Answer:
- The procedure is: solve the quadratic equation for
.
Explanation:
This question assumes uniformly accelerated motion, for which the distance d a particle travels in time t is given by the general equation:
That is a quadratic equation, where the independent variable is the time .
Thus, the procedure that will find the time t at which the distance value is known to be D is to solve the quadratic equation for .
To solve it you start by changing the equation to the general form of the quadratic equations, rearranging the terms:
Some times that equation may be solved by factoring, and always it can be solved by using the quadratic formula:
Where:
That may have two solutions. Some times one of the solution makes no physical sense (for example time cannot be negative) but others the two solutions are valid.
Answer:
option C
Explanation:
given,
mass of water = 4 Kg
Water is heated to = 800 W
time of immersion = 10 min
= 10 x 60 = 600 s
using equation of specific heat
Q = m S ΔT
S is the specific heat capacity of water which is equal to 4182 J/kg°C.
and another formula of heat
Q = Pt
now,
P t = m S ΔT
800 x 600 = 4 x 4182 x ΔT
ΔT = 29° C
temperature increased is equal to ΔT = 29° C
Hence, the correct answer is option C
Answer:
the charge generated in the circuit is 240 C.
Explanation:
Given;
current flowing in the circuit, I = 2A
time of current flow, t = 2 minutes = 2 x 60s = 120 s
The current flowing through a given circuit is defined as the quantity of charge flowing through the circuit in a given time.
where;
Q is the charge flowing in the circuit
Q = 2 x 120
Q = 240 C
Therefore, the charge generated in the circuit is 240 C.
<span>A red ladybug appears red in white light,red in red light, and black in blue light.</span>
Answer:
23.1 N/C
Explanation:
OP = 3 m , OQ = 4 m
q = - 8 nC, Q = 75 nC
Electric field at P due to the charge Q is
Electric field at P due to the charge q is
According to the diagram, tanθ = 3/4
Resolve the components of E1 along x axis and along y axis.
So, Electric field along X axis, Ex = - E1 Cos θ
Ex = - 27 x 4 / 5 = - 21.6 N/C
Electric field along y axis, Ey = E1 Sinθ - E2
Ey = 27 x 3 /5 - 8 = 8.2 N/C
The resultant electric field at P is given by
