Answer:
1. Either larger or smaller than the displacement of either wave acting alone, depending on the signs of the displacements of the two waves.
The simplest way to do this is to set up equivalent fractions, like this-
=
Solve for x by using cross multiplication.
40*2.2= 88
1*x=88
x=88
Therefore, the boy weighs 88lbs.
Answer:
25.71 kgm/s
Explanation:
Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.
Given that K₂ = 0.7K₁
1/2mv₂² = 0.7(1/2mv₁²)
v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s
Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.
Th magnitude of object A's momentum change is thus 25.71 kgm/s
Complete question:
Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor? If the potential difference due to electric field between the two ends of the resistor is 10 V.
Answer:
The electric field inside this metal resistor is 3125 V/m
Explanation:
Given;
length of the wire, L = 3.2 mm = 3.2 x 10⁻³ m
diameter of the wire, d = 0.4 mm = 0.4 x 10⁻³ m
the potential difference due to electric field between the two ends of the resistor, V = 10 V
The electric field inside this metal resistor is given by;
ΔV = EL
where;
ΔV is change in electric potential
E = ΔV / L
E = 10 / (3.2 x 10⁻³ )
E = 3125 V/m
Therefore, the electric field inside this metal resistor is 3125 V/m
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
