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3 years ago

8

Humans first appeared during the _____.

2 answers:

Alinara [238K]3 years ago

8 0

<span>Humans first appeared during the Quaternary period. <span>The Quaternary or Neozoic period develops in the Cenozoic era, which extends from 2,588 million years ago to the present. The Quaternary is divided into two periods. The Pleistocene, the first and longest period of the period, was characterized by the cycles of glaciations and the appearance of the first humans (homos habilis); this period extended from 2,588 million years to 10,000 years BC The Holocene, the second epoch of the period, was characterized by the end of the glaciation and the emergence of human civilization; this period extends from 10,000 years BC to the present.</span></span>

sergeinik [125]3 years ago

6 0

your answer is Quaternary period

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A 500kg car is driving through a neighborhood at 20m/s. Then the car hits a speed bump which causes it to slow to 15m/s. What wa

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Based on the formula for calculating impulse, the impulse of the speed bump to the car is 2500 Ns.

<h3>What is the impulse of the speed bump?</h3>

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5 0

3 years ago

How much work is required to accelerate a proton from rest up to a speed of 0.993 c? Express your answer with the appropriate un

Keith_Richards [23]

Answer:

(a). The work done is 7001 MeV.

(b). The momentum of this proton is $4.20\times10^{8}\ kg-m/s$ .

Explanation:

Given that,

Speed = 0.993 c

We need to calculate the work done

Using work energy theorem

The work done is equal to the kinetic energy relative to the proton

$W=K.E$

$W=\dfrac{m_{p}c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-m_{p}c^2$

Put the value into the formula

$W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(\dfrac{0.993c}{c})^2}}-1.67\times10^{-27}\times(3\times10^{8})^2$

$W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(0.993)^2}}-1.67\times10^{-27}\times(3\times10^{8})^2$

$W=1.122\times10^{-9}\ J$

$W=7001\ MeV$

(b). We need to calculate the momentum of this proton

Using formula of momentum

$p=\dfrac{m_{0}v}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Put the value into the formula

$p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(\dfrac{0.993c}{c})^2}}$

$p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(0.993)^2}}$

$p=1.404\times10^{-26}c$

$p=4.20\times10^{8}\ kg-m/s$

Hence, (a). The work done is 7001 MeV.

(b). The momentum of this proton is $4.20\times10^{8}\ kg-m/s$ .

4 0

3 years ago