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2 years ago

7

You move to Mars and as a momento from Earth take your Ma-maw's mercury barometer. You place it outside in the Martian atmospher

e and read a very low pressure. Does the pressure reading accurately reflect the atmospheric on the surface of Mars

1 answer:

oksano4ka [1.4K]2 years ago

8 0

Answer:

Explanation:

No, the reading is not expected to be accurate. This is because Relative to Earth, the air on Mars is extremely thin. The Martian atmosphere is primarily carbon dioxide with a much lower surface pressure, and Mars does not have oceans and an Earthlike hydrological cycle so latent heat release is not as important as it is for Earth.

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Calculate the kinetic energy of a dog,

Vanyuwa [196]

B: 20 j

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Please need help on this one

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The second answer from the top, no the energy in the wave pushed the water particles from above the earthquake in the opposite direction.

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3 years ago

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A satellite with a mass of 110 kg and a kinetic energy of 3.0 ´ 109 J must be moving at a speed of

djverab [1.8K]

Answer: A satellite with a mass of 110 kg and a kinetic energy of 3.08×10^9 J must be moving at a speed of 7483 m/s.

Explanation: To find the answer we need to know about the kinetic energy of a body.

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We have the expression for kinetic energy of a body as,

$KE=\frac{1}{2}mv^2$

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$m=110kg\\KE=3.08*10^9J\\$

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$v=\sqrt{\frac{2KE}{m} } =\sqrt{\frac{2*3.08*10^9}{110} } =7.483*10^3 m/s$

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2 years ago

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A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

a

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

b

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$

$F_{31 +32} = 2.16 *10^{-5} (4i + 3j) - 12*10^{-5} j$

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of $F_{31 +32}$ is mathematically evaluated as

$|F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}$

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction is obtained as

$tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}$

$\theta = tan ^{-1} [-0.63889]$

$\theta = - 32.57 ^o$

$\theta = 360 - 32.57$

$\theta =327.43 ^o$

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2 years ago