From the relationship between acceleration a and g on an inclined plane, the angle of the slope is 13 degrees
Given that a solid sphere starts from rest and rolls down a slope that is 6.4 m long. The speed at the bottom of the slope is 5.3 m/s, the distance travelled is 6.4 m. That is,
Initial velocity U = 0 ( since it starts from rest)
Final velocity V = 5.3 m/s
distance S = 6.4 m
Let us first calculate its acceleration by using third equation of motion.
[tex]V^{2}[/tex] = [tex]U^{2}[/tex] + 2aS
[tex]5.3^{2}[/tex] = 0 + 2 x 6.4a
28.09 = 12.8a
a = 28.09 / 12.8
a = 2.2 m / [tex]s^{2}[/tex]
To calculate the angle of the slope, let us use the relationship between acceleration a and g on an inclined plane.
acceleration a = gsin∅
substitute all the relevant parameters
2.2 = 9.8 sin∅
sin∅ = 2.2/9.8
sin∅ = 0.224
∅ = [tex]Sin^{-1}[/tex](0.224)
∅ = 12.97 degrees
∅ = 13 degrees (approximately)
Therefore, the angle of the slope is 13 degrees
Learn more about inclined plane here: https://brainly.com/question/25845680
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
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
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.
