The full question reads;
Jason is employed at a moving company. A wooden crate weighing 75 kg is positioned on the wooden ramp of his truck, inclined at an angle of 11°.
What is the force magnitude, directed parallel to the ramp, that he needs to apply to initiate the upward movement of the crate?
Answer:
F = 501.5 N
Explanation:
We have the following information;
Mass of the wooden crate; m = 75 kg
Incline angle; θ = 11°
To move the wooden crate up, we must consider that friction is acting in the opposite direction of the movement along the inclined surface. Therefore, the force required can be expressed by;
F = mgsin θ + μmg cos θ
Using online resources, the coefficient of friction between wooden surfaces is μ = 0.5
Thus;
F = (75 × 9.81 × sin 11) + (0.5 × 75 × 9.81 × cos 11)
F = 501.5 N
Explanation:
Please refer to the attachment for the solution.
A. A car moving at a constant speed on a flat, straight road. B. A vehicle traveling at a steady speed on a 10-degree incline. An object operates within an inertial reference frame if there is no net force acting upon it. According to Newton's second law, this implies that the object's acceleration also equals zero. Assessing the scenarios yields: A. A car moving at a constant speed on a flat road qualifies as an inertial reference frame, since its velocity and direction remain unchanged; thus, acceleration is zero. B. A car moving steadily up a 10-degree incline still constitutes an inertial reference frame, for similar reasons. C. A car accelerating after departing a stop sign does not represent an inertial frame due to its change in speed. D. A car driving at a steady speed around a curve cannot be considered an inertial reference frame since its direction is changing, resulting in a change in velocity and thus acceleration. Therefore, options A and B are correct.
Answer:
Explanation:
Provided:
The trolley, with mass M, is allowed to roll freely without friction.
The coefficient of friction between the trolley and mass m is 
.
A force F is applied to mass m.
The acceleration of the system is
The frictional force will counterbalance the weight of the block.
The frictional force is 
Answer:
a)106.48 x 10⁵ kg.m²
b)144.97 x 10⁵ kgm² s⁻¹
Explanation:
a)Given
m = 5500 kg
l = 44 m
The moment of inertia for one blade
= 1/3 x m l²
where m denotes the mass of the blade
l represents the length of each blade.
Substituting the necessary values, the moment of inertia for one blade is
= 1/3 x 5500 x 44²
= 35.49 x 10⁵ kg.m²
Total moment of inertia for 3 blades
= 3 x 35.49 x 10⁵ kg.m²
= 106.48 x 10⁵ kg.m²
b) The angular momentum 'L' is calculated using
L = x ω
where,
= the moment of inertia of the turbine i.e 106.48 x 10⁵ kg.m²
ω= angular velocity =2π f
f represents the frequency of rotation of the blade i.e 13 rpm
f = 13 rpm=>= 13 / 60 revolutions per second
ω = 2π f => 2π x 13 / 60 rad / s
L= x ω =>106.48 x 10⁵ x 2π x 13 / 60
= 144.97 x 10⁵ kgm² s⁻¹
