Answer:
3.1216 m/s.
Explanation:
Given:
M1 = 0.153 kg
v1 = 0.7 m/s
M2 = 0.308 kg
v2 = -2.16 m/s
M1v1 + M2v2 = M1V1 + M2V2
0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2
= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2
0.153V1 + 0.308V2 = -0.55818. i
For the velocities,
v1 - v2 = -(V1 - V2)
0.7 - (-2.16) = -(V1 - V2)
-(V1 - V2) = 2.86
V2 - V1 = 2.86. ii
Solving equation i and ii simultaneously,
V1 = 3.1216 m/s
V2 = 0.2616 m/s
The ratio of the turns to the voltage should be equal
i.e: 200/120 = t/12
so the secondary coil should have 20 turns
Answer:
Magnetic field experienced = 4.5 × 10⁻⁴ T
Explanation:
The magnetic field around an infinite straight current-carrying wire at a distance r from the wire is given by
B = (μ₀I)/(2πr)
B = ?
I = 20 KA = 20000 A
r = 8.9 m
μ₀ = magnetic permeability = 1.257 × 10⁻⁶ T.m/A
B = (1.257 × 10⁻⁶ × 20000)/(2π×8.9) = 4.5 × 10⁻⁴ T
Answer:
a) True. The image of the mite is virtual
e) True. The image must be within the focal length of the eyepiece len
Explanation:
Let's review the general characteristics of compound microscopes
Formed by two converging lenses
Magnification is
M = -L/fo 0.25/fe
Where fo is the focal length of the objective lens and fe is the focal length of the ocular lens, L is the tube length
Let's review the claims
a) True. The image of the mite is virtual
b) False. The effect is the opposite of the magnification equation
c) False. The objective lens forms a real image
d) False. As the seal distance increases the magnification decreases
e) True. The image must be within the focal length of the eyepiece len