The <span>force that is needed to accelerate an object 5 m/s if the object has a mass of 10kg 50N because you multiply 5 and 10</span>
the wavelength is to be 56.67 cm and I know that i should be applying the double slit equation for destructive interference.
You draw 3 circles around the stations with the size of the circle equal to the distance from the earthquake. Then you simply find where the edge circles all overlap.
The main formula to be used here is
Force = (mass) x (acceleration).
We'll get to work in just a second. But first, I must confess to you that I see
two things happening here, and I only know how to handle one of them. So
my answer will be incomplete, but I believe it will be more reliable than the
first answer that was previously offered here.
On the <u>right</u> side ... where the 2 kg and the 3 kg are hanging over the same
pulley, those weights are not balanced, so the 3 kg will pull the 2kg down, with
some acceleration. I don't know what to do with that, because . . .
At the <em>same time</em>, both of those will be pulled <u>up</u> by the 10 kg on the other side
of the upper pulley.
I think I can handle the 10 kg, and work out the acceleration that IT has.
Let's look at only the forces on the 10 kg:
-- The force of gravity is pulling it down, with the whatever the weight of 10 kg is.
-- At the same time, the rope is pulling it UP, with whatever the weight of 5 kg is ...
that's the weight of the two smaller blocks on the other end of the rope.
So, the net force on the 10 kg is the weight of (10 - 5) = 5 kg, downward.
The weight of 5 kg is (mass) x (gravity) = (5 x 9.8) = 49 newtons.
The acceleration of 10 kg, with 49 newtons of force on it, is
Acceleration = (force) / (mass) = 49/10 = <em>4.9 meters per second²</em>
It's not in motion when the line straight and flat . there's #9
