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den301095 [7]
3 years ago
7

Ceres, Pluto, and Eris are all round in shape and classified as:_________ A) Leftover planetesimals that formed inside the frost

line are known as . B) Pieces of asteroids that have fallen to Earth are called . C) The objects that are sometimes visible in the night sky with long tails are . D) Leftover planetesimals that formed in the region of the solar system now occupied by the jovian planets are called . E) Meteor showers are associated with debris from
Physics
1 answer:
motikmotik3 years ago
5 0

Answer

Ceres, Pluto, and Eris are classified as DWARF PLANET.

A) Leftover planetesimals inside the frost line are known as ASTEROIDS.

B) METEORITES are the pieces of Asteroids which are fallen on the earth's surface.                                                

C) COMETS are the objects which are visible with long tails.            

D) COMETS are also the leftover planetesimals that are occupied by the jovian planets and are formed in the solar system.                

E) Meteor showers are associated with debris from COMETS                      

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Tim’s cow is anemic. The cow is lacking which type of nutrient?
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Answer:

Iron deficiency

Explanation:

or more scientifically explained as decreased hemoglobin levels in your blood but still caused by lack of iron.

6 0
3 years ago
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Velocity is speed in a certain direction.<br><br> True<br> False
lara [203]
True if you look up the question Is velocity speed in a certain direction you would’ve gotten the answer but I’m pretty sure it’s true
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The head of a rattlesnake can accelerate at 49 m/s2 in striking a victim. If a car could do as well, how long would it take to r
Anettt [7]
<h2>Time taken is 0.459 seconds</h2>

Explanation:

We have equation of motion v = u + at

     Initial velocity, u = 0 m/s

     Final velocity, v = 81 km/hr = 22.5 m/s    

     Time, t = ?

     Acceleration, a = 49 m/s²

     Substituting

                      v = u + at  

                      22.5 = 0 + 49 x t

                      t = 0.459 seconds

   Time taken is 0.459 seconds

3 0
3 years ago
An L-R-C series circuit, R = 160 Ω , L = 0.790 H , and C = 1.30×10−2 μF . The source has a voltage amplitude of 140 V and a freq
wolverine [178]

Answer: a) 1 b) 61 W c) 61 W

Explanation:

a) The  Power Factor (also known as cos φ), is defined by the difference in phase between current and voltage, in a RLC series circuit, and is expressed as follows:

cos φ = R / Z = R / \sqrt{(R)^{2} + (Xl -Xc)^{2} }

In resonance, XL =XC, so the circuit behaves like it were purely resistive, so Z=R.

Replacing in the expression for power factor, we have:

cos φ = R/Z = R/R = 1

This means that in resonance, current and voltage are in phase each other.

b) The average power delivered by the source, in resonance, is simply the power dissipated at the resistance R, as follows:

Pavg = V² rms / R = V₀² / √2 / R = 61 W

c) If the circuit remains in resonance, the average power , which does not depends on frequency provided this condition remains, keeps the same, i.e. , 61 W.

7 0
3 years ago
A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur
morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\

\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] =  4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\

Substitute the given parameters and solve for the angular speed;

\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\  \ +\  \ 0.35(0.02^2\  + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

5 0
2 years ago
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