A washing machine spin dryer has a period T. In order to triple the centripetal
acceleration of the clothes in the dryer the period should be ? 3T
A cube of mass 2.0 kg is kept pressed against a vertical wall by applying a horizontal
force of 60N. If the coefficient of friction between the cube and the wall is 0.50, the
frictional force between the cube and the wall is ? 19.6 N
The angle which vector J + (root3)K makes with the y axis is? Pi/3 radians?
Thanks.
Ac = v^2/R =
c = 2 pi R
Hey, you need to go faster, not slower !
T = c/v = 2 pi R/v
so
v^2 = (2 pi R)^2/T^2
v^2/R = {(2pi)^2/R T^2]
3 v^2/R = {(2pi)^2/R T^2] *3
so 3/T^2 = 1/T'^2
3 T'^2 = T^2
T'^2 = T^2/3
T' = T /sqrt 3
A cube of mass 2.0 kg is kept pressed against a vertical wall by applying a horizontal
force of 60N. If the coefficient of friction between the cube and the wall is 0.50, the
frictional force between the cube and the wall is ? 19.6 N
----------------------------
If the cube is not accelerating down, the frictional force is equal to the weight (and less than the coef times the normal force)
m g = 2* 9.81 = 19.6 right
tan theta = sqrt 3/1
theta = 60 deg = pi/3 right
Thank you Damon.
You are welcome :)
Thanks Damon
To triple the centripetal acceleration of the clothes in the washing machine spin dryer, we need to understand that the centripetal acceleration is given by the formula a = (4π^2r) / T^2, where a is the centripetal acceleration, r is the radius, and T is the period.
Since we want to triple the centripetal acceleration, we can solve the equation a' = 3a, where a' is the new centripetal acceleration.
Substituting the values into the equation, we get:
(4π^2r) / (3T^2) = 3[(4π^2r) / T^2]
Simplifying the equation, we can cancel out the terms (4π^2r) and T^2:
1 / (3T^2) = 3 / T^2
Cross-multiplying and solving for T, we get:
3T^2 = 9
T^2 = 9/3
T^2 = 3
T = √3
Therefore, the period should be √3T to triple the centripetal acceleration of the clothes in the washing machine spin dryer.
In the case of the cube pressed against a vertical wall, the frictional force can be determined using the equation f = μN, where f is the frictional force, μ is the coefficient of friction, and N is the normal force.
Since the cube is pressed against the wall in a vertical direction, the normal force is equal to the weight of the cube, given by N = mg. Therefore, N = 2.0 kg * 9.8 m/s^2 = 19.6 N.
Substituting the values into the equation, we get:
f = (0.50)(19.6 N)
f = 9.8 N
Therefore, the frictional force between the cube and the wall is 9.8 N.
Lastly, to determine the angle which vector J + √3K makes with the y-axis, we can use the dot product between the two vectors. The dot product is calculated as follows:
J · (J + √3K) = |J| |J + √3K| cos(θ)
Since vector J lies only in the y direction, we have:
J · (J + √3K) = |J| |J + √3K| cos(θ)
J · (J + √3K) = 0 + |J||J + √3K| cos(θ)
0 + |J||J + √3K| cos(θ) = (1)(|J + √3K|) cos(θ)
cos(θ) = (|J + √3K|) / |J|
Taking the magnitudes of the vectors, we can substitute the values:
cos(θ) = (√(1^2 + (√3)^2)) / 1
cos(θ) = √(1 + 3) / 1
cos(θ) = √4 / 1
cos(θ) = 2 / 1
cos(θ) = 2
Therefore, the angle which vector J + √3K makes with the y-axis is π/3 radians.