Answer:

Explanation:
To solve the problem, the concepts related to the magnetic field and the current produced in a lightning bolt are necessary.
The current is defined by the load due to time, that is to say

Where,


So the current can be expressed as:


Once the current is found it is now possible to find the magnetic field, as this is given by the equation,

Where,
Permeability Constant
I= Current
r= radius
Replacing the values we have


Answer:
petroleum and Natural gas are fossil fuels
Explanation:
The current in the circuit is 5 A
Explanation:
The intensity of current is given by the equation:

where
I is the current
q is the amount of charge passing through a given point of the circuit in a time interval of t
For the cell in this problem, we have
q = 150 C is the charge
t = 30 s is the time interval
Substituting into the equation, we f ind

Learn more about current:
brainly.com/question/4438943
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#LearnwithBrainly
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:
a)KE=878.8 J
b)W=2636.4 J
Explanation:
Given that
mass ,m = 65 kg
Initial speed ,u = 5.2 m/s
a)
We know that kinetic energy KE is given as follows

m=mass
u=velocity
Now by putting the values in the above equation we get

KE=878.8 J
b)
We know that
Work done by all forces = Change in the kinetic energy
The final velocity , v= 2 u = 2 x 5.2 m/s
v= 10.4 m/s

Now by putting the values in the above equation we get

W=2636.4 J
a)KE=878.8 J
b)W=2636.4 J