Answer:
Resultant is 152 N at 28.5 degrees south to the 100 N force
Explanation:
Answer:
It represents the change in charge Q from time t = a to t = b
Explanation:
As given in the question the current is defined as the derivative of charge.
I(t) = dQ(t)/dt ..... (i)
But if we take the inegral of the equation (i) for the time interval from t=a to
t =b we get
Q =∫_a^b▒〖I(t) 〗 dt
which shows the change in charge Q from time t = a to t = b. Form here we can say that, change in charge is defiend as the integral of current for specific interval of time.
If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
The solution you should use is Hooke's law: F=-kx
It should have the same signs because they repel due to the stretch of the spring.
a. Since there is a constant energy within the spring, then Hooke's law will determine the possible algebraic signs. The solution should be
<span>F = kx
270 N/m x 0.38 m = 102.6 N
</span>
b. Then use Coulomb's law; F=kq1q2/r^2 to find the charges produced in the force.
