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seropon [69]
3 years ago
7

The rate (in cubic feet per hour) that a spherical snowball melts is proportional to the snowball's volume raised to the 2/3 pow

er. (This assumes that the rate snow melts is proportional to the area exposed to the air -- i.e. the surface area of the snowball.) If a snowball of radius 1 foot (volume 4/3 \[Pi] cubic feet) melts in 6 hours, how long will it take a snowball of raduis 3 feet (volume 36 cubic feet) to melt?
Physics
1 answer:
Darina [25.2K]3 years ago
6 0

Answer:

A 3 feet radius snowball will melt in 54 hours.

Explanation:

As we can assume that the rate of  snowball takes to melt is proportional to the surface area, then the rate for a 3 feet radius will be:

T= A(3 ft)/A(1 ft) * 6 hr

A is the area of the snowballs. For a spherical geometry is computing as:

A=4.pi.R^2

Then dividing the areas:

A(3 feet)/A(1 foot) = (4 pi (3 ft)^2)/(4 pi (1 ft)^2) =  (36pi ft^2)/(4pi ft^2)= 9

Finally, the rate for the 3 feet radius snowball is:

T= 9 * 6 hr = 54 hr

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IRINA_888 [86]

Initial height: 66.5 m

Explanation:

The problem can be solved by using the principle of conservation of energy.

If we neglect air resistance, the total mechanical energy of the car is conserved during the fall, therefore we can write:

K_i + U_i = K_f + U_f

where :

K_i = 0 is the kinetic energy of the car at the top (it starts from rest)

U_i = mgh is the gravitational potential energy of the car at the top, with:

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h = the heigth of the car

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v = 130 km/h final speed of the car

U_f = 0 is the gravitational potential energy of the car at the bottom

Re-arranging the equation,  we find

mgh=\frac{1}{2}mv^2

and we have:

g=9.8 m/s^2\\v = 130 km/h = 36.1 m/s

Solving for h, we find the initial height of the car:

h=\frac{v^2}{2g}=\frac{36.1^2}{2(9.8)}=66.5 m

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3 years ago
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What is the velocity of a wave with a wavelength of 3.7 m and a frequency of 22.5 Hz?
ExtremeBDS [4]

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2 years ago
In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? pls help
Alexxx [7]

Answer:

Below

Explanation:

First draw the vectors that represent both electric fields.

E1 is the elictric field created by q1, E2 is the one created by q2.

● q1 is negative so E1 will point from P.

● q2 is positive so E2 will point out of P

(Picture below)

■■■■■■■■■■■■■■■■■■■■■■■■■■

The resulting electric field is equal to the sum of the two fields since both vectors are colinear.

Let E be the total field.

● E = E1 + E2

The formula of the electric field intensity is:

● E = K ×(q/d^2)

-K is Coulomb's constant

-d is the distance between the charge and the object ( here P)

-q is the charge

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E1 = K × (q1/d1^2)

The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)

Coulombs constant is 9×10^9 m^2/C^2

● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]

● E1 = -359.43 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E2 = K ×(q2/d^2)

The distance between q2 and P is 0.25 m.

● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]

● E2 = 463.68 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E = E1 + E2

● E = -359.43+463.68

● E = 105.25 N/C

4 0
3 years ago
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