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4 years ago

14

HELP MEEEEE!!!!!!!!!!A student make a drawing of an aluminum atom.Which of these should the student show in the nucleus of the a

tom?

1 answer:

TEA [102]4 years ago

8 0

I would say B. Neutrons

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Usain Bolt ran the 100m dash in 9.69 seconds. How fast does he run in the 100m dash?

Alla [95]

The answer is in the question, it’s 9.69

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3 years ago

Explain how the density of a gas is different from the density of a liquid.

zavuch27 [327]

Answer

The same number of particles in a gas spread further apart than in the liquid or solid states.

Explanation:

The same number of particles in a gas spread further apart than in the liquid or solid states. The same mass takes up a bigger volume. This means the gas is less dense. Density also depends on the material.

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3 years ago

A rock with a mass of 1000 kg falls to the earth. what would the force be?

andrew11 [14]

F=ma = 1000kg * 9.8 m/s^2 = 9800 N

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3 years ago

I need help with #40 need answer and show me what you multiplied please

ale4655 [162]

(0.5)×(0squared)×(3)=(1.5j)

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3 years ago

A 81 kg man is riding on a 40 kg cart traveling at a speed of 2.3 m/s. He jumps off with zero horizontal speed relative to the g

Alexus [3.1K]

Answer:

$\Delta v= 4.66\frac{m}{s}$

Explanation:

In this case we have to use the Principle of conservation of Momentum:

<em>This principle says that in a system the total momentum is constant if no external forces act in the system. The formula is:</em>

$m_1v_1+m_2v_2=m_1u_1+m_2u_2$

<em>Where:</em>

$m_1:$ Mass of the first object.

$m_2:$ Mass of the second object.

$v_1:$ Initial velocity of the first object.

$v_2:$ Initial velocity of the second object.

$u_1:$ Final velocity of the first object.

$u_2:$ Final velocity of the second object.

In <u>this problem</u> we have:

$m_1=81kg\\m_2=40kg\\v_1_2=2.3\frac{m}{s}$

$u_1=0\frac{m}{s}$

Observation: $v_1_2:$ Is because the system has the same initial velocity.

First we have to find $u_2$ ,

$m_1v_1+m_2v_2=m_1u_1+m_2u_2$

We can rewrite it as:

$(m_1+m_2)v_1_2=m_1u_1+m_2u_2$

Replacing with the data:

$(m_1+m_2)v_1_2=m_1u_1+m_2u_2\\\\(81kg+40kg)2.3\frac{m}{s}=81kg(0\frac{m}{s})+40kg(u_2)\\\\(121kg)2.3\frac{m}{s}=40kg(u_2)\\\\\frac{(121kg)2.3\frac{m}{s}}{40kg}=u_2\\\\\frac{278.3}{40}\frac{m}{s}=u_2\\\\6.96\frac{m}{s}=u_2$

We found the final velocity of the cart, but the problem asks for the resulting change in the cart speed, this means:

$\Delta v=u_2-v_2\\\Delta v=6.96\frac{m}{s}-2.3\frac{m}{s}\\\Delta v= 4.66\frac{m}{s}$

Then, the resulting change in the cart speed is:

$\Delta v= 4.66\frac{m}{s}$

5 0

3 years ago