Answer: 5 gm/cc
Explanation:
200 gm/40 cc
= 5 gm/cc
The emf induced in the second coil is given by:
V = -M(di/dt)
V = emf, M = mutual indutance, di/dt = change of current in the first coil over time
The current in the first coil is given by:
i = i₀
i₀ = 5.0A, a = 2.0×10³s⁻¹
i = 5.0e^(-2.0×10³t)
Calculate di/dt by differentiating i with respect to t.
di/dt = -1.0×10⁴e^(-2.0×10³t)
Calculate a general formula for V. Givens:
M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)
Plug in and solve for V:
V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))
V = 320e^(-2.0×10³t)
We want to find the induced emf right after the current starts to decay. Plug in t = 0s:
V = 320e^(-2.0×10³(0))
V = 320e^0
V = 320 volts
We want to find the induced emf at t = 1.0×10⁻³s:
V = 320e^(-2.0×10³(1.0×10⁻³))
V = 43 volts
Given parameters:
First velocity = 2.50m/s
Time of travel = 3s
Second velocity = 1.50m/s
Unknown:
The displacement during the first interval = ?
Velocity is the displacement of a body with time. Displacement is a distance move in a specific direction by a body.
Velocity = 
So;
Displacement = Velocity x Time taken
Now input the parameter for the first velocity and time of travel;
Displacement = 2.5 x 3 = 7.5m
The displacement id 7.5m
<span>The expected boiling point of a substance increases with an increasing molar mass of the substance. So I would expect the lighter compounds to boil at lower temperature. Also more polar molecules tend to boil at higher temperature.
Pentane is the lightest of all three compounds, so it should boil the soonest.
Heptane is a lot heavier than Pentane, but slightly lighter than Heptanol. So it should boil next.
Heptanol is the heaviest of all of the molecules and is also quite polar compared to heptane. So it should have the highest of all the boiling points.</span>
Answer:
a) v₂ = 4.2 m/s
b) v₂ = 5 m/s
Explanation:
a)
We will use the law of conservation of momentum here:

where,
m₁ = m₂ = mass of bowling pin = 1.8 kg
u₁ = speed of first pin before collsion = 5 m/s
u₂ = speed of second pin before collsion = 0 m/s
v₁ = speed of first pin after collsion = 0.8 m/s
v₂ = speed of second after before collsion = ?
Therefore,

<u>v₂ = 4.2 m/s</u>
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b)
We will use the law of conservation of momentum here:

where,
m₁ = m₂ = mass of bowling pin = 1.8 kg
u₁ = speed of first pin before collsion = 5 m/s
u₂ = speed of second pin before collsion = 0 m/s
v₁ = speed of first pin after collsion = 0 m/s
v₂ = speed of second after before collsion = ?
Therefore,

<u>v₂ = 5 m/s</u>