I need help on two problems :P
1. A spherical balloon is partially blown up and its surface area is measured. More air is added increasing the volume of the balloon. If the surface area of the balloon expands by a factor of 2.0 during this procedure, by what factor does the radius of the balloon change?
2. A poster advertising a student election candidate is too large according to the election rules. The candidate is told she must reduce the length and width of the poster by 20.0%. By what percentage will the area of the poster be reduced?
1. Area is proportional ot the square of radius. Therefore, to double the area, muliply the radius by sqrt2 = 1.414
2. (0.8)^2 = 0.64 is (new area)/(old area)
The reduction percent is 1-0.64 = 36%
This is really not physics.
Sure, I can help you with both problems! Let's take them one at a time.
1. To find the change in radius, we can use the fact that the surface area of a sphere is given by the equation:
Surface Area = 4πr^2
Let's denote the initial radius as r1 and the final radius as r2. We are given that the surface area expands by a factor of 2.0, so we can write:
2 * Surface Area_initial = Surface Area_final
Now let's substitute the equations for the surface areas using the radii:
2 * (4πr1^2) = 4πr2^2
Simplifying the equation:
8πr1^2 = 4πr2^2
Dividing both sides by 4π:
2r1^2 = r2^2
Taking the square root of both sides:
√(2r1^2) = √(r2^2)
√2 * r1 = r2
Therefore, the final radius (r2) is √2 times the initial radius (r1). So, the radius of the balloon changes by a factor of √2.
2. To find the percentage reduction in the area of the poster, we need to use the formula for the area of a rectangle:
Area = length * width
Let's denote the initial length as L1, the initial width as W1, and the reduction percentage as P. We are given that both the length and width will be reduced by 20.0%, so we can write:
Reduced Length = L1 - (P/100) * L1
Reduced Width = W1 - (P/100) * W1
The final area, denoted as A2, is given by:
A2 = Reduced Length * Reduced Width
= (L1 - (P/100) * L1) * (W1 - (P/100) * W1)
Expanding the equation:
A2 = (L1 - (P/100) * L1) * (W1 - (P/100) * W1)
= L1 * W1 - ((P/100) * L1 * W1) - ((P/100) * W1 * L1) + ((P/100)^2 * L1 * W1)
= L1 * W1 - 2*(P/100) * L1 * W1 + (P/100)^2 * L1 * W1
Simplifying the equation:
A2 = L1 * W1 * ( 1 - 2(P/100) + (P/100)^2 )
The initial area, denoted as A1, is given by:
A1 = L1 * W1
Therefore, the percentage reduction in the area is:
((A1 - A2)/A1) * 100
= ((L1 * W1 - L1 * W1 * (1 - 2(P/100) + (P/100)^2))/ (L1 * W1) ) * 100
= ((2(P/100) - (P/100)^2) * 100
So, the area of the poster will be reduced by (2(P/100) - (P/100)^2) * 100 percent.
I hope this helps! Let me know if you have any further questions.
Sure! Let's solve these two problems step by step.
1. To find the factor by which the radius of the balloon changes, we need to understand the relationship between the radius, surface area, and volume of a sphere.
For a sphere, the surface area (A) is given by the formula: A = 4πr^2, where r is the radius.
The volume of a sphere (V) is given by the formula: V = (4/3)πr^3.
In this problem, the surface area expands by a factor of 2.0. Let's call the initial radius of the balloon r1, and the final radius after expanding as r2.
We can write the equation for the change in surface area as follows: (4πr2^2) / (4πr1^2) = 2.0.
Now, canceling out the common terms, we have: r2^2 / r1^2 = 2.0.
To find the factor by which the radius changes, take the square root of both sides: (r2 / r1) = √2.0.
Therefore, the radius increases by a factor of √2.0, which is approximately 1.414.
2. To calculate the percentage reduction in the area of the poster, we need to understand the relationship between the length, width, and area of a rectangle.
Let's assume the initial length of the poster is L1, the initial width is W1, and the final length and width after reducing by 20% are L2 and W2, respectively.
The area of a rectangle (A) is given by the formula: A = L * W.
In this problem, the dimensions reduce by 20%. This means L2 = 0.8 * L1 and W2 = 0.8 * W1.
To find the percentage reduction in the area, we need to compare the initial area with the final area.
The initial area is A1 = L1 * W1, and the final area is A2 = L2 * W2.
Substituting the values, we have: A2 = (0.8 * L1) * (0.8 * W1) = 0.64 * (L1 * W1).
Therefore, the final area is 0.64 times the initial area. To find the percentage reduction, we subtract 1 from the fraction and multiply by 100.
Percentage reduction in the area = (1 - 0.64) * 100 = 36%.
So, the area of the poster will be reduced by 36% when the length and width are reduced by 20%.