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3 years ago

Jupiter is composed almost entirely of gases and liquids, mostly

Physics

2 answers:

Shkiper50 [21]3 years ago

6 0

<span>Jupiter is composed almost entirely of gases and liquids, mostly C. hydrogen and helium.</span>

Crazy boy [7]3 years ago

3 0

Answer:

The answer is C. hydrogen and helium hopefully this helps!

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You are looking down on a N = 9 turn coil in a magnetic field B = 0.5 T which points directly down into the screen. If the diame

Novay_Z [31]

Answer:

The magnitude of the voltage is $2.27\times10^{-4}\ V$ and the direction of the current is clockwise.

Explanation:

Given that,

Number of turns = 9

Magnetic field = 0.5 T

Diameter = 3 cm

Time t = 0.14 s

We need to calculate the flux

Using formula of flux

$\phi=NAB$

Put the value into the formula

$\phi=9\times\pi\times(1.5\times10^{-2})^2\times0.5$

$\phi=0.003180$

We need to calculate the emf

Using formula of emf

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{0.003180}{0.14}$

$\epsilon =-0.000227\ V$

$\epsilon=-2.27\times10^{-4}\ V$

Negative sign shows the direction of current.

Hence, The magnitude of the voltage is $2.27\times10^{-4}\ V$ and the direction of the current is clockwise.

8 0

3 years ago

A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax

Colt1911 [192]

Answer:

From the question we are told that

The length of the rod is $L_o$

The speed is v

The angle made by the rod is $\theta$

Generally the x-component of the rod's length is

$L_x = L_o cos (\theta )$

Generally the length of the rod along the x-axis as seen by the observer, is mathematically defined by the theory of relativity as

$L_xo = L_x \sqrt{1 - \frac{v^2}{c^2} }$

=> $L_xo = [L_o cos (\theta )] \sqrt{1 - \frac{v^2}{c^2} }$

Generally the y-component of the rods length is mathematically represented as

$L_y = L_o sin (\theta)$

Generally the length of the rod along the y-axis as seen by the observer, is also equivalent to the actual length of the rod along the y-axis i.e $L_y$

Generally the resultant length of the rod as seen by the observer is mathematically represented as

$L_r = \sqrt{ L_{xo} ^2 + L_y^2}$

=> $L_r = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}$

=> $L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}$

=> $L_r = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}$

=> $L_r = \sqrt{L_o^2 * cos^2(\theta) [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}$

=> $L_r = \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }$

=> $L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Hence the length of the rod as measured by a stationary observer is

$L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Generally the angle made is mathematically represented

$tan(\theta) = \frac{L_y}{L_x}$

=> $tan {\theta } = \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }$

=> $tan(\theta ) = \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }$

Explanation:

7 0

3 years ago

Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the mag

nadya68 [22]

Answer:

F = 147,78*10⁻⁹ [N]

Explanation:

By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .

The angle β ( angle between the line running through one of the charges in y axis and the charge in x axis) is

tan β = 0,5/0,7

tan β = 0,7142 then β = arctan 0,7142 ⇒ β = 35 ⁰

cos β = 0,81

d = √ (0,5)² + (0,7)² d1stance between charges

d = √0,25 + 0,49

d = √0,74 m

d = 0,86 m

Now Foce between two charges is:

F = K* q₁*q₂/ d² (1)

Where K = 9*10⁹ N*m²/C²

q₁ = 2,5* 10⁻⁹C

q₂ = 3,0*10⁻⁹C

d² = 0,74 m²

Plugging these values in (1)

F = 9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74 [N*m²/C²]*C*C/m²

F = 91,21 * 10⁻⁹ [N]

And Fx = F*cos β

Fx = 91,21 * 10⁻⁹ *0,81

Fx =73,89*10⁻⁹ [N]

Then total force acting on charge located at x = 0,7 m is:

F = 2* Fx

F = 2*73,89*10⁻⁹ [N]

F = 147,78*10⁻⁹ [N]

8 0

3 years ago

Pls help I really need help w this one Tnks

lyudmila [28]

Sorry I do not know srry

6 0

3 years ago

Fg =G m1m2/r2 solver for G

Blababa [14]

G = $\frac{F_{g} r^{2} }{m_{1} m_{2} }$

Explanation:

Solving for G simply implies that we make G the subject of the formula:

Given equation:

F $_{g}$ = G $\frac{m_{1} m_{2} }{ r^{2} }$

To make G the subject of this expression follow these steps:

Multiply both sides of the equation by $r^{2}$

F $_{g}$ x $r^{2}$ = G $\frac{m_{1} m_{2} }{ r^{2} }$ x $r^{2}$

This gives:

F $_{g}$ $r^{2}$ = G $m_{1} m_{2}$

Multiply both sides by $\frac{1}{m_{1} m_{2} }$

F $_{g}$ $r^{2}$ x $\frac{1}{m_{1} m_{2} }$ = $\frac{1}{m_{1} m_{2} }$ x G $m_{1} m_{2}$

Therefore:

G = $\frac{F_{g} r^{2} }{m_{1} m_{2} }$

Learn more:

Solving formula brainly.com/question/2998489

#learnwithBrainly

6 0

3 years ago