vf^2 = 2(473040000 * 10^6 ) / 1200

∴ Vf = 887918.92 m/s

i) Determine how fast the rocket is ( acceleration of the rocket )

a = Vf / t

= 887918.92 / ( 1 year )

= 0.0282 m/s^2

ii) determine distance travelled by rocket

Vf^2 - Vi^2 = 2as

Vi = 0

hence ; Vf^2 = 2as

s ( distance ) = Vf^2 / ( 2a )

= ( 887918.92 )^2 / ( 2 * 0.0282 )

= 13.98 * 10^12 m

8 0

2 years ago

A coil with an inductance of 2.3 H and a resistance of 14 Ω is suddenly connected to an ideal battery with ε = 100 V. At 0.13 s

klemol [59]

Given Information:

Resistance = R = 14 Ω

Inductance = L = 2.3 H

voltage = V = 100 V

time = t = 0.13 s

Required Information:

(a) energy is being stored in the magnetic field

(b) thermal energy is appearing in the resistance

Answer:

(a) energy is being stored in the magnetic field ≈ 219 watts

(b) thermal energy is appearing in the resistance ≈ 267 watts

Explanation:

The energy stored in the inductor is given by

$U = \frac{1}{2} Li^{2}$

The rate at which the energy is being stored in the inductor is given by

$\frac{dU}{dt} = Li\frac{di}{dt} \: \: \: \: eq. 1$

The current through the RL circuit is given by

$i = \frac{V}{R} (1-e^{-\frac{t}{ \tau} })$

Where τ is the the time constant and is given by

$\tau = \frac{L}{R}\\ \tau = \frac{2.3}{14}\\ \tau = 0.16$

$i = \frac{110}{14} (1-e^{-\frac{t}{ 0.16} })\\i = 7.86(1-e^{-6.25t})\\\frac{di}{dt} = 49.125e^{-6.25t}$

Therefore, eq. 1 becomes

$\frac{dU}{dt} = (2.3)(7.86(1-e^{-6.25t}))(49.125e^{-6.25t})$

At t = 0.13 seconds

$\frac{dU}{dt} = (2.3) (4.37) (21.8)\\\frac{dU}{dt} = 219.11 \: watts$

(b) thermal energy is appearing in the resistance

The thermal energy is given by

$P = i^{2}R\\P = (7.86(1-e^{-6.25t}))^{2} \cdot 14\\P = (4.37)^{2}\cdot 14\\P = 267.35 \: watts$

The energy delivered by battery is

$P = Vi\\P = 110\cdot 4.37\\P = 481 \: watts$

4 0

3 years ago