Given :
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.
To Find :
The coefficient of static friction between the box and the plane.
Solution :
Vertical component of force :

Horizontal component of force(Normal reaction) :

Since, box is on the verge of slipping :

Therefore, the coefficient of static friction between the box and the plane is 1.07.
Hence, this is the required solution.
b. 460.8 m/s
Explanation:
The relationship between the speed of the wave along the string, the length of the string and the frequency of the note is

where v is the speed of the wave, L is the length of the string and f is the frequency. Re-arranging the equation and substituting the data of the problem (L=0.90 m and f=256 Hz), we can find v:

c. 18,000 m
Explanation:
The relationship between speed of the wave, distance travelled and time taken is

where
v = 6,000 m/s is the speed of the wave
d = ? is the distance travelled
t = 3 s is the time taken
Re-arranging the formula and substituting the numbers into it, we find:

Answer:
The reason is because both are exposed to a virtually infinite heat sink, due to the virtually infinite mass and of the surrounding environment, compared to the sizes of either the cup or the kettle such that the equilibrium temperature,
reached is the same for both the cup and the kettle as given by the relation;

Due to the large heat sink, T₂ - T₁ ≈ 0 such that the temperature of the kettle and that of the cup will both cool to the temperature of the environment
Explanation:
Answer:
42.58kg
Explanation:
By Newton's second law, F = ma.
F is the force being applied, in this case 112N. a is the acceleration, in this case 2.63 m/s^2.
Thus, with some simple algebraic manipulation, we get the mass to equal:
m = F/a = 112N / 2.63 m/s^2 = 42.58kg