Answer:
86.51° North of West or 273.49°
Explanation:
Let V' = velocity of boat relative to the earth, v = velocity of boat relative to water and V = velocity of water.
Now, by vector addition V' = V + v'.
Since v' = 6.10 m/s in the north direction, v' = (6.10 m/s)j and V = 100 m/s in the east direction, V = (100 m/s)i. So that
V' = V + v'
V' = (100 m/s)i + (6.10 m/s)j
So, we find the direction,Ф the boat must steer to from the components of V'.
So tanФ = 6.10 m/s ÷ 100 m/s
tanФ = 0.061
Ф = tan⁻¹(0.061) = 3.49°
So, the angle from the north is thus 90° - 3.49° = 86.51° North of West or 270° + 3.49° = 273.49°
Answer:
(d) a net external force must be acting on the system
Explanation:
Momentum is given as the product of mass and velocity.
P = MV
According to Newton's second law of motion, " Force applied to a body (system) is directly proportional to the rate of change of momentum of the body (system) which takes place in the direction of the applied force (external force).
F ∝ΔMV
Therefore, If the total momentum of a system is changing, a net external force must be acting on the system.
(d) a net external force must be acting on the system
Answer:
v = 10 m/s
Explanation:
Let's assume the wheel does not slip as it accelerates.
Energy theory is more straightforward than kinematics in my opinion.
Work done on the wheel
W = Fd = 45(12) = 540 J
Some is converted to potential energy
PE = mgh = 4(9.8)12sin30 = 235.2 J
As there is no friction mentioned, the remainder is kinetic energy
KE = 540 - 235.2 = 304.8 J
KE = ½mv² + ½Iω²
ω = v/R
KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²
v = √(2KE / (m + I/R²))
v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6
v = 10.07968...
Answer:
The acceleration of the wagon is 3 m/s².
To calculate the acceleration of the wagon, we use the formula below.
Formula:
F = ma............. Equation 1
Where:
F = horizontal Force
m = mass of the wagon
a = acceleration of the wagon.
make a the subject of the equation
a = F/m.............. Equation 2
From the question,
Given:
F = 30 N
m = 10 kg
Substitute these values into equation 2
a = 30/10
a = 3 m/s²
Hence, the acceleration of the wagon is 3 m/s².
