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3 years ago

What causes the tail of a comet to become visible?

Physics

2 answers:

nlexa [21]3 years ago

5 0

Answer:

The correct answer is option C. Sun.

Explanation:

The tail of a comet as well as the comma are visible thanks to the influence of sunlight on these objects.

If there is a case that these objects cross the inner Solar System, then they will be visible from Earth. The reason why this occurs is that as the comet approaches the internal solar system, the materials that are inside the comet vaporize and flow out of the nucleus thanks to solar radiation, and while they do carry dust with them.

This dust is the one that reflects sunlight directly, while the gases shine because of ionization.

Usually we will need a telescope to see a comet, although there are some that can be seen with the naked eye.

zavuch27 [327]3 years ago

4 0

Answer:

it is the sun

Explanation:

i just did the test

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Buoyancy increases with the increase in the density of a) Submerged body b) Fluid

AlekseyPX

Answer:

Submerged body

Explanation:

If buoyancy is greater than weight then object will float.
If buoyancy is less then weight then object will sink.
If buoyancy=weight then objects remains stable

4 0

3 years ago

Given the height of rod,length of shadow of tree and length of shadow of the rod, estimate the height of the tree? Given:height

diamong [38]

Answer:

1000 cm.

Explanation:

To obtain the estimated tree height :

(Height of rod / length of rod shadow) = (height of tree / length of tree shadow)

Substituting values into the formula :

(150cm / 120 cm) = (height of tree / 800 cm)

Using cross multiplication :

Height of tree * 120 = 150 * 800

Height of tree = (150 * 800) / 120

Height of tree = 120,000 / 120

Height of tree = 1000

Hence, estimate height of tree = 1000 cm

5 0

2 years ago

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

anastassius [24]

Explanation:

Forces on Block A:

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 $m_B$

Forces on Block B:

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 N 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 T 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

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

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

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, 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 $N = 50.9\:\text{N}$