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3 years ago

PLEASE HELP :)

1 answer:

6 0

1) in the opposite direction always
2) drag force is the force that acts in the opposite direction of the applied force (air resistance if a drag force for example)
3) frictions forces between the sole and the ground as the owner walks
4) chalk would glide very easily, so a creaking sound could never be produced, but also one could not write anything on the board!
5) I apologize for my lack of familiarity with the kind of sport but the probable answer is to raise the friction of their hands in order to decrease slipping and get a tighter, more firm grip

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Answer:

It tells us that the element has 6 electrons in it's shells

Explanation:

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3 years ago

What is the force required to stop a 20 kg man traveling at 12.3 m/s East in 0.9 s? *Round to TWO decimal points and include you

abruzzese [7]

F = ma

We have mass = 20kg

And we need to solve for acceleration

So acceleration is change in velocity over time, in this case we have one velocity and we can assume the man started from rest so

12.3 / 0.9 = a

a = 13.6667

Now we can plug that into F = ma

F = (20)(13.6667)

F = 273.334

Rounding

F = 273.33

Now he is traveling east so we need a force towards the rest, or in the opposite direction to stop his motion.

If we assume east is the positive direction then we need a force of

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4 years ago

Assuming there are no accidents or delays, the distance that a car travels down the interstate

Ganezh [65]

Explanation:

The distance that a car travels down the interstate can be calculated with the following formula:

Distance = Speed x Time

(A) Speed of the car, v = 70 miles per hour = 31.29 m/s

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D = 31.29 m/s × 21600 s

D = 675864 meters

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3 years ago

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Snowcat [4.5K]

Answer:

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3 years ago

A skateboarder is trying to skate horizontally off a building and jump over a pool as shown! The pool is 25.0m wide and the roof

umka21 [38]

Answer:

a) t = 2.55s

b) $v_{0x} = 9.80 m/s$

c) yes

Explanation:

In order to solve this problem, we can start by drawing a sketch of the situation so we can better visualize what the problem is about (see attached picture).

For the first question. We are talking about a movement in two dimensions. So on the first question they are asking us for vertical movement. It will be uniformly accelerated, so we can use the following formula:

$y_{f}=y_{0}+V_{0y}t+\frac{1}{2}at^{2}$

We know the following:

$y_{f}=0$

$y_{0} = 32m$

$V_{0y}=0$

t=?

$a=-9.81 m/s ^{2}$

With this data, we can simplify our equation, so we end up with:

$y_{0}+\frac{1}{2}at^{2}=0$

so we can now substitute the data we know and solve for t:

$32m-\frac{1}{2}(9.81 m/s^{2})t^{2}=0$

$(-4.905 m/s^{2})t^{2}=-32m$

$t^{2} = \frac{-32m}{-4.905 m/s^{2}}$

$t=\sqrt{6.52s^{2}}$

t = 2.55 s

For part b, since we are talking about horizontal movement and we are neglecting drag, this means that his horizontal speed will be constant. So we can use the following formula:

$V_{x}=\frac{x}{t}$