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3 years ago

Students make the claim that Jupiter is the warmest gaseous planet.

Physics

2 answers:

mart [117]3 years ago

5 0

Answer:D. Jupiter is the closest outer planet to the Sun.

Anuta_ua [19.1K]3 years ago

5 0

$\tt \: \pink {Option \: D}$

$:\implies$ Jupiter is the closest outer planet to the Sun.

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Can someone help me please

makvit [3.9K]

Answer:

Time, I believe. Pretty sure it's time lol

3 0

2 years ago

Two uncharged, conducting spheres are separated by a distance d. When charge −Q is moved from sphere A to sphere B, the Coulomb

rosijanka [135]

Answer: a) the force will be repulsive

b) the ratio of the new force to the old force will be 2

c) O

Explanation:

a) since charge -Q is moved from A to B, this implies that sphere A is negatively charged. The two spheres are now negatively charged and will repel themselves.

b) initial force will be -q(-Q)/d2

Adding extra charge -Q will cause change on B to become -2Q

The new force will be - 2Q(-q)/d2

Dividing new force by old force will give 2

C) if B is neutralized, the net charge becomes 0 and there will be no force on it.

3 0

2 years ago

How does the pupillary response prevent injury? What would happen without it?

Aleksandr [31]

Pupils dilate and constrict in order to allow an adequate amount of light to pass through the retina and vision. If there is not enough light and the pupils do not dilate, a small amount of light will pass to the retina and the vision will be damaged.

6 0

3 years ago

Read 2 more answers

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

Dmitry [639]

Answer:

Part a)

$F = 135.7 N$

Part b)

$F = 62.5 N$

Explanation:

Part a)

If block is sliding up then net force must be zero and friction will be in opposite to the direction of motion of the block

$Fcos\theta = mg + F_f$

$Fsin\theta = F_n$

so we have

$Fcos\theta = mg + \mu(Fsin\theta)$

$F(cos\theta - \mu sin\theta) = mg$

$F = \frac{mg}{cos\theta - \mu sin\theta}$

$F = \frac{55}{cos50 - 0.310(sin50)}$

$F = 135.7 N$

Part b)

If block is sliding down then net force must be zero and friction will be in opposite to the direction of motion of the block

$Fcos\theta = mg - F_f$

$Fsin\theta = F_n$

so we have

$Fcos\theta = mg - \mu(Fsin\theta)$

$F(cos\theta + \mu sin\theta) = mg$

$F = \frac{mg}{cos\theta + \mu sin\theta}$

$F = \frac{55}{cos50 + 0.310(sin50)}$

$F = 62.5 N$

6 0

3 years ago

Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$

8 0

3 years ago