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3 years ago

(6.0210^23) (8.6510^4)

1 answer:

5 0

To solve this problem with have to follow the order of operations. (PEMDAS) Parenthesis, Exponents, Multiply, Divide, Add, Subtract.

We'll start with the first set of parenthesis.

(6.02*10^23)

10^23 = 1e+23 (1 followed by 23 zeroes)

6.02*1e+23 = 6.02

Now we'll solve the second set of parenthesis.

(8.65*10^4)

10^4 = 10,000

8.65 * 10,000 = 86,500

Now onto the final step. We multiply together both numbers.

6.02 * 86,500 = 520,730<span>
</span>

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A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/

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At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

$\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}$ .

At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

In other words, at $t = 20\; {\rm s}$ , the truck was $400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}$ ahead of the car. The velocity of the car is greater than that of the truck by $35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}$ . It would take another $(50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s}$ before the car catches up with the truck.

Hence, the car would catch up with the truck at $t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}$ .

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3 years ago

(6.02*10^23) (8.65*10^4)

(6.0210^23) (8.6510^4)