Answer:
a) 166.4 s
b) (2.155 × 10⁷) s
Explanation:
15600 KWh for a year,
1 year consists of 365 × 24 hours = 8760 hours.
So, the power consumed in a year for an average household = (Energy/time)
= (15600/8760) = 1.781 KW = 1781 W
a) If the average rate of energy consumed by the house was instead diverted to lift a 1.80 × 10 3 kg car 16.8 m into the air, how long would it take
The power required for this lifting = (mgh/t)
m = 1800 kg
g = 9.8 m/s²
h = 16.8 m
t = ?
P = 1781 W
1781 = (1800×9.8×16.8)/t
t = (296,352/1781)
t = 166.4 s
b) how long would it take to lift a loaded Boeing 747 airplane, with a mass of 4.05 × 10 5 kg , to a cruising altitude of 9.67 km
The power required for this lifting = (mgh/t)
m = 405000 kg
g = 9.8 m/s²
h = 9.67 km = 9670 m
t = ?
P = 1781 W
1781 = (405000×9.8×9670)/t
t = (38,380,230,000/1781)
t = 21,549,820 s = (2.155 × 10⁷) s
Hope this Helps!!!
Answer:
Explanation:
a )
Radius of the sun = .69645 x 10⁹ m .
600 times = 600 x .69645 x 10⁹ m
= 4.1787 x 10¹¹ m .
surface area A = 4π (4.1787 x 10¹¹)²
= 219.317 x 10²²
energy radiated E = σ A Τ⁴
= 5.67 x 10⁻⁸ x 219.317 x 10²² x (3000)⁴
= 100695 x 10²⁶ J
To know the wavelength of photon emitted
= 2.89777 x 10⁻³ / 3000
= 966 nm
= 1275 /966 eV
1.32 x 1.6 x 10⁻¹⁹ J
= 2.112 x 10⁻¹⁹ J
No of photons radiated = 100695 x 10²⁶ / 2.112 x 10⁻¹⁹
= 47677.5 x 10⁴⁵
= .476 x 10⁵⁰ .
b )
energy radiated by our sun per second
E₂ = σ A 5800⁴
energy radiated by Betelgeuse per second
E₁ = σ x 600²A x 3000⁴
E₁ / E₂ = σ x 600²A x 3000⁴ / σ A 5800⁴
= 36 X 10⁴ x 3⁴ x 10¹² / 58⁴ x 10⁸
= 25.76 x 10⁸ x 10⁻⁵
= 25760 times .
The surface tension acts to hold the surface intact. Capillary action occurs when the adhesion to the surface material is stronger than the cohesive forces between the water molecules. ... Water wants to stick to the glass and surface tension will push the water up, until the force of gravity prevents further rise.
Narrower tube openings allow capillary action to pull water higher
Answer:
0.056 miles away
Explanation:
From sound wave,
v = 2x/t .................................. Equation 1
Where v = velocity of sound in air, x = distance of echo, t = time.
making x the subject of the equation,
x = 2v/t........................... Equation 2.
Given: v = 344 m/s, t = 7.6 s.
Substituting into equation 2
x = 2(344)/7.6
x = 90.53 m.
x = 90.53/1609.344
x = 0.056 mile.
Thus the lighting strike 0.056 miles away
The calorie was originally defined as the amount of heat required at a pressure of 1 standard atmosphere to raise the temperature of 1 gram of water 1° Celsius. ... Since 1925 this calorie has been defined in terms of the joule, the definition since 1948 being that one calorie is equal to approximately 4.2 joules.
