The new sweet potato plant grew from the root of the original through

The freezer compartment in a conventional refrigerator can be modeled as a rectangular cavity 0.3 m high and 0.25 m wide with a

Mkey [24]

Answer:

Thickness of Styrofoam insulation is 0.02741 m.

Explanation:

Given that,

Height = 0.25 m

Depth = 0.5 m

Power = 400 W

Temperature = 33°C

We need to calculate the area of Styrofoam

Using formula of area

$A=2(lb+bh+hl)$

Put the value into the formula

$A=2(0.3\times0.5+0.25\times0.5+0.5\times0.3)$

$A=0.85\ m^2$

Inner surface temperature of freezer

$T_{i}=-10°C=263\ K$

Outer surface temperature of freezer

$T_{o}=33+273=306\ K$

We need to calculate the thickness of Styrofoam insulation

Using Fourier law,

$q=\dfrac{kA}{L}(T_{o}-T_{i})$

$L=\dfrac{kA}{q}(T_{o}-T_{i})$

Put the value into the formula

$L=\dfrac{0.30\times0.85}{400}(306-263)$

$L=0.02741\ m$

Hence, Thickness of Styrofoam insulation is 0.02741 m.

3 0

3 years ago

The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5

ValentinkaMS [17]

The average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.

<h3>Acceleration of the box</h3>

The acceleration of the box is calculated as follows;

vf² = vi² + 2as

a = (vf² - vi²)/2s

a = (11.5² - 13²) / (2 x 8.5)

a = -2.16 m/s²

<h3>Time of motion of the box</h3>

The time taken for the box to travel is calculated as follows;

a = (vf - vi)/t

t = (vf - vi) / a

t = (11.5 - 13) / (-2.16)

t = 0.69 s

<h3>Average power supplied by the friction</h3>

P = Fv

P = (ma)(vf - vi)

P = (1 x -2.16) x (11.5 - 13)

P = 3.24 W

Thus, the average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.

Learn more about average power here: brainly.com/question/19415290

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7 0

1 year ago

What is the wavelength of an earthquake that shakes you with a frequency of 10.0 Hz and gets to another city 84.0 km away in 12.

Harrizon [31]

Answer:

700 m.

Explanation:

Wavelength: This can be defined as the distance in meter required by a wave to complete one oscillation. The S.I unit of of wavelength is meter (m).

the expression connecting velocity, frequency and wavelength is given as,

v = λf......................... Equation 1

Where v = velocity of the earthquake, λ = wavelength of the earthquake, f = frequency of the earthquake.

make λ the subject of the equation

λ = v/f ................. Equation 2

recall,

v = distance/time

v = d/t ................. Equation 2

Where d = distance, t -= time.

Given: d = 84 km = 84000 m, t = 12 s.

Substitute into equation 2

v = 84000/12

v = 7000 m/s.

Also given: f = 10 Hz.

Substitute into equation 2

λ = 7000/10

λ = 700 m.

Hence the wavelength of the earthquake = 700 m.

5 0

3 years ago

Read 2 more answers

Determine the number of ways to throw two die and get the number 11, as well as the probability of getting 11.

Ivenika [448]

Answer:

1/18

Explanation:

The number of ways in which two die can be thrown is the sample space of the experiment

(1,1), (1,2) (1,3), (1,4) (1,5) (1,6)

(2,1), (2,2) (2,3), (2,4) (2,5) (2,6)

(3,1), (3,2) (3,3), (3,4) (3,5) (3,6)

(4,1), (4,2) (4,3), (4,4) (4,5) (4,6)

(5,1), (5,2) (5,3), (5,4) (5,5) (5,6)

(6,1), (6,2) (6,3), (6,4) (6,5) (6,6)

From here it can be seen that there are only two cases where the sum on the two die is 11 i.e., (6,5) and (5,6)

$\text{Probability of getting 11}=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\\\Rightarrow \text{Probability of getting 11}=\frac{2}{36}\\\Rightarrow \text{Probability of getting 11}=\frac{1}{18}=0.056$

∴ Probability of getting 11 is<u> 1/18</u>

8 0

3 years ago

parallel-plate capacitor is made of two square plates 25 cm on a side and 1.0 mmapart. The capacitor is connected to a 50.0-V ba

noname [10]

Answer:

6.9 x 10^-7 J

3.5 x 10^-7 J

Explanation:

<u>Identify the unknown: </u>

The energies stored in the capacitor before and after the plates are pulled farther apart

<u>List the Knowns: </u>

Voltage of the battery: V = 50 V

Area of the plates: A = 0.25 x 0.25 = 0.0625 m^2

Original distance between the plates: d = 1 mm = 10^-3 m

New distance between the plates: d = 2 mm = 2 x 10^-3 m

Permittivity of free space: ∈o = 8.85 x 10^-12 C^2/Nm^2-

<u>Set Up the Problem: </u>

Capacitance of a parallel-plate capacitor:

C=∈o*A/d

Energy stored in a capacitor:

U_c=(1/2)*V^2*C

<u>Solve the Problem: </u>

<u>Before the plates are pulled farther apart: </u>

C = 8.85 x 10^-12 x 0.0625/10^-3 = 5•53 x 10^-10 F

U_c = (1/2) x (50)^2 x 5.53 x 10^-10= 6.9 x 10^-7 J

<u>After the plates are pulled farther apart: </u>

C = 8.85 x 10^-12 x 0.0625/2*10^-3 = 2•77 x 10^-10 F

U_c = (1/2) x (50)^2 x 2•77 x 10^-10 = 3.5 x 10^-7 J

The energy decrease because the capacitance decrease, so the stored charge decrease and transferred to the battery

3 0

3 years ago