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Oduvanchick [21]
2 years ago
6

A 41.0 g marble moving at 2.30 m/s strikes a 25.0 g marble at rest. What is the speed of each marble immediately after the colli

sion? Note that the collision is elastic and that it is a "head-on" collision so all motion is along a line.
Physics
1 answer:
Gre4nikov [31]2 years ago
6 0

Answer:

speed of each marble after collision will be 1.728 m/sec

Explanation:

We have given mass of the marble m_1=41gram=0.041kg

Velocity of marble v_1=2.30m/sec

Its collides with other marble of mass 25 gram

So mass of other marble m_2=25gram=0.025kg

Second marble is at so v_2=0m/sec

We have to find the velocity of second marble

From momentum conservation we know that

m_1v_1+m_2v_2=(m_!+m_2)v, here v is common velocity of both marble after collision

So 0.041\times 2.30+0.025\times 0=(0.041+0.025)v

v = 1.428 m /sec

So speed of each marble after collision will be 1.728 m/sec

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Answer:

\bold {1s^22s^22p^63s^23p^64s^23d^2}

Explanation:

electronic configuration of an atom is the spatial arrangement of the electrons around the nucleus in the energy orbits

We know that the atomic number of Titanium is 22

The electronic configuration of Titanium is

1s^22s^22p^63s^23p^64s^23d^2

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3 years ago
n outer space, a constant net force with a magnitude of 140 N is exerted on a 32.5 kg probe initially at rest. A) What accelerat
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Answer:

a) 4.31 m/s²

b) 215.5 m

Explanation:

a) According to Newton's first law of motion

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m = 32.5 kg

a = ?

140 = 32.5 × a

a = 140/32.5 = 4.31 m/s²

b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s

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3 years ago
Frequency, period and wavelength<br> 11th grade high school physics
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2 years ago
Two identical small metal spheres with q1 &gt; 0 and |q1| &gt; |q2| attract each other with a force of magnitude 72.1 mN when se
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1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

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r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

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We can solve the equation for Q:

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Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

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