Answer:
5.56 A
Explanation:
From the question,
Q = it.............. Equation 1
Where Q = charges, i = current, t = time.
Make i the subject of the equation
i = Q/t.............. Equation 2
Given: Q = 200 coulombs, t = 0.6 minutes = (0.6×60) seconds
Substitite these values into equation 2
i = 200/(0.6×60)
i = 5.56 A
Hence the magnitude of the current flowing through the circuit is 5.56 A
Answer:
First Law
Explanation:
I learned this in 6th grade
Answer:
<em>The net force acting on the object is 0 N</em>
Explanation:
<u>Newton's Second Law of Forces</u>
The net force acting on a body is proportional to the mass of the object and its acceleration.
The net force can be calculated as the sum of all the force vectors in each rectangular coordinate separately.
The image shows a free body diagram where four forces are acting: two in the vertical direction and two in the horizontal direction.
Note the forces in the vertical direction have the same magnitude and opposite directions, thus the net force is zero in that direction.
Since we are given the acceleration a =0, the net force is also 0, thus the horizontal forces should be in equilibrium.
The applied force of Fapp=10 N is compensated by the friction force whose value is, necessarily Fr=10 N in the opposite direction.
The net force acting on the object is 0 N
Answer:
- 0.09 % of the original radioactive nucllde its left after 10 half-lives
- It will take 241,100 years for 10 half-lives of plutonium-239 to pass.
Explanation:
The equation for radioactive decay its:
,
where N(t) its quantity of material at time t, 
its the initial quantity of material and 
its the mean lifetime of the radioactive element.
The half-life 
its the time at which the quantity of material its the half of the initial value, so, we can find:
so:
So, after 10 half-lives, we got:
So, we got that a 0.09 % of the original radioactive nucllde its left.
Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass
It will take 241,100 years for 10 half-lives of plutonium-239 to pass.
