Answer:
8.136×10⁻⁵ J
Explanation:
Applying,
Q = Cv................ Equation 1
Where Q = Charge on the capacitor, v = voltage of the battery, C = capacitance of the capacitor.
From the question,
Given: C = 6.78μF = 6.78×10⁻⁶ F, v = 12 V
Substitute these values into equation 1
Q = (6.78×10⁻⁶ )(12)
Q = 8.136×10⁻⁵ J
Hence the charge on the capacitor is 8.136×10⁻⁵ J
Answer:
C) the magnitude of the acceleration is a minimum.
Explanation:
As we know that ,the general equation of the simple harmonic motion given as
The displacement x given as
x=X sinω t
Then the velocity v will become
v= X ω cosωt
The acceleration a
a= - X ω² sinω t
The speed of the particle will be maximum when cosωt will become 1 unit.
It means that sinωt will become zero.So acceleration and displacement will be minimum.
Therefore when speed is maximum then acceleration will be minimum.
At the mean position the speed of the particle is maximum that is why kinetic energy also will be maximum and the potential energy will be minimum.
Therefore option C is correct.
Answer:
10N right
Explanation:
First, we need to find the base of the triangle to find the force acting on the right of the object.
- We have been given the hypotenuse and we need to find the adjacent, so we are using cos
- SOH, CAH, TOA
- cos(θ) = adjacent ÷ hypotenuse
Next, sub in the values (we don't know the adjacent so i've called it x):
To find x, we need to multiply both sides by 60
Now we know all the forces, we can work out the net force:
- Both of the 40N cancel out because they are opposite forces
- The force on the right of the object is 10N stronger than the force on the left (30N - 20N), so the object would move to the right
Answer:
A bar chart is orientated horizontally, whereas a column chart is arranged vertically. Sometimes “bar chart” refers to both forms. These types of charts are usually used for comparison purposes (unlike line charts, which describe change).
Answer:
0.54
Explanation:
Draw a free body diagram. There are 5 forces on the desk:
Weight force mg pulling down
Applied force 24 N pushing down
Normal force Fn pushing up
Applied force 130 N pushing right
Friction force Fnμ pushing left
Sum of the forces in the y direction:
∑F = ma
Fn − mg − 24 = 0
Fn = mg + 24
Fn = (22)(9.8) + 24
Fn = 240
Sum of the forces in the x direction:
∑F = ma
130 − Fnμ = 0
Fnμ = 130
μ = 130 / Fn
μ = 130 / 240
μ = 0.54
