To solve the exercise it is necessary to take into account the concepts of wavelength as a function of speed.
From the definition we know that the wavelength is described under the equation,
Where,
c = Speed of light (vacuum)
f = frequency
Our values are,
Replacing we have,
<em>Therefore the wavelength of this wave is 
</em>
Answer;
Q = 359.2-J
Explanation;
Given that;
Constants for mercury at 1 atm
Heat capacity of Hg(l) is 28.0 J/(mol*K)
melting point is 234.32 K
Enthalpy of fusion is 2.29 kJ/mol
17.7-g Hg / 200.6g/mol = 0.0882 mol Hg;
°C + 273 = 298 K;
2.29-kJ/mol = 2290-J/mol
Q = (m x ΔT x Cp) + (m x Hf)
Q = 0.0882-mol x (298 - 234.32) x 28.0-J/mol*K) + (0.0882-mol x 2290-J/mol)
Q = 157.26-J + 201.978-J
Q = 359.2-J
Q=359-J (3 sig fig allowed due to 17.7-g given in problem)
Answer:
0.19
Explanation:
mass of block, m = 40 kg
F = 150 N
Angle make with the horizontal, θ = 60 degree
Let μ be the coefficient of kinetic friction
The component of force along horizontal direction is F Cos θ
= 150 cos 60 = 75 N
As it is moving with constant velocity it mean the acceleration of the block is zero.
Applied force in horizontal direction = friction force
75 = μ x Normal reaction
75 = μ x m x g
75 = μ x 40 x 9.8
μ = 0.19
Thus, the coefficient of kinetic friction is 0.19.
Answer:
250,000 Joules
Explanation:
The work required is the amount of energy added, so the answer is the difference in kinetic energy before and after the acceleration of the car. (The problem doesn't say it's flat where the car is, but well assume it since there is no information about an elevation change that would require us to concern ourselves with potential energy!)
W = E_final - E_initial
= 1/2 m v_final^2 - 1/2 m v_initial^2
= 1/2 m (v_final^2 - v_initial^2)
= 1/2 (1000 kg) ( (30 m/s)^2 - (20 m/s^2) )
= 1/2 (1000 kg) (500 m^2/s^2)
= 250,000 kg m^2/s^2
= 250,000 Joules
