Answer:
The motion of the block is downwards with acceleration 1.7 m/s^2.
Explanation:
First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.
Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.
Now, let’s investigate the free-body diagram of the block.
Along the x-direction, there are two forces: The x-component of the block’s weight and the kinetic friction force. Therefore,
As for the static friction, we will consider the angle 31.8, but just before the block starts the move.
Look at your speedometer for say, a couple of seconds. Depends on whether or not you are moving on average at a constant speed (speedo won't change much) or whether you're in a polluting traffic jam/queue in which case the speedo will go up and down like a yo yo. to determine the speed, you'd probably need to plot the speed on the speedo against the times at which the speedo speeds were read from the speedo.
Cups
teaspoon
tablespoon
liters
milliliters
gallons
pints
tons
inches
Answer:
X₃₁ = 0.58 m and X₃₂ = -1.38 m
Explanation:
For this exercise we use Newton's second law where the force is the Coulomb force
F₁₃ - F₂₃ = 0
F₁₃ = F₂₃
Since all charges are of the same sign, forces are repulsive
F₁₃ = k q₁ q₃ / r₁₃²
F₂₃ = k q₂ q₃ / r₂₃²
Let's find the distances
r₁₃ = x₃- 0
r₂₃ = 2 –x₃
We substitute
k q q / x₃² = k 4q q / (2-x₃)²
q² (2 - x₃)² = 4 q² x₃²
4- 4x₃ + x₃² = 4 x₃²
5x₃² + 4 x₃ - 4 = 0
We solve the quadratic equation
x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2 5
x₃ = [-4 ± 9.80] 10
X₃₁ = 0.58 m
X₃₂ = -1.38 m
For this two distance it is given that the two forces are equal
