Answer:
radial acceleration is 41.8 m / s²
Explanation:
The acceleration for circular motion is
a = v² / r
They also give us the X and Y position where the body falls when the rope breaks, let's write the projectile launch equations
x = vox t
y = v₀ₓ t - ½ g t2
Since the circle is horizontally the v₀ₓ is zero (v₀ₓ = 0)
x = v₀ₓ t
t = x / v₀ₓ
y = - ½ g t²
Let's replace and calculate the initial velocity on the X axis
y = - ½ g (x / vox)²
v₀ₓ = √ (g x² / 2 y)
v₀ₓ = √ [- (-9.8) 1.6² / (2 1.00)]
v₀ₓ = 3.54 m / s
This is the horizontal velocity, but since it circle is in horizontal position it is also the velocity of the body at the point of rupture.
Now we can calculate the radial acceleration
a = v² / r
a = 3.54² / 0.300
a = 41.8 m / s²
Well you have to think of it like electricity go through your answer closes to that and figure it out
To solve this problem we will apply the concepts related to the Magnetic Force, this is given by the product between the current, the body length, the magnetic field and the angle between the force and the magnetic field, mathematically that is,
Here,
I = Current
L = Length
B = Magnetic Field
= Angle between Force and Magnetic Field
But 
Rearranging to find the Magnetic Field,
Here the force per unit length,
Replacing with our values,
Therefore the magnitude of the magnetic field in the region through which the current passes is 0.0078T
the relatively thick part of the earth's crust that forms the large landmasses. It is generally older and more complex than the oceanic crust.
