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White raven [17]
3 years ago
15

How do I explain the laws of motion when I am walking down the street?

Physics
2 answers:
nataly862011 [7]3 years ago
7 0

-- If the person to whom you're explaining them is alongside you, then
you should adjust your walkiing rate to match his, so that he remains
within earshot for the duration, and receives the benefit of the complete
explanation.

-- If you're presenting the explanation on the phone, then your walking
rate doesn't matter.  But you still must adjust it to avoid running out of
breath as you speak, and you must also maintain your own awareness
of your surroundings, so that your explanation is not inadvertently
truncated.


Darina [25.2K]3 years ago
4 0
Well one way that you could explain is well by just walking and just say that one of the laws is an object in motion tends to stay in motion. The you could punch the palm of your hand ad then say that and object will stop if hit with an unmeasured force. You could also drag your feet on the sidewalk and talk about how friction stops an object from moving.

I hope that this was helpful to you.
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If the magnitude of a charge is half as much as another charge, but the force experienced is the same, then the electric field s
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Answer:

the electric field strength of this charge is two times the strength of the other charge

Explanation:

Using the relationship between electric field and the charge, which is inversely proportionality. Let the the magnitude of the first charge be Q and the respective electric field be E. It implies that;

E1/E2 = Q2/Q1

E2 = E1 x Q1/Q2

      = E x Q/ (Q/2)

       = 2E

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3 years ago
A horizontal applied force of magnitude 25 N acts on a block sliding on a horizontal surface. The force of friction between the
kompoz [17]

' A ' and ' D ' are both correct statements.

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3 years ago
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<span>The correct answer is: Mechanical Energy

Explanation:
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3 years ago
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A ball is thrown vertically upward from the ground. Its distance in feet from the ground in t seconds is s equals negative 16 t
marishachu [46]

Answer:

t₁ = 3 s

Explanation:

In this exercise, the vertical displacement equation is not given

        y = 240 t + 16 t²

Where y is the displacement, 240 is the initial velocity and 16 is half the value of the acceleration

Let's replace

      864 = 240 t + 16 t²

Let's solve the second degree equation

    16 t² + 240 t - 864 = 0

Let's divide by 16

    t² + 15 t - 54 = 0

The solution of this equation is

     t = [-15 ± √(15 2 - 4 1 (-54)) ] / 2 1

     t = [-15 ±√(225 +216)] / 2

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We have two solutions.

     t₁ = [-15 +21] / 2

     t₁ = 3 s

     t₂ = -18 s

Since time cannot have negative values, the correct t₁ = 3s

4 0
3 years ago
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