Answer:
Mercury is a bad conductor of heat but a fair conductor of electricity
By the way PURE SILVER is the best conductor of electricity
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Answer:
7200 kg.m/s
Explanation:
According the law of conservation of linear momentum, the sum of momentum before and after collision are equal.
Using this principle, the sum of initial momentum will be given as p=mv where p is momentum, m is mass and v is velocity
Initial momentum
Mass of whale*initial velocity of whale + mass of seal*initial seal velocity
Since the seal is initially stationary, its velocity is zero. By substitution and taking right direction as positive
Initial momentum will be
1200*6+(280*0)=7200 kg.m/s
Since both initial and final momentum should be equal, hence the final momentum will also be 7200 kg.m/s
Answer:
Option C
Explanation:
We have to check range of all options first
For A:
Largest Value: 5
Smallest Value: 1
So range = Largest value - smallest value
5-1 = 4
For B:
Largest Value: 6
Smallest Value: 4
Range = 6-4 = 2
For C:
Largest Value: 9
Smallest Value: 1
Range = 9-1 = 8
For D:
Largest Value = 9
Smallest Value = 3
Range = 9-3=6
So, the data set in option C has the largest range
Answer:
F = GMmx/[√(a² + x²)]³
Explanation:
The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is
dF = GmdM/L²
Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.
So, the horizontal components add from two symmetrically opposite mass elements dM,
Thus, the horizontal component of the force is
dF' = dFcosФ where Ф is the angle between L and the x axis
dF' = GmdMcosФ/L²
L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.
L = √(a² + x²)
cosФ = x/L
dF' = GmdMcosФ/L²
dF' = GmdMx/L³
dF' = GmdMx/[√(a² + x²)]³
Integrating both sides we have
∫dF' = ∫GmdMx/[√(a² + x²)]³
∫dF' = Gm∫dMx/[√(a² + x²)]³ ∫dM = M
F = GmMx/[√(a² + x²)]³
F = GMmx/[√(a² + x²)]³
So, the force due to the sphere of mass m is
F = GMmx/[√(a² + x²)]³
The answer is 2.49 x 10^5 KJ. This was obtained (1) use the formula for specific heat to achieve Q or heat then (2) get the energy to melt the copper lastly (3) Subtract both work and the total energy required to completely melt the copper bar is achieved.
