The maximum magnitude of the friction force exerted on the box by the ramp is 44.41 N.
The given parameters;
Mass of the box, m = 10 kgInclination of the ramp, θ = 25⁰Coefficient of static friction, μ = 0.5 Coefficient of kinetic friction, μk = 0.35The normal force on the wooden box is calculated as follows;
[tex]F_n = mg \times cos(\theta)\\\\F_n = 10 \times 9.8 \times cos(25)\\\\F_n = 88.8 2 \ N[/tex]
The maximum magnitude of the friction force exerted on the box by the ramp is calculated as follows;
[tex]F_f = \mu \times F_n\\\\F_f = 0.5 \times 88.82 \\\\F_f = 44.41 \ N[/tex]
Learn more about static frictional forces here: https://brainly.com/question/4515354
A pumpkin is launched in the air and travels at a horizontal velocity of 25 meters per second for 5 seconds. How far does it travel horizontally?
Answer:
30.3 meters, 172 degrees
Explanation:
To insure the most accurate solution, this problem is best solved using a calculator and trigonometric principles. The first step is to determine the sum of all the horizontal (east-west) displacements and the sum of all the vertical (north-south) displacements.
Horizontal: 2.0 meters, West + 31.0 meters, West + 3.0 meters, East = 30.0 meters, West
Vertical: 12.0 meters, North + 8.0 meters, South = 4.0 meters, North
The series of five displacements is equivalent to two displacements of 30 meters, West and 4 meters, North. The resultant of these two displacements can be found using the Pythagorean theorem (for the magnitude) and the tangent function (for the direction). A non-scaled sketch is useful for visualizing the situation.
Applying the Pythagorean theorem leads to the magnitude of the resultant (R).
R2 = (30.0 m)2 + (4.0 m)2 = 916 m2
R = Sqrt(916 m2)
R = 30.3 meters
The angle theta in the diagram above can be found using the tangent function.
tangent(theta) = opposite/adjacent = (4.0 m) / (30.0 m)
tangent(theta) = 0.1333
theta = invtan(0.1333)
theta = 7.59 degrees
This angle theta is the angle between west and the resultant. Directions of vectors are expressed as the counterclockwise angle of rotation relative to east. So the direction is 7.59 degrees short of 180 degrees. That is, the direction is ~172 degrees.
Calculate the torque produced by a 50.0 N perpendicular force at the end of a 0.300 m long wrench.
Answer:
Torque = 50N x 0.3m = 15Nm
Explanation:
Torque = Force x length of lever arm. To obtain the torque simply multiply the two given values.
Please write a paragraph explaining the bible verse below in your own words.
Exodus 16:19-20
Answer:
Moses had told them to not keep the food till morning but some kept some anyways because they probably thought they were going to starve or not have food the next morning but what I think it means is that you have to trust in God that he will provide for you and so when the people kept the food cuz he thought they were probably going to start of the next day it got maggots
Which light is most sensitive to the eyes?
Answer:
Our eyes are most sensitive to the wavelengths corresponding to the yellow and green colors of the spectrum. Flashy signs and some fire engines are painted in a yellowish-green color to attract our attention.
What is the effect of erosion?
A. New land forms at the mouth of a river.
B. New land forms at the top of a mountain.
C. A mountain forms.
D. A fossil is created.
2.(01.01 LC)
Which of the following is true for gravitational force? (3 points)
Decreases with increase in mass
Increases with increase in mass
Increases with increase in distance
Decreases with decrease in distance
Answer:
Increases with increase in mass
Explanation:
gravity is proportional to mass and inversely proportional to the square of the distance between them
F = GMm/d²
5) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal while you move the wagon 10.0 m forward. The coefficient of friction between the wagon and road is 0.500. Calculate the work down by you and the work done by friction.
Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.
By Newton's second law, the net vertical force is
• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0
where a is the acceleration of the wagon.
Solve for n (the magnitude of the normal force) :
n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N
Then
f = 0.500 (58.0 N) = 29.0 N
Meanwhile, the horizontal component of the applied force has magnitude
(80.0 N) cos(30.0°) ≈ 69.3 N
Now calculate the work done by either force.
• friction: -(29.0 N) (10.0 m) = -290. J
• pull: (69.3 N) (10.0 m) = 693 J
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Answer:
Ok thanks for the points though
Explanation:
