PLEASE HELP HDJFJDJDJJCJDJDJFJD

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

Darina [25.2K]

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

6 0

2 years ago

Your oven has a power rating of 5000 watts. How many kilowatts is this ?

Gekata [30.6K]

1,000 watts = 1 kilowatt
2,000 watts = 2 kilowatts
3,000 watts = 3 kilowatts
4,000 watts = 4 kilowatts
<em>5,000 watts = 5 kilowatts</em>

8 0

2 years ago

Find the frequency of a wave with a period of 0.5 seconds

strojnjashka [21]

Frequency of the wave is 2 per second

Explanation:

Frequency is the number of times waves pass at a particular point of time. Here, time period = 0.5 s

Frequency is given by the formula

f = 1/T, where f is the frequency and T is the time period

⇒ f = 1/0.5 = 2 per second

8 0

2 years ago

The radius of the thinner wire is 0.22 mm and the radius of the thick wire is 0.55 mm. There are 4.0 x 1028 mobile electrons per

Natasha_Volkova [10]

Answer:

0.2631 N/C

Explanation:

Given that:

The radius of the wire r = 0.22 mm = 0.22 × 10⁻³ m

The radius of the thick wire r' = 0.55 mm = 0.55 × 10⁻³ m

The numbers of electrons passing through B, N = 6.0 × 10¹⁸ electrons

Electron mobility μ = 6.0 x 10-4 (m/s)/(N/C)

= 0.0006

The number of electron flow per second is calculated as follows:

$I = \frac{q}{t}$

$I = \frac{Ne}{t}$

$I = \frac{6*10^{18}(1.6*10^{-18})C}{1 \ s}$

$I = 0.96 \ A$

The magnitude of the electric field is:

E = $\frac{I}{ \mu n eA}$

E = $\frac{I}{ \mu n e(\pi r^2)}$

E = $\frac{0.96}{(0.0006 m/s N/C ) (4*10^{28})(1.6*10^{-19}C)(0.55*10^{-3}m)^2}$

E = 0.2631 N/C

8 0

3 years ago

A cylindrical rod of length L is connected across a fixed potential difference, creating a current I through the rod. What would

Inessa05 [86]

Answer:

The value of current through the rod becomes half

$i' = \frac{i}{2}$

Explanation:

As per Ohm's law we know that the current through a resistor is given as

$i = \frac{V}{R}$

here we know that

$R = \rho \frac{L}{A}$

here we know that the length of the cylinder is L and area is A so the value of current through the rod is given as

$i = \frac{V A}{\rho L}$

now we have change the length of the conductor to twice of initial value and rest all parameters will remain the same

so we will have

$i' = \frac{VA}{\rho (2L)}$

now from above two equations we have

$\frac{i}{i'} = 2$

so new current will become

$i' = \frac{i}{2}$

3 0

3 years ago