The calculated magnitude is 6.73 x 10³ V/m.
AMU is described as being one-twelfth the mass of a carbon-12 atom (12C). C makes up more than 98% of the carbon that can be found in nature, making it the most prevalent isotope. The magnitude of the field is the change in potential across a small distance in the indicated direction divided by that distance.
Potential difference = 8.20 kV= 8.20 x 10³ V
radius= 19.4/100=0.194 m
total distance that is circumference of the circle= 2πr =2 x 3.14 x 0.194
= 1.218 m
therefore Magnitude= 8.20 x 10³ / 1.218
=6.73 x 10³ V/m
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To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.
The angular velocity can be described as
Where,
Final Angular Velocity
Initial Angular velocity
Angular acceleration
t = time
The relation between the tangential acceleration is given as,
where,
r = radius.
PART A ) Using our values and replacing at the previous equation we have that
Replacing the previous equation with our values we have,
The tangential velocity then would be,
Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation
Replacing with our values and re-arrange to find 
That is equal in revolution to
The linear displacement of the system is,
Answer:
x(t) = - 6 cos 2t
Explanation:
Force of spring = - kx
k= spring constant
x= distance traveled by compressing
But force = mass × acceleration
==> Force = m × d²x/dt²
===> md²x/dt² = -kx
==> md²x/dt² + kx=0 ------------------------(1)
Now Again, by Hook's law
Force = -kx
==> 960=-k × 400
==> -k =960 /4 =240 N/m
ignoring -ve sign k= 240 N/m
Put given data in eq (1)
We get
60d²x/dt² + 240x=0
==> d²x/dt² + 4x=0
General solution for this differential eq is;
x(t) = A cos 2t + B sin 2t ------------------------(2)
Now initially
position of mass spring
at time = 0 sec
x (0) = 0 m
initial velocity v= = dx/dt= 6m/s
from (2) we have;
dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)
put t =0 and dx/dt = v(0) = -6 we get;
-2A sin 2(0)+2Bcos(0) =-6
==> 2B = -6
B= -3
Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get
x(t) = - 6 cos 2t
==>
Answer:10
Explanation:
You have to do speed divided by time so your answers 10
the neutrons through the nucleus?
Explanation:
