Answer:
2.96 × 10^4 N
Explanation:
1 atm = 101325 N/m², pressure inside the airtight room = 1.02 atm, pressure outside due to hurricane = 0.91 atm
net pressure directed outward = P inside - P outside
net pressure = 1.02 - 0.91 = 0.11 atm
where 1 atm = 101325N/m²
0.11 atm = 0.11 × 101325 N/m² = 11145.75 N/m²
area of the square wall = l × l where l is the length of the wall in meters = 1.63 × 1.63 = 2.6569
net pressure = net force / area
make net force subject of the formula
net force = net pressure × area = 11145.75 × 2.6569 = 2.96 × 10 ^4 N
Well, as the waves move it moves the rope as if its trying to take shape of it. Since the rope it light it will move along the ocean and the ocean will keep pushing up on the rope. (even without the waves the water is pushing the rope up so it can take its shape)
Maybe that'll help
Answer:
<h2>A. Nearsightedness</h2>
Explanation:
A nearsightedness is an eye defect that occurs when someone is only able to see close ranged object but not far distance object. According to the question, if the length of my eye decreases slightly as I age, this means there is a possibility that I will find it difficult to view a far distance object as I age.
At 70, once my eyes had decreased slightly in length, this means I will only be able to see close ranged object but not far distant object, showing that I am now suffering from nearsightedness according to its definition above.
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.
