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3 years ago

13

Mercury, Mars, and Earth’s moon all show evidence of heavy cratering from meteor impacts while Earth shows much less evidence of

meteor impacts. The most reasonable conclusion to make from this data is that the early solar system had a much higher density of rocky material, but–
a) Earth has been resurfaced by ongoing processes.
b) the rocky material was less likely to reach Earth.
c) Earth formed much later, when the solar system had fewer meteors.
d) the protection from impact provided by Earth’s atmosphere is decreasing.

1 answer:

algol [13]3 years ago

4 0

Answer:

D)

Explanation:

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Which of the following is an example of Newton's Third Law?* O A stack of pennies will not move unless you flick them over. O Fa

Morgarella [4.7K]

Answer:

A ball hits the ground and the ground pushes up on it

Explanation:

Newton's third law basically states that for every action, there's a reaction.

a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.

Hope this Helps!!! :)

5 0

3 years ago

A long straight cylindrical shell has an inner radius R_i and an outer radius R_0. It carries a current i, uniformly distributed

Irina-Kira [14]

Answer:

Correct answer is option D

- Wire is on the cylinder axis and carries current i in the direction opposite to that of the current in the shell

Explanation:

- It cannot be Option E, because the magnetic field outside the wire would not be 0 due to the current carried by the conductor

-Also, the parallel wire cannot carry current in the same direction because, that would amplify the magnetic field created by the outer cylinder (since B is dir. proportional to the current) -and now, that leaves only option C and D. If, it is Option C, then that means one side of the cylinder would be more closer to the parallel wire than the other, so there would be different B fields on the two opposite sides of the cylinder. So, that means the answer is option D.

7 0

3 years ago

~~~NEED HELP ASAP~~~<br>Please solve each section and show all work for each section.

anastassius [24]

Explanation:

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for <em>N</em> as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

or

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension <em>T</em>, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

3 years ago

An airplane starts from A and flies to B at a constant speed. After reaching B it returns to A at the same speed. There was no w

Dafna1 [17]

Answer:

When there is wind it takes longer

Explanation:

With no wind, the round trip time is

$t_1=\frac{d}{v}+\frac{d}{v}=\frac{2d}{v}$

When we have a constant wind speed w

$t_2=\frac{d}{v-w} +\frac{d}{v+w} =\frac{2vd}{v^{2} -w^{2}}$

comparing the reciprocal times;

$\frac{1}{t_2}=\frac{v^{2}-w^{2} }{2vd}=\frac{v}{2d}-\frac{w^{2}}{2vd} \leq \frac{v}{2d}=\frac{1}{t_1}$

This means that t1 is smaller than t2, ergo, it takes longer with wind

4 0

2 years ago

Hector drew the diagram below to illustrate projectile motion.

Serjik [45]

<u>Answer </u>

He should make the arrow for “Path” curve downward.

<u>Explanation </u>

The force of gravity is usually directly downward. So from the diagram, it is correctly labelled.

If the object was given a horizontal force, the direction of the inertial is also correct. Inertial is the force the resist the change of state of motion.

What Hector should change is the path followed by the object. It will be a curve not a straight line as it is drawn.

The correct answer is He should make the arrow for “Path” curve downward.

6 0

3 years ago

Read 2 more answers