Answer:
θ = 14.27°
Explanation:
The only force acting on the puck is the gravitational force. Since the track is banked with an angle θ, we have to separate the components of the weight.
For the sake of simplicity, I will denote the perpendicular direction to the truck as the y-direction, and the direction along the radius as the x-direction.
So, the free-body diagram of the puck is as follows:
1- x-component of the weight of the puck: mgsinθ
2- y-component of the weight of the puck: mgcosθ
3- Normal force in the y-direction perpendicular to the track.
Since there is no motion on the y-direction, normal force is equal to the y-component of the weight of the puck.
The x-component of the weight of the puck is equal to the centripetal force according to Newton's Second Law:
Substituting the variables given in the question, the angle of the track can be found:
If a baseball dropped from the roof of a tall building takes 3.1 seconds to hit the ground, the height of the building would be 47.1 m.
According to Newton's second law of motion,
Where, s is the distance covered by the body, u is its initial velocity, t is the time taken by the body, and a is the acceleration of the body.
Here, the distance covered will be equal to the height of the building.
Since the ball was dropped, its initial velocity will be zero.
Also, the acceleration of the ball is the acceleration due to gravity, .
Substituting the values,
Therefore, the height of the building would be 47.1 m.
Learn more about Newton,s laws of motion here:
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Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.