Calculations made using Celsius or Fahrenheit will not work for gas law calculations

Derive an expression for the gravitational potential energy of a system consisting of Earth and a brick of mass m placed at Eart

Arlecino [84]

Answer:

The gravitational potential energy of a system is -3/2 (GmE)(m)/RE

Explanation:

Given

mE = Mass of Earth

RE = Radius of Earth

G = Gravitational Constant

Let p = The mass density of the earth is

p = M/(4/3πRE³)

p = 3M/4πRE³

Taking for instance,a very thin spherical shell in the earth;

Let r = radius

dr = thickness

Its volume is given by;

dV = 4πr²dr

Since mass = density* volume;

It's mass would be

dm = p * 4πr²dr

The gravitational potential at the center due would equal;

dV = -Gdm/r

Substitute (p * 4πr²dr) for dm

dV = -G(p * 4πr²dr)/r

dV = -G(p * 4πrdr)

The gravitational potential at the center of the earth would equal;

V = ∫dV

V = ∫ -G(p * 4πrdr) {RE,0}

V = -4πGp∫rdr {RE,0}

V = -4πGp (r²/2) {RE,0}

V = -4πGp{RE²/2)

V = -4Gπ * 3M/4πRE³ * RE²/2

V = -3/2 GmE/RE

The gravitational potential energy of the system of the earth and the brick at the center equals

U = Vm

U = -3/2 GmE/RE * m

U = -3/2 (GmE)(m)/RE

5 0

3 years ago

When power and current are known use R = P/I^2 to calculate the resistance.

inna [77]

<h2> $\bf{ \underline{Given:- }}$ </h2>

$\sf• \: Power \:( P) = 300 \: watts$

$\sf• \: Current \: (I) = 30 \: amps$

$\\$

<h2> $\bf{ \underline{To \: Find:- }}$ </h2>

$\sf• \: The \: Resistance$

$\\$

$\huge\bf{ \underline{Solution :-:- }}$

$\bf \red {\bigstar{ Formula \: of \: Power \: (P) = \frac{ {I}^{2} }{R}}}$

$\sf \rightarrow 300= \frac{ {30}^{2} }{ R}$

$\sf \rightarrow 300 = \frac{900}{ R}$

$\rightarrow \sf R = \frac{300}{900}$

$\rightarrow \sf R = \frac{1}{3}$

$\rightarrow \sf R = 0.33$

$\bf \purple{Hence, \: the \: resistance \: is \: 0.33 \: Ω \: . }$

7 0

2 years ago

X-ray diffraction is used to study the structure of crystallized proteins, nucleic acids, and other biological macromolecules. I

stepladder [879]

Answer:

PART A: option b. .43nm

PART B: option d. 0.11nm

PART C: option c. The wavelengths of visible light are too long compared to the atomic spacing.

Explanation:

Given data

Wavelength λ = 0.20 nm

Angle θ = 0.8 rad

(a)

wavelength of x-ray to give maximum at the same location

λ₂ = m λ

Here, m = 2 is the interference fringe order.

Substitute the values in the above equation.

λ₂ = 2 × 0.2

= 0.4 nm

Hence, the wavelength of x-ray to give maximum at the same location is 0.4nm

(b)

The crystal plane separation is equal to d

The value of θ is equal to 0.8 rad.

Convert rad into degree as follows:

0.8 rad = $\frac{180^0}{\pi rad} (\frac{0.8rad}{1})$ = 144°/π = 45.86°

Solve for d, using equation (1) as follows:

2dsinθ = mλ

d = mλ / 2sinθ

d = (1) 0.17 / 2Sin45.86°

d = 0.17 / 1.9065

d = 0.089 nm

(c)

The visible light can not be used to study the structure of proteins because of the high wavelength of the visible light.

7 0

2 years ago

This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The mag

sweet-ann [11.9K]

Answer:

Explanation:

Given

coefficient of kinetic friction $(\mu _k)$ =0.34

inclination $\theta =44$

weight of block=51 N

(a) When block is moving upward friction force acts downward

thus

$Fsin\theta -W-f_r=0$

as block is moving with constant velocity thus $F_{net}$ is zero

$f_r=\mu _kN=0.34\times Fcos\theta$

$F\left ( \sin \theta -\mu \cos \theta \right )=W$

$F=\frac{51}{0.45}=113.31 N$

(b)When Block slides down the wall friction changes its direction to oppose the block

$Fsin\theta -W+f_r=0$

$F\left ( \sin \theta +\mu \cos \theta \right )=W$

$F=\frac{W}{\left ( \sin \theta +\mu \cos \theta \right )}$

$F=\frac{51}{0.939}=54.299 N$

3 0

3 years ago

Which type of mater conduct heat?

Bingel [31]

Answer:

Copper is the best conductor

6 0

2 years ago

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