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2 years ago

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

1 answer:

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:

Hello there!!!

You just need to use simple formula for force and momentum, 

F= m.a

and momentum (p)= m.v

where m= mass

v= velocity.

a= acceleration .

And the solutions are in pictures. 

Hope it helps..

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