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tigry1 [53]
2 years ago
15

Why is it important on diagrams to represent a force with an arrow? Why should mass NOT be represented by an arrow?

Physics
1 answer:
Cloud [144]2 years ago
6 0

It is important on diagrams to represent a force with an arrow because force is a vector quantity but it is not important to represent a mass since it is scalar quantity.

<h3>Representing a force with arrow and not mass</h3>

Vector quantities are always represented with arrows because they have magnitude and direction.

Scalar quantities have only magnitude, and cannot be represented with arrows.

Thus,  it is important on diagrams to represent a force with an arrow because force is a vector quantity but it is not important to represent a mass since it is scalar quantity.

Learn more about vector quantities here: brainly.com/question/26044328

#SPJ1

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Suppose you have 20% of 20/100 or 1/5. Describe how you can solve the problem using 1/5 instead of 20%.
grin007 [14]

Answer:

To calculate Percent in fraction You need to do following steps.

Divide number by 100.

=20/100

=1/5

6 0
3 years ago
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If an object has a large momentum, how easy is it to stop, and why?
podryga [215]
It is dificult to stop an object with a lot of momentum becuase of inertia with means that a moving object will want to keep moving and a stationary object wil want to stay stationary and the larger the momentum of the moving object the harder it will be to stop
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1) When making a digital animation of a person running on a sidewalk in a scene, which parameter would be an initial condition?
lbvjy [14]

Answer:

The position of the person's feet. a running shoe company studying how different surfaces affect the life of a shoe tread

Explanation:

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3 years ago
A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly
Mnenie [13.5K]

Answer:

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The acceleration of the tortoise is 0.28 m/s²

Explanation:

The equations that describe the position and velocity of the hare and the tortoise are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

To find the acceleration of the hare once it begins to slow down, we have to find how much time the hare traveled during the deceleration and what was its initial speed.

First, the hare moves with constant acceleration for 4.7 s. Then, its velocity at  t = 4.7 s will be:

v = v0 + a · t    (v0 = 0 because the hare starts form rest)

v = a · t = 0.9 m/s² · 4.7 s = <u>4.2 m/s</u>

<u />

The distance traveled by the hare while accelerating can be calculated using the equation of the position:

x = x0 + v0 · t + 1/2 · a · t²      (x0 = 0 and v0 = 0)

x = 1/2 · a · t² = 1/2 · 0.9 m/s² · (4.7)² = <u>9.9 m</u>

<u />

Then, the hare runs at a constant speed of 4.2 m/s for 11.7 s. The distance traveled at constant speed will be:

x =  v · t

x = 4.2 m/s · 11.7 s = <u>49.1 m</u>

<u />

Then, the distance traveled by the hare while slowing down was:

Distance traveled while slowing down = 72 m - 49.1 m - 9.9 m = 13 m

Let´s find how much time it took the hare to come to stop, so we can calculate the acceleration. We know that when the position is 13 m, the velocity is 0.

v = v0 + a · t

0 = 4.2 m/s + a · t

-4.2 m/s / t = a

Replacing in the equation of the position:

x = v0 · t + 1/2 · a · t²      (considering x0 as the point at which the hare started to slow down)

13 m = 4.2 m/s · t - 1/2 · 4.2 m/s / t · t²

13 m = 4.2 m/s · t - 2.1 m/s · t

13 m = 2.1 m/s · t

t = 13 m / 2.1 m/s

t = 6.2 s

Then, the acceleration of the hare while slowing down will be:

-v0/t = a

-4.2 m/s / 6.2 s = a

a = -0.68 m/s²

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The hare traveled 72 m in (6.2 s + 11.7 s + 4.7 s) 22.6 s. The tortoise reaches the final position of the hare at the same time, so, using the equation of the position we can calculate the acceleration of the tortoise:

x = x0 + v0 · t + 1/2 · a · t²     (x0 = 0 and v0 = 0)

x = 1/2 · a · t²

72 m = 1/2 · a · (22.6 s)²

144 m / (22.6 s)² = a

a = 0.28 m/s²

The acceleration of the tortoise is 0.28 m/s²

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An object of mass m is initially at rest and free to move without friction in any direction in the xy-plane. A constant net forc
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Answer:3/5

Explanation:

6/6

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