Answer:
Explanation:
According to first law of thermodynamics:
∆U= q + w
= 10kj+(-70kJ)
-60kJ
, w = + 70 kJ
(work done on the system is positive)
q = -10kJ ( heat is given out, so negative)
∆U = -10 + (+70) = +60 kJ
Thus, the internal energy of the system decreases by 60 kJ.
Answer:
F₄ = 29.819 N
Explanation:
Given
F₁ = (- 25*Cos 50° i + 25*Sin 50° j + 0 k) N
F₂ = (12*Cos 50° i + 12*Sin 50° j + 0 k) N
F₃ = (0 i + 0 j + 4 k) N
Then we have
F₁ + F₂ + F₃ + F₄ = 0
⇒ F₄ = - (F₁ + F₂ + F₃)
⇒ F₄ = - ((- 25*Cos 50° i + 25*Sin 50° j) N + (12*Cos 50° i + 12*Sin 50° j) N + (4 k) N) = (13*Cos 50° i - 37*Sin 50° j - 4 k) N
The magnitude of the force will be
F₄ = √((13*Cos 50°)² + (- 37*Sin 50°)² + (- 4)²) N = 29.819 N
Answer:
V = λ f (wavelength * frequency)
λ = V / f = 343 m/s / 262 / s = 1.3 m
Answer:
The coefficient of static friction between the box and floor is, μ = 0.061
Explanation:
Given data,
The mass of the box, m = 50 kg
The force exerted by the person, F = 50 N
The time period of motion, t = 10 s
The frictional force acting on the box, f = 30 N
The normal force on the box, η = mg
= 50 x 9.8
= 490 N
The coefficient of friction,
μ = f/ η
= 30 / 490
= 0.061
Hence, the coefficient of static friction between the box and floor is, μ = 0.061
