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sergij07 [2.7K]
2 years ago
8

In the levers lab when you changed the position of the fulcrum, what else changed?

Physics
1 answer:
dimulka [17.4K]2 years ago
3 0
A. Weight
because take an example of a lever, say you have a ball on a lever. if you roll the ball to the left, the WEIGHT will bring down the opposite end.
hoped this helped :)
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3. A sprinter leaves the starting blocks with an acceleration of 4.5 m/s2. What is the
UkoKoshka [18]

Hi there! :)

\large\boxed{v_{f} = 18 m/s}

Use the following kinematic equation to solve for the final velocity:

v_{f} = v_{i} + at

In this instance, the runner started from rest, so the initial velocity is 0 m/s. We can rewrite the equation as:

v_{f} = at

Plug in the given acceleration and time:

v_{f} = 4.5 * 4 = 18 m/s

5 0
3 years ago
A force of 20N changes the position of a body. If mass of the body is 2kg, find the acceleration produced in the body.2. A ball
shepuryov [24]

Explanation:

<em>Hello</em><em> </em><em>there</em><em>!</em><em>!</em><em>!</em>

<em>You</em><em> </em><em>just</em><em> </em><em>need</em><em> </em><em>to</em><em> </em><em>use</em><em> </em><em>simple</em><em> </em><em>formula</em><em> </em><em>for</em><em> </em><em>force</em><em> </em><em>and</em><em> </em><em>momentum</em><em>, </em>

<em>F</em><em>=</em><em> </em><em>m.a</em>

<em>and</em><em> </em><em>momentum</em><em> </em><em>(</em><em>p</em><em>)</em><em>=</em><em> </em><em>m.v</em>

<em>where</em><em> </em><em>m</em><em>=</em><em> </em><em>mass</em>

<em>v</em><em>=</em><em> </em><em>velocity</em><em>.</em>

<em>a</em><em>=</em><em> </em><em>acceleration</em><em> </em><em>.</em>

<em>And</em><em> </em><em>the</em><em> </em><em>solutions</em><em> </em><em>are</em><em> </em><em>in</em><em> </em><em>pictures</em><em>. </em>

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>

5 0
3 years ago
PLEASE HELP IF U PASS TESTS
bezimeni [28]
I think iron? i’m not 100% sure
5 0
3 years ago
Read 2 more answers
a roller coaster moves on a certain section of it's track with an average speed of 13 m/s. how much distance does it cover in 5.
viva [34]
  • Speed=13m/s
  • Time=5.8s

\\ \sf\longmapsto Speed=\dfrac{Distance}{Time}

\\ \sf\longmapsto Distance=Speed\times Time

\\ \sf\longmapsto Distance=13(5.8)

\\ \sf\longmapsto Distance=75.4m

7 0
3 years ago
Read 2 more answers
A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each
ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

E_{1}+E_{3}-E_{2}=0

We know that the electric field is:

E=k\frac{q}{r^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge
  • r is the distance from the charge to the point

So, we have:

k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0

Let's solve it for r(3).

\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0

r_{3}=0.0743\:  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

3 0
2 years ago
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