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2 years ago

This sphere contains all of the frozen water on Earth

1 answer:

7 0

Cryosphere is the sphere that contains all of the frozen water on Earth. The correct option among all the options that are given in the question is the fourth option or the last option. Cryosphere is actually those parts of the surface of the earth where the water is in a frozen state, like in seas and oceans.

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Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

Science Write two examples of heating appliances

bearhunter [10]

Answer:

a} Infrared radiant heaters.

b} Fan heaters.

Explanation:

I hope this is okay, should I give more.

4 0

2 years ago

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

4 0

2 years ago

The pressure in a traveling sound wave is given by the equation ΔP = (1.78 Pa) sin [ (0.888 m-1)x - (500 s-1)t] Find (a) the pre

frozen [14]

Answer:

a) $P_m=1.78\ Pa$

b) $f=79.5775\ Hz$

c) $\lambda=7.076\ m$

d) $v=563.06\ m.s^{-1}$

Explanation:

Given equation of pressure variation:

$\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]$

We have the standard equation of periodic oscillations:

$\Delta P=P_m\ sin\ (kx-\omega.t)$

By comparing, we deduce:

(a)

amplitude:

$P_m=1.78\ Pa$

(b)

angular frequency:

$\omega=2\pi.f$

$2\pi.f=500$

∴Frequency of oscillations:

$f=\frac{500}{2\pi}$

$f=79.5775\ Hz$

(c)

wavelength is given by:

$\lambda=\frac{2\pi}{k}$

$\lambda=\frac{2\pi}{0.888}$

$\lambda=7.076\ m$

(d)

Speed of the wave is gives by:

$v=\frac{\omega}{k}$

$v=\frac{500}{0.888}$

$v=563.06\ m.s^{-1}$

8 0

2 years ago

Hi please, I Have An attachment on Waves, Just two Objective Questions Whoever Answers Will be Marked Brainliest thank you.

kifflom [539]

Answer:

The first answer is W and Z, since they appear to be a period apart. Dont know the second question. I did what I could, hope someone can answer the second.

5 0

2 years ago