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ivanzaharov [21]

3 years ago

5

Summarize the Venn diagram.

1 answer:

Elena L [17]3 years ago

5 0

Answer:

If this is a multiple choice question, then the answer is D.

Explanation:

Compounds are composed of atoms, which are composed of subatomic particles and consist of matter. Since they are composed of atoms rather than vice-versa, compounds cannot be found inside atoms and are not the most basic form of matter (ruling out A and C). A pure substance contains atoms, which are each composed of subatomic particles. Therefore, a pure substance must have atoms if it contains subatomic particles (ruling out B). The only answer left is D.

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A weight lifter lifts a 345 N set of weights from ground level to a position over his head, a vertical distance of 1.89 m. How m

Pavlova-9 [17]

Answer:652.05 J

Explanation:

Given

Weight of lifter $W=345 N$

vertical distance move $h=1.89 m$

Work done in lifting the weight is equal change in Potential Energy of weight

Change in Potential Energy $=m g h$

$\Delta PE=345 \times 1.89=652.05 J$

therefore work done is equal to $652.05 J$

3 0

3 years ago

A cue ball of mass m1 = 0.325 kg is shot at another billiard ball, with mass m2 = 0.59 kg, which is at rest. The cue ball has an

Roman55 [17]

Answer:

$v_{2f} = \frac{2vm_1}{m_2 + m_1}$

Explanation:

If the collision is elastic and exactly head-on, then we can use the law of momentum conservation for the motion of the 2 balls

Before the collision

$P_i = m_1v$

After the collision

$P_f = m_1v_{1f} + m_2v_{2f}$

So using the law of momentum conservation

$P_i = P_f$

$m_1v = m_1v_{1f} + m_2v_{2f}$

We can solve for the speed of ball 1 post collision in terms of others:

$v_{1f} = v - v_{2f}\frac{m_2}{m_1}$

Their kinetic energy is also conserved before and after collision

$m_1v^2/2 = m_1v_{1f}^2/2 + m_2v_{2f}^2/2$

$m_1v^2 = m_1v_{1f}^2 + m_2v_{2f}^2$

From here we can plug in $v_{1f} = v - v_{2f}\frac{m_2}{m_1}$

$m_1v^2 = m_1\left(v - v_{2f}\frac{m_2}{m_1}\right)^2 + m_2v_{2f}^2$

$m_1v^2 = m_1\left(v^2 - 2vv_{2f}\frac{m_2}{m_1} + v_{2f}^2\frac{m_2^2}{m_1^2}\right) + m_2v_{2f}^2$

$m_1v^2 = m_1v^2 - 2vv_{2f}m_2 + v_{2f}^2\frac{m_2^2}{m_1} + m_2v_{2f}^2$

$v_{2f}^2(m_2 + \frac{m_2^2}{m_1}) - 2vm_2v_{2f} = 0$

$v_{2f}(1 + \frac{m_2}{m_1}) = 2v$

$v_{2f} = \frac{2v}{1 + \frac{m_2}{m_1}} = \frac{2v}{\frac{m_1 + m_2}{m_1}} = \frac{2vm_1}{m_2 + m_1}$

8 0

3 years ago

Read 2 more answers

How to find gravitational potential energy?

Nikitich [7]

Mass x height x gravity

4 0

3 years ago

Please help im being timed on this and need to get my grade up

nadya68 [22]

Answer:

a I think hope this helps

8 0

2 years ago

A circuit is constructed with six resistors and two batteries as shown. the battery voltages are v1 = 18 v and v2 = 12 v. the po

VladimirAG [237]

Answer:

V4=9.197v

Explanation:

Given:

V1= 18v ,V2= 12v ,r1=r5=58ohms ,r2=r6=124ohms , r3=47ohms ,r4= 125ohms

V4= I4R4 = V2/(R4 + R5)×R4

V4= 12×125 /(125 + 58)

V4=1500/183 =9.197v

5 0

3 years ago