Answer:
(A) She needs to move the decimal point by 3 places
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:
The field gets weaker
Explanation:
I’m taking the test right now, hope this helps!!
Answer:
The charge on each plate is 0.0048 nC
Explanation:
for the distance between the plates d and given the area of plates, A, and ε = 8.85×10^-12 C^2/N.m^2, the capacitance of the plates is given by:
C = (A×ε)/d
=[(0.2304×10^-2)(0.2304×10^-2)×(8.85×10^-12))/(0.5974×10^-3)
= 7.86×10^-14 F
then if the plates are connected to a battery of voltage V = 61 V, the charge on the plates is given by:
q = C×V
= (7.86×10^-14)×(61)
= 4.80×10^-14 C
≈ 0.0048 nC
Therefore, the charge on each plate is 0.0048 nC.
Answer:
The current in the circuit increases
Explanation:
The ohm's law states that the potential across a circuit is proportional to the current in the circuit.
V ∝ I
Where 'V' is the potential difference across the circuit and 'I' is the current in the circuit.
The proportionality constant present in the equation is the resistance of the circuit. Hence, the equation becomes
V = IR
According to the equation, when V is directly proportional to 'I' where 'R' remains as constant, then the change in 'V is brings change in 'I' to make the equation valid.
So, when there is an increase in the voltage, the current on the circuit increases.
