Answer: A = y/cos32
That is the amplitude A in terms of the displacement y.
Explanation: Since the displacement in the question In the question is the same in both direction, it is a Simple Harmonic Motion problem. In s.h.m the amplitude of displacement A is related to the displacement itself y by this simple equation
y = A* cos(theta)
So, A = y/cos(theta)
A = y/cos32.
If the magnitude of the displacement y is given, you just substitute in.
Answer:
Explanation:
As given, the student has three balloons and rubs two of them on a piece of wool. The rubbing of balloon on wool is the independent variable as it was done on two and not on the third as control.
Explanation:
Fgravity = G*(mass1*mass2)/D².
G is the gravitational constant, which has the same value throughout our universe.
D is the distance between the objects.
so, if you triple one of the masses, what does that do to our equation ?
Fgravitynew = G*(3*mass1*mass2)/D²
due to the commutative property of multiplication
Fgravitynew = 3* G*(mass1*mass2)/D² = 3* Fgravity
so, the right answer is 3×12 = 36 units.
Answer:So, the difference between charging by induction and conduction comes down to the contact of the neutral object and the object used to charge it. Conduction requires direct contact, while induction does not.
Explanation:
Answer:
0.94 m³/s
Explanation:
From the question given above, the following data were obtained:
Air flow (in ft³/min) = 2×10³ ft³/min
Air flow (in m³/s) =.?
Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:
35.315 ft³/min = 1 m³/min
Therefore,
2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min
2×10³ ft³/min = 56.63 m³/min
Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:
1 m³/min = 1/60 m³/s
Therefore,
56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min
56.63 m³/min = 0.94 m³/s
Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.
