Ok so the mass of a proton is 0.000000000000000000000000001672622 how do I make this in scientific notation

Mathematics

1 answer:

Soloha48 [4]3 years ago

3 0

Answer:

1.7 x 10^-27

Step-by-step explanation

Move the decimal to the right until it's a single digit number, because you are moving it to the right, it would be a negative number. If you are able to round up then please do. Remember to check your significant figures if you're in chemistry.

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Fill in the blank to make the expression a perfect square. v^2 - 18v + ?

Oliga [24]

Answer: (v - 9)^2 - 81

Step-by-step explanation: Use the formula (b/2)^2 in order to create a new term to complete the square.

Hope this helps! :) ~Zane

5 0

3 years ago

A rocket is shot straight up into the air with an initial velocity of 500 ft per second and from a height of 20 feet above the g

iragen [17]

The height of the rocket above the ground after $t$ seconds is given by the equation : $H= -16t^2+Vt+h$ , where $V$ is the initial velocity and $h$ is the initial height.

Given that, $V= 500 ft/second$ and $h= 20 ft$

So, the equation will become: $H= -16t^2 +500t+20$

A) For finding the height of the rocket 3 seconds after the launch, we will plug $t=3$ into the above equation. So....

$H= -16(3)^2+500(3)+20\\ \\ H= -16(9)+1500+20\\ \\ H= -144+1500+20=1376$

So, the height of the rocket 3 seconds after the launch is 1376 feet.

B) When the rocket at a height of 400 feet, then we will plug $H= 400$

$400=-16t^2+500t+20\\ \\ 16t^2-500t-20+400=0\\ \\ 16t^2-500t+380=0\\ \\ 4(4t^2-125t+95)=0\\ \\ 4t^2-125t+95=0$

Using quadratic formula, we will get......

$t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(95)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{14105}}{8}\\ \\ t= 30.4705... \\ and \\ t= 0.7794...$

So, after 0.7794...seconds and 30.4705...seconds the rocket is at a height of 400 feet above the ground.

C) The time duration that the rocket remains in the air means we need to find the time taken by the rocket to reach the ground. When it reaches the ground, then $H=0$ . So.....