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3 years ago

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According to the Big Bang theory, when the universe first formed, the temperature was extremely hot. As it began to cool, proton

s and neutrons began to form and come together. Which would be the first element created directly after the Big Bang?

1 answer:

alex41 [277]3 years ago

5 0

hydrogen would be the first Element

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What is the potential energy of the system composed of the three charges q1, q3, and q4, when q1 is at point R? Define the poten

Ad libitum [116K]

Answer:

Incomplete question check attachment for complete question

Also it is given that

q1=-0.7uC

q4=-1.7uC

q3=-1.7uC

Also the distance are given as

a=2.2cm=0.022m

d2=3.6cm=0.036m

Explanation:

The potential energy due to point R is given as

The potential energy due to charge q1 and q3 plus the potential energy due to charge q4 and q1 plus the potential energy due to charge q3 and q4

So, let take it one after the other

Potential energy is give as

P.E=kq1q2/r

Therefore,

Potential energy due to charge q1 and q3

U¹³=kq1q3/r

To get the distance between charge q1 and q3, we will apply Pythagoras theorem

r=√(d2²+a²)

r=√(0.036²+0.022²)

r=0.0422m

k is a constant =9×10^9Nm²/C²

Then,

U¹³=kq1q3/r

U¹³=9×10^9×0.7×10^-6×1.7×10^-6/0.0422

U¹³=0.254J

Potential energy due to charge q1 and q4

U¹⁴=kq1q4/r

To get the distance between charge q1 and q4, we will apply Pythagoras theorem

r=√(d2²+a²)

r=√(0.036²+0.022²)

r=0.0422m

k is a constant =9×10^9Nm²/C²

Then,

U¹⁴=kq1q4/r

U¹⁴=9×10^9×0.7×10^-6×1.7×10^-6/0.0422

U¹⁴=0.254J

Potential energy due to charge q3 and q4

U³⁴=kq3q4/r

r=2a=2×0.022=0.044m

k is a constant =9×10^9Nm²/C²

Then,

U³⁴=kq3q4/r

U³⁴=9×10^9×1.7×10^-6×1.7×10^-6/0.044

U¹⁴=0.591J

Then, the total energy is

U= U¹³+ U¹⁴ + U³⁴

U=0.254+0.254+0.591

U=1.099J

Then also, the potential energy is zero because at infinity both U¹³ and U¹⁴ will have infinite potential because their distance apart will be infinite.

6 0

3 years ago

A ball on an extremely low-friction, tilted surface will slide downhill without rotating. If the surface is rough, however, the

spayn [35]

It will go left or right

6 0

3 years ago

A 2300 kg sailboat is moving west at 5.5 m/s when an eastward wind

pogonyaev

The boat is initially at equilibrium since it seems to start off at a constant speed of 5.5 m/s. If the wind applies a force of 950 N, then it is applying an acceleration <em>a</em> of

950 N = (2300 kg) <em>a</em>

<em>a</em> = (950 N) / (2300 kg)

<em>a</em> ≈ 0.413 m/s²

Take east to be positive and west to be negative, so that the boat has an initial velocity of -5.5 m/s. Then after 11.5 s, the boat will attain a velocity of

<em>v</em> = -5.5 m/s + <em>a</em> (11.5 s)

<em>v</em> = -0.75 m/s

which means the wind slows the boat down to a velocity of 0.75 m/s westward.

5 0

3 years ago

An object located near the surface of Earth has a weight of a 245 N

marusya05 [52]

Answer:

The mass of the object is 24.5 kg and weight of the object on Mars is 91.14 N.

Explanation:

Weight of the object on the surface of Earth, W = 245 N

On the surface of Earth, acceleration due to gravity, g = 10 m/s²

Weight of an object is given by :

W = mg

m is mass

$m=\dfrac{W}{g}\\\\m=\dfrac{245\ N}{10\ m/s^2}\\\\=24.5\ kg$

So, the mass of the object is 24.5 kg

Acceleration due to gravity on Mars, g' = 3.72 m/s²

Weight of the object on Mars,

W' =mg'

W' = 24.5 kg × 3.72 m/s²

= 91.14 N

So, the weight of the object on Mars is 91.14 N.

4 0

3 years ago

There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is

Mademuasel [1]

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

4 0

3 years ago