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3 years ago

The mass of an electron is 9.11 x 10-31 kg. If you want to increase the speed of the

1 answer:

5 0

I'm not sure of this question, yet i know you may get angry but, just know your loved and if going through a hard time i'm so so sorry, i know how it feels it sucks. take care of yourself, drink water, listen to that one song that makes you dance in your room and go do what YOU love. your beautiful your perfect and dont listen to what anyone says. i love u, everyone loves u, i know its scary but it happens to everyone life isnt a peice of cake unfortunatly and it sucks but woahh your strong! if you need someone to talk to leave a comment below, happy hoildays everyone!

P.S, im looking into the answer of this question. I may not answer it right on time (since i'm not the smartest...) but i promise ill try my best! Also, i am not doing this for points im doing this to make others smile. I hate how no one checks up on anyone, lots of people are going through a stressful time (including me) and i just want to let everyone know that its okay not to be okay! I hope that you all find happiness in every possible situation, once again HAPPY HOLIDAYS!

You might be interested in

The tallest sequoia sempervirens tree in California’s redwood national parks is 111 m tall. Suppose an object is thrown downward

ch4aika [34]

The object's <u>initial velocity</u> is equal to $-10.59\frac{m}{s}$

Why?

From the statement we know the height of the tree and the time it takes to reach the ground, so, if we need to calculate its initial velocity, we can use the following formula:

$y=y_o-v_{o}*t-\frac{1}{2}g*t^{2}$

Where,

y, is the final height (0 meters in this case)

yo, is the initial height (111 meters in this case)

t, is the time elapsed (3.8 seconds in this case)

vo, is the initial speed.

g, is the acceleration due to gravity (-9.81 m/s2)

Now, let's set the origin at the top of the tree, so, rewriting the formula, we have:

$y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}$

So, isolating the initial velocity, we have:

$y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}$

$y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}\\\\v_{o}*t=y-y_o-\frac{1}{2}g*t^{2}\\\\v_{o}=\frac{y-y_o-\frac{1}{2}g*t^{2}}{t}$

Finally, substituting and calculating, we have:

$v_{o}=\frac{-111m-0-\frac{1}{2}(-9.8\frac{m}{s^{2}}) *(3.8s)^{2})}{3.8s}\\\\v_{o}=\frac{-111m-\frac{1}{2}(-9.8\frac{m}{s^{2}})*(14.44s^{2})}{3.8s}\\\\v_{o}=\frac{-111m-\frac{1}{2}(-141.51m)}{3.8s}=\frac{-111m+70.75m}{3.8s}\\\\v_{o}=\frac{-40.25m}{3.8s}=-10.59\frac{m}{s}$

Hence, we have that the <u>initial velocity</u> of the object is $-10.59\frac{m}{s}$

Have a nice day!

7 0

4 years ago

What is the relevant similarity in the following argument?

Tasya [4]

Your answer would be C

7 0

3 years ago

An astronaut is moving in space when a big explosion occurs about 50 meters behind him. How will the astronaut come to know abou

LUCKY_DIMON [66]

Answer:

The correct answer is B.

The astronaut will know due to the light from the explosion.

Explanation:

Sound and vibrations require a medium such as air to travel through. Space, there is no air. Only a vacuum. So sound and vibrations are unable to travel. Light requires no medium to travel. It can go through a vacuum.

Therefore the Astronaut will see a bright flash of light as it travels from the explosion to outer space. It is also important to note that light can travel very far because nothing else interacts with its wave particles and as such, it cannot be impeded.

Cheers!

7 0

3 years ago

Which of these observations would show a chemical

11Alexandr11 [23.1K]

Answer: I'm pretty sure it'd be that lime juice burns your tongue

also where is choices a and b??

6 0

4 years ago

A horizontal line on a distance-time graph means the object is

IgorLugansk [536]

At rest because if the distance is not changing, then it is not moving any further, so it must not be moving! The time keeps going no matter what, so the distance, whether it is 0 m or 10,000 km, if the y is horizontal the distance does not change.

6 0

4 years ago