<span>(a) E = ½ Q²/C, so ..
(b) E(max) = ½Li² (i=current), so .</span>
Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Using the principle of floatation.
u = w............(a)
Upthrust of fluid is equal to the weight of the object.
Let the volume of the wood be V.
The upthrust u, is related to the volume submerged in water, and that is 1/5 of it volume, that is (1/5)V = 0.2V
Formula for upthrust, u = vdg
where v = volume of fluid displaced
d = density of fluid
g = acceleration due to gravity
weight, w = mg
where m = mass
g = acceleration due to gravity
From (a)
u = w
vdg = mg Cancel out g
vd = m
The v is equal to 0.2V, which is the submerged volume. Notice that the small letter v is volume of fluid displaced, and capital V is the volume of the solid.
d is density of fluid which is water in this case, 1000 kg/m³
0.2V * 1000 = m
200V = m
Hence the mass of the object is 200V kg.
But Density of solid = Mass of solid / Volume of solid
= 200V / V
= 200 kg/m³
Density of solid = 200 kg/m³
2) Unbalanced. Mike will push the box with a force of 20 N. The forces would be balanced if the box responded with 30 N.
3) Balanced. Both boys are pulling with the same force. Neither is winning.
4) Unbalanced. The rope will move with 10 N to the west. The teachers are winning.
5) Unbalanced. The kids are pulling 220 N to the east. The kids are winning.
6) Balanced. You and the dog are pulling with the same force.
