Complete question is;
Jason works for a moving company. A 75 kg wooden crate is sitting on the wooden ramp of his truck; the ramp is angled at 11°.
What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving UP the ramp?
Answer:
F = 501.5 N
Explanation:
We are given;
Mass of wooden crate; m = 75 kg
Angle of ramp; θ = 11°
Now, for the wooden crate to slide upwards, it means that the force of friction would be acting in an opposite to the slide along the inclined plane. Thus, the force will be given by;
F = mgsin θ + μmg cos θ
From online values, 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
Carbon is found in the solid form in geosphere of our earth. Coal and oil are some of the examples of materials containing carbon in the geosphere. when the coal or oil is burnt, carbon dioxide is formed and released in atmosphere. This carbon dioxide is absorbed by the water of the hydrosphere with the help of algae and plankton. The water turns acidic in nature. This way carbon is transferred from geosphere to hydrosphere.
Answer:
The net force on the skater is zero. (
)
Explanation:
According to Newton's First Law, an object is at equilibrium when either it is at rest or moves at constant velocity, which means a net force of zero. Based on the given statement, there are no external forces acting on skate and, therefore, the net force on the skater is zero. (
)
Answer:
simple
Explanation:
<h3>CONCAVE MIRRORS AND LENSES</h3>
<h3>f= negative</h3>
<h3>CONVEX MIRRORS AND LENSES</h3><h3 /><h3>f= positive</h3>
<h3>PLEASE FOLLOW ME AND MARK IT BRAINLIEST</h3>
Answer:
0.98kW
Explanation:
The conservation of energy is given by the following equation,


Where
Mass flow
Specific Enthalpy (IN)
Specific Enthalpy (OUT)
Gravity
Heigth state (In, OUT)
Velocity (In, Out)
Our values are given by,




For this problem we know that as pressure, temperature as velocity remains constant, then


Then we have that our equation now is,


