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2 years ago

What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.

Physics

1 answer:

Volgvan2 years ago

4 0

Answer:

Density = 1.1839 kg/m³

Mass = 227.3088 kg

Specific Gravity = 0.00118746 kg/m³

Explanation:

Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³

Now, from tables, density of air at 25°C is 1.1839 kg/m³

Now formula for density is;

ρ = mass(m)/volume(v)

Plugging in the relevant values to give;

1.1839 = m/192

m = 227.3088 kg

Formula for specific gravity of air is;

S.G_air = density of air/density of water

From tables, density of water at 25°C is 997 kg/m³

S.G_air = 1.1839/997 = 0.00118746 kg/m³

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10 months ago

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

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

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

6 0

2 years ago

A circular loop of flexible iron wire has an initial circumference of 165cm , but its circumference is decreasing at a constant

ArbitrLikvidat [17]

Answer:

emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise

Explanation:

Given data

initial circumference = 165 cm

rate = 12.0 cm/s

magnitude = 0.500 T

tome = 9 sec

to find out

emf induced and direction

solution

we know emf in loop is - d∅/dt ........1

here ∅ = ( BAcosθ)

so we say angle is zero degree and magnetic filed is uniform here so that

emf = - d ( BAcos0) /dt

emf = - B dA /dt ..............2

so area will be

dA/dt = d(πr²) / dt

dA/dt = 2πr dr/dt

we know 2πr = c,

r = c/2π = 165 / 2π

r = 26.27 cm

c is circumference so from equation 2

emf = - B 2πr dr/dt ................3

and

here we find rate of change of radius that is

dr/dt = 12/2π = 1.91 $10^{-2}$ cm/s

so when 9.0s have passed that radius of coil = 26.27 - 191 (9)

radius = 9.08 $10^{-2}$ cm

so now from equation 3 we find emf

emf = - (0.500 ) 2π(9.08 $10^{-2}$ ) 1.91 $10^{-2}$

emf = - 0.005445

and magnitude of emf = 0.005445 V

emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise

4 0

2 years ago