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butalik [34]
4 years ago
15

What is the changing relationship between volume and surface area as an object get bigger?

Physics
1 answer:
soldier1979 [14.2K]4 years ago
8 0

As the cube size increases or the cell gets bigger , then the surface area to volume ratio - SA:V ratio decreases. When an object/cell is very small, it has a large surface area to volume ratio, while a large object/ cell has a small surface area to volume ratio.


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A window in a house has a rectangular shape of 2.0 m by 1.0 m. The glass in the window is 0.5 cm thick, with a thermal conductiv
KonstantinChe [14]

Answer:

 P = 5280 W

Explanation:

The conductivity of the materials determines that heat flows from the hot part to the cold part, the equation for thermal conductivity transfer is

    P = Q / t = k A (T_{h} -T_{c}) / L

Where k is the thermal conductivity of the glass 0.8 W / ºC, A the area of ​​the window, T the temperature and L is glass thickness

Let's calculate the window area

    A = l * a

    A = 2.0 1.0

    A = 2.0 m²

Let's replace

    L = 0.5 cm (1 m / 100 cm) = 0.005 m

    P = 0.8 2 (20.5 - 4) / 0.005

    P = 5280 W

7 0
3 years ago
PLEASE HELP ASAP!!!<br> GIVING BRAINLIEST!!!<br> 40 points!!
Nana76 [90]
W = Fd = 4(2100) = 8400 J

So the answer is A) 8400 J


I was just rewriting my notes on the work lesson I did in class today, so I saw this question at the perfect time!! :)

Hope it helps!! :)
4 0
3 years ago
Read 2 more answers
A circular copper loop is placed perpendicular to a uniform magnetic field of 0.50 T. Due to external forces, the area of the lo
pogonyaev

Answer:

6.3\cdot 10^{-4} V

Explanation:

According to Faraday-Newmann-Lenz, the induced emf in the loop is given by:

\epsilon=-\frac{\Delta \Phi}{\Delta t} (1)

where \frac{\Delta \Phi}{\Delta t} is the rate of variation of the magnetic flux through the loop.

We know that the magnetic flux through the loop is given by

\Phi = BA

where B is the magnetic field and A is the area of the loop. Since the magnetic field is constant, we can write the variation of flux as

\Delta \Phi = B \Delta A

So eq.(1) becomes

\epsilon=-B\frac{\Delta A}{\Delta t}

and the problem gives us:

B=0.50 T is the magnetic field

\frac{\Delta \Phi}{\Delta t}=-1.26\cdot 10^{-3} m^2/s is the rate at which the area changes

Substituting into the equation, we find

\epsilon=-(0.50 T)(-1.26\cdot 10^{-3} m^2/s)=6.3\cdot 10^{-4} V=0.63 mV

8 0
3 years ago
A 3.00 kg cart on a frictionless track is pulled by a string so that it accelerates at 2.00 m/s/s. what is the tension in the st
AnnZ [28]
Maybe the picture helps. The blue block represents the cart with a mass of 3 kg. The person(black block) is pulling the cart to the right with a force F so that the acceleration a is 2 m/s². According to Newton's 2nd law: F = m*a.

5 0
3 years ago
Read 2 more answers
One mole of an ideal gas does 2100 J of work as it expands isothermally to a final pressure 1.00 x 10^5Pa and volume of 0.025m^3
elena-s [515]

Answer:

(A) 0.004 m^3

(B) 48.11 K

Explanation:

Work, W = 2100 J

Pressure, P = 1 x 10^5 Pa

V2 = 0.025 m^3

(A) let the initial volume of the gas is V1.

W = P (V2 - V1)

2100 = 1 x 10^5 (0.025 - V1)

0021 = 0.025 - V1

V1 = 0.025 - 0.021

V1 = 0.004 m^3

Thus, the initial volume of the gas is 0.004 m^3.

(B) Let the temperature of the gas is T

Use ideal gas equation

P1 x V1 = n x R T

1 x 10^5 x 0.004 = 1 x 8.314 x T

T = 48.11 K

Thus, the initial temperature of the gas is 48.11 K.

4 0
3 years ago
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