Answer:
A glass flask whose volume is 1000 cm ^3 at 0.0 ^oC is completely filled with mercury at this. Every substance when heat energy is supplied, expands due to the Rate of thermal expansion will be different for different materials. Volume of the glass flask and mercury at 0 degree Celsius V0=1000cm3=1×10−3m3 V 0
Explanation:
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Answer:
Explanation:
Before the dialectic was inserted the capacitor is Co
When the slab is inserted,
The capacitor becomes
C=kCo
The charge Q is given as
Q=CV
Then, when C=Co
Qo=CoV
Then, when C=kCo
Q=kCoV
Then, the change in charges is given as
Q-Qo= kCoV - CoV
∆Q= kCoV - CoV
Current is given as
I=dQ/dt
I= (kCoV - CoV) / dt
I=Co(kV-V)/dt
Note Co is the value capacitor
So, Capacitance of parallel plates capacitor is given as
Co=εoA/d
Then,
I=εoA(kV-V)/d•dt
I=VεoA(k-1)/d•dt
Where A=πr²
I = V•εo•πr²•(k-1) / d•dt
This is the required expression for current is in the required term
Answer:
b) Document lessons learned.
Explanation:
First he should do documentation
then C
then D
then A
Using the following given values:
Object 1:
Mass = M1 = 2 kg
Velocity before collision = Vb1 = 20 m/s
Velocity after collision = Va1 = -5 m/s
Object 2:
Mass = M2 = 3 kg
Velocity before collision = Vb2 = -10 m/s
Velocity after collision = Va2 = ? m/s<span>
</span>
Obtaining Va2 via law of conservation of momentum:
total momentum after collision = total momentum before collision
M1 * Va1 + M2 * Va2 = M1 * Vb1 + M2 * Vb2
2*-5 + 3Va2 = 2*20 + 3*-10
Va2 = 6.67
Total kinetic energy before collision:
KE1 = (1/2)*M1*Vb1^2 + (1/2)*M2*Vb2^2
<span>KE1 = (1/2)*2*(20)^2 + (1/2)*3*(-10)^2
KE1 = 550 J
</span>Total kinetic energy after collision:
KE2 = (1/2)*M1*Va1^2 + (1/2)*M2*Va2^2
<span>KE2 = (1/2)*2*(-5)^2 + (1/2)*3*(6.67)^2
KE2 = 91.73 J
</span>
Total kinetic energy lost:
Energy lost = KE1 - KE2 = 550 - 91.73 = 458.27 J
Answer:
The variation is inversely proportional to the decrease density
Explanation:
This is an exercise where we will use the Archimedes principle that states that the thrust is equal to the weight of the dislodged liquid
B = m g
ρf = m / Vf
m = ρf Vf
B = ρf Vf g
If we use Newton's second law for equilibrium
B - W = 0
B = m g
ρ Vf g = ρb Vb g
ρb / ρf = Vf / Vb (1)
Let's apply this expression to our case
In water
Vf / Vg = 0.90
replace in equation 1
ρb / ρf = 0.90
ρb = ρf 0.90
ρb = 1000 0.90
ρb = 900 kg / m3
Now we change the liquid to one with lower density, let's calculate the volume ratio
Vf / Vg = ρb / ρf
The density of the body (ρb) remains constant if the density of the fluid decreases, as in the denominator the volume fraction increases, whereby the submerged part decreases
The variation is inversely proportional to the decrease density
