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KIM [24]
3 years ago
9

Walk in a straight line. Now stop. Did you accelerate? Explain

Physics
2 answers:
sasho [114]3 years ago
7 0
YES. If you where not moving and then suddenly started moving forward to walk in the straight line you did indeed accelerate at one point in your journey. And then when you stopped you where decelerating
Brums [2.3K]3 years ago
4 0

Answer;

It's about acceleration, right?

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Answer:

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Explanation:

gagadsgadgagdagadagfdagdgadgaag

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) The centre of a mass, m, is at a distance, r, from the centre of a larger mass M. Assuming there are no other masses present,
ivanzaharov [21]

Answer:

U = - G m M / r

Explanation:

The gravitational potential energy is given by the expression

         U = - G m₁ m₂ / r

dodne G is the gravitational cosntnate (G = 6.67 10⁻¹¹¹), m and m are the mass of the bodies involved

subtype the given values

         U = - G m M / r

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4 years ago
What effect does the Sun have on the ocean tides?
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Answer:

B

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3 years ago
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Answer:

A) Burning fossil fuels pollutes the environment

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3 years ago
Read 2 more answers
(a) If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you ca
Colt1911 [192]

a) 52.5 m

b) 16.0 m/s

Explanation:

a)

The motion of a car slowed down by friction is a uniformly accelerated motion, so we can use the following suvat equation:

v^2-u^2=2as

where

v = 0 is the final velocity (the car comes to a stop)

u = 28.7 m/s is the initial velocity of the car

a is the acceleration

s is the stopping distance

For a car acted upon the force of friction, the acceleration is given by the ratio between the force of friction and the mass of the car, so:

a=\frac{-\mu mg}{m}=-\mu g

where:

\mu=0.80 is the coefficient of friction

g=9.8 m/s^2 is the acceleration due to gravity

Substituting and solving for s, we find:

s=\frac{v^2-u^2}{-2\mu g}=\frac{0-(28.7)^2}{-2(0.80)(9.8)}=52.5 m

b)

In this case, the car is moving on a wet road. Therefore, the coefficient of kinetic friction is

\mu=0.25

Here we want the stopping distance of the car to remain the same as part a), so

s=52.5 m

We can use again the same suvat equation:

v^2-u^2=2as

And since the final velocity is zero

u = 0

We can find the initial velocity of the car:

v=\sqrt{2as}=\sqrt{2\mu gs}=\sqrt{2(0.25)(9.8)(52.5)}=16.0 m/s

7 0
3 years ago
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