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3 years ago

If Mars were 10 times closer to the Sun, then the Sun would attract Mars with

2 answers:

7 0

The velocity does not change greatly between Mars and Venus —explained Dong— but Venus's closer proximity to the sun boosts the density by almost a factor of 4.5. This would mean that atmosphere on Mars would be lost even more rapidly than at its current position.

Goshia [24]3 years ago

4 0

Answer:

Explanation:

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The airplane is flying with a constant velocity. Which force acting on the airplane below represents the friction from air resis

tankabanditka [31]

The correct answer to the question is C i.e C represents the friction from air resistance.

EXPLANATION:

Before coming into any conclusion, first we have to understand friction.

The friction is the opposing force which acts tangentially between two bodies in contact when there is a relative motion between them.

The air resistance is that frictional force which is provided by the air to the moving body through it. Hence, the friction from air resistance will be directed opposite to the motion of the body.

In the given diagram, the airplane is going horizontally. The force A acts in forward direction while force C acts in backward direction. The forces B and D are acting vertically. There is no motion in vertical direction. Hence, the net force of A and C will cause the airplane to move.

As the plane is moving along the direction of A, the frictional force must act along the direction of C.

8 0

3 years ago

Read 2 more answers

Emma is the sole earning member of her family. She wants to study further but is forced to give up on her dreams to support her

xz_007 [3.2K]

The theory that helps to explain Emma's actions will is Determinism. Option C is corect.

<h3>What is determinism?</h3>

The idea is that all of our actions are the result of some uncontrollable internal or external force or cause.

Similar to the scenario given above, Emma's decision to abandon her studies was driven by an external factor—the fact that she is the family's sole provider of income, over which she has no influence.

The theory that helps to explain Emma's actions will is Determinism.

Hence option C is corect.

To learn more about determinism refer;

brainly.com/question/13369636

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3 0

2 years ago

Read 2 more answers

Suppose that you are headed toward a plateau 4040 m high. If the angle of elevation to the top of the plateau is 2020degrees°,

Lelu [443]

Answer :

The distance is 109.89 m.

Explanation :

Given that,

Height = 40 m

Angle = 20°

We need to calculate the base of the plateau

Using formula of angle

$\tan\thata=\dfrac{h}{x}$

Where, h = height

x = base

$\theta$ = angle

Put the value into the formula

$\tan20^{\circ}=\dfrac{40}{x}$

$x=\dfrac{40}{tan20^{\circ}}$

$x =109.89\ m$

Hence, The distance is 109.89 m.

3 0

3 years ago

Please follow I will follow you

klio [65]

Answer:

aight

Explanation:

5 0

3 years ago

Read 2 more answers

12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L

Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

6 0

4 years ago