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Goryan [66]
2 years ago
15

What is the central idea for stacking rocks vertically to test the effects of forces

Physics
1 answer:
otez555 [7]2 years ago
8 0

Answer:

You can see how forces can stack on one another by comparing the normal force of the bottom most rock to the normal force of the upper most rock.

Explanation:

Say you have 3 rocks all with mass m=3kg

These 3 rocks will all have the gravitational force of -29.4N (multiply 3kg by -9.8m/s^2)

If none of the rocks are stacked, then they all have the same normal force as well (29.4N)

However, stack them, and see how their normal forces change.

m1 will continue to have a normal force of 29.4N

m2 will now have a normal force of 58.8

m3 will now have a normal force of 88.2

Though they all retain the same gravitational force, their normal force changes depending on the forces acting down on them.

Mass 1 have no forces acting downwards other than gravity.

Mass 2 has the force of mass 1 and gravity acting down on it.

Mass 3 has the force of mass 1, mass 2, and gravity acting down on it.

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What is frequency? <br><br> (P.S. please help)
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Explanation:

the rate at which something occurs or is repeated over a particular period of time or in a given sample

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This one too dude help me
Mrac [35]

Answer:

Vas happenin!!

atomic number: 5

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Number of protons: 5

Number of neutrons: 5

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Hope this helps!

-Zayn Malik

Explanation:

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3 years ago
5. A man has a weight of 100 Newtons. How much work is done if he climbs 4 meters up a ladder? Plug numbers under the equation.
Virty [35]

Answer:

<h2>400 J</h2>

Explanation:

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workdone = force × distance

From the question we have

workdone = 100 × 4

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<h3>400 J</h3>

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5 0
3 years ago
A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car ent
Paha777 [63]

Answer:

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

a2 = 0.8712 m/s²

3 0
3 years ago
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