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4 years ago

Please someone help me lol

Physics

1 answer:

Makovka662 [10]4 years ago

7 0

From Largest to Smallest:

Earth > Biosphere > Biome > Ecosystem > Community > Population > Organism > Organelles

Hope this helps!

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The density of an object is 1.23 g/cm3. In which of the following materials will it float

Leviafan [203]

Citric Acid is the correct answer because it contains a density of 1.66 g/cm3, whereas water= 1.00 g/cm3, Olive oil= 0.93 g/cm3, Ethyl alcohol= 0.79 g/cm3

5 0

3 years ago

Read 2 more answers

What do you guys know about emotional intelligence? And how is it used in the real world?

tester [92]

Emotional Intelligence is our own abilities of being able to identify and regulate our own emotions. This is used by individuals in the real world in order to regulate crying in front of everyone for example. Calming ourselves down by breathing is another good example of regulating emotion which plays into emotional intelligence.

Hope this helps mate :D

7 0

3 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$

5 0

3 years ago

Your little sister is building a radio from scratch. Plans call for a 500 μH inductor wound on a cardboard tube. She brings you

gayaneshka [121]

Answer:

N = 195 turns

Explanation:

The inductance of the inductor, L = 500 μH = 500 * 10⁻⁶H

The length of the tube, l = 12 cm = 0.12 m

The diameter of the tube, d = 4 cm = 0.04 m

Radius, r = 0.04/2 = 0.02 m

Area of the tube, A = πr² = 0.02²π = 0.0004π m²

$\mu_{0} = 4\pi * 10^{-7}$

The inductance of a solenoid is given by:

$L = \frac{\mu_{0}N^{2} A }{l}$

$500 * 10^{-6} = \frac{4\pi *10^{-7} N^{2} *4\pi *10^{-4} }{0.12}\\500 * 10^{-6} = 0.00000001316N^{2} \\N^{2} = \frac{500 * 10^{-6}}{0.00000001316}\\N^{2} = 37995.44\\N = \sqrt{37995.44} \\N = 194.92 turns$

8 0

3 years ago

1. In Newton’s ring experiment, the diameter of the 5th ring is 0.30 cm and diameter of 15th the ring is 0.62 cm. Find the diame

IgorC [24]

Answer:

Diameter of Newton’s 5th ring = 0.30 cm

Diameter of Newton’s 15th ring = 0.62 cm

Diameter of Newton’s 25th ring = ?

From Newton’s rings experiment we infer that

D2n+m − D2n = 4λmR

For the 5th and 15th rings we have

D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)

For 15th and 25th rings

D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)

We equate the two derivatives

Equation (2) = Equation (1)

D225 − D215 = D215 − D25

D225 = 2D215 – D25

Substituting the values into the equation

D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2

D25 = 0.8239 cm

4 0

3 years ago