Answer:
Given:
1$ = 4q
To Find:
How many quarters are in 20$
Explanation:
To find out how many quarters are in 20$ we need to multiple 4 × 20.
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2 answers:
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Answer:
Explanation:
Let the length of the string is L.
Let T be the tension in the string.
Resolve the components of T.
As the charge q is in equilibrium.
T Sinθ = Fe ..... (1)
T Cosθ = mg .......(2)
Divide equation (1) by equation (2), we get
tan θ = Fe / mg
As θ is very small, so tanθ and Sinθ is equal to θ.
Answer:
58.27 N
Explanation:
the data we have is:
mass:
coefficient of friction:
and we also know the acceleration of gravity is
We need to do an analysis of horizontal and vertical forces acting on the object:
-------
Vertically the forces acting on the object:
- Normal force
(acting up from the object)
- weight:
(acting down from)
so the sum of forces in the vertical axis "y" are:
from Newton's second Law we know that , so:
and since the object is not accelerating in the vertical direction (the movement is only horizontal) , and:
-----------
now let's analyze the horizontal forces
- frictional force:
and since
-->
- force to move the object:
and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:
and we are told that the crate moves at a steady speed, thus there is no acceleration:
and we get:
substituting known values:
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
- Replacing by the givens in (5), we can solve for μsmín, as follows: