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Gelneren [198K]
3 years ago
7

The specific heat of soil is 0.20 kcal/kg*C and the specific heat of water is 1.00 kcal/kg*C. This means that if 1 kg of soil an

d 1 kg of water each receive 1 kcal of energy, ideally.
Physics
1 answer:
stiv31 [10]3 years ago
8 0

Answer: The soil will be 4\°C warmer than the water.

Explanation:

The heat (thermal energy) absorbed can be found using the following equation:

Q=m.C.\Delta T

Where:

Q is the heat  

m is the mass of the element

C is the specific heat capacity of the material.

\Delta T is the variation in temperature

<u>In the case of soil we have:</u>

Q_{soil}=m_{soil}.C_{soil}.\Delta T_{soil} (1)

Where:

Q_{soil}=1 kcal

m_{soil}=1 kg

C_{soil}=0.2 kcal/kg \°C

\Delta T_{soil}

<u>In the case of water we have:</u>

Q_{water}=m_{water}.C_{water}.\Delta T_{water} (2)

Where:

Q_{water}=1 kcal

m_{water}=1 kg

C_{water}=1 kcal/kg \°C

\Delta T_{water}

Isolating \Delta T from both equations:

\Delta T_{soil}=\frac{Q_{soil}}{m_{soil}.C_{soil}} (3)

\Delta T_{soil}=\frac{1 kcal}{1 kg(0.2 kcal/kg \°C)}

\Delta T_{soil}=5\°C (4)

\Delta T_{water}=\frac{Q_{water}}{m_{water}.C_{water}} (5)

\Delta T_{water}=\frac{1 kcal}{1 kg(1 kcal/kg \°C)}

\Delta T_{water}=1\°C (6)

Comparing (4) and (6) we can find the soil will be 4\°C warmer than the water.

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4 0
3 years ago
You need to design a 60.0-Hz ac generator that has a maximum emf of 5200 V. The generator is to contain a 180-turn coil that has
OlgaM077 [116]

Answer:

0.082 T

Explanation:

Given:

Frequency of the generator (f) = 60.0 Hz

Maximum emf of the generator (E) = 5200 V

Number of turns (N) = 180

Area per turn (A) = 0.94 m²

Magnetic field magnitude (B) = ?

We know that, the angular frequency of the generator is given as:

\omega=2\pi f

Plug in the value of 'f' and solve for 'ω'. This gives,

\omega=2\pi\times 60.0=120\pi\ rad/s

Now, the maximum emf of the generator is given by the formula:

E=NAB\omega

Rewriting in terms of magnetic field, 'B', we get:

B=\frac{E}{NA \omega}

Plug in the given values and solve for 'B'. This gives,

B=\frac{5200\ V}{180\times 0.94\ m^2\times 120\pi\ rad/s}\\\\B=\frac{5200\ V}{63786.897}\\\\B=0.082\ T

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8 0
4 years ago
A helium balloon ride lifts up passengers in a basket. Assume the density of air is 1.28 kg/m3 and the density of helium in the
max2010maxim [7]

Answer:

V = 381.70 m³

Explanation:

ρ air = 1.28 kg / m³

ρ helium = 0.18 kg / m³

R = 4.5 m

Vb = 0.068 m³

mb = 123 kg

To determine the volume of helium in the balloon when fully inflated

V = 4 / 3 π * R ³

V = 4 * π / 3 ( 4.5 m )³

V = 381.70 m³

To determine the mass total

m = ρ helium * V

m = 0.18 kg / m³  * 381.70  m³

m = 68.70 kg

mt = ( 68.70 + 123 )kg

mt = 191.70 kg

3 0
4 years ago
A box with the mass of 20 kg at 5 m is lifted to 20 m. How much work was 7 points<br> done?
kirza4 [7]

Answer:

2,900\: \mathrm{J}

Explanation:

Work is given by the equation W=F\Delta x where F is force and \Delta x is displacement.

Displacement is defined as change in position. In this case, the box moves from 5m to 20m. Its displacement is 20-5=15\:\mathrm{m}.

The force acting on the box is the force of gravity, given as F_g=mg.

Therefore, the total work done is:

W=F\Delta x=mg\Delta x = 20\cdot 9.81\cdot 15=2,943=\fbox{$2,900\:\mathrm{J}$}(two significant figures).

3 0
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