Answer:
D = 527.31 Km
Explanation:
given,
angle of ship, θ = 23.5° N of W
distance travel in the direction = 575 Km
Distance of ship in west from harbor = ?
now,
Distance of the ship in the west direction
D = d cos θ
d = 575 Km
θ = 23.5°
inserting all the values
D = 575 x cos 23.5°
D = 575 x 0.91706
D = 527.31 Km
Hence, the distance travel by the ship in west from harbor is equal to D = 527.31 Km
W work
F force
s distance
If F = constant:
W₁ = F·s
If you triple the force and the distance:
W₂ = 3F · 3s = 9 F·s = 9 W₁
Answer:
Explanation:
We shall apply law of conservation of momentum in space to know the velocity of combination after the impact
m₁v₁ = m₂v₂
.1 x 4 = ( 1 + .1 ) v₂
v₂ = .3636 m /s
1 )
Kinetic energy of the combination
= 1/2 x 1.1 x ( .3636)²
= 7.3 x 10⁻² J
2 )
Initial kinetic energy of the system
= 1/2 x 0.1 x 4²
= 0.8 J
Final kinetic energy of the system = 7.3 x 10⁻²
Loss of energy = .8 - .073
= .727 J
This energy was converted into internal energy of the system .
3 )
increase in entropy = dQ / T
Here dQ = .727 J
T = 300 ( Constant )
dQ / T = 2.42 X 10⁻³ J/K