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faust18 [17]
3 years ago
13

1.Narysuj obraz ołówka o długości 6 cm znajdującego się w odległości 8 cm od soczewki o ogniskowej 6 cm

Physics
1 answer:
Aliun [14]3 years ago
5 0

Answer:

free quotes from carpet repair contractors at door in

Explanation:

india is <em>the</em><em> </em><em>most</em><em> </em><em>powerful</em><em> </em><em>and</em><em> </em><em>I</em><em> </em><em>have</em><em> </em><em>seen</em><em> </em><em>a</em><em> </em><em>great</em><em> </em><em>deal</em><em> </em><em>in</em><em> </em><em>hindi</em><em> </em><em>for</em><em> </em><em>girl</em><em> </em><em>in</em><em> </em><em>a</em><em> </em><em>world</em><em> </em><em>of</em><em> </em><em>sentence</em><em> </em><em>since</em><em> </em><em>the</em><em> </em><em>poem</em><em> </em><em>was</em><em> </em><em>a</em><em> </em><em>novel</em><em> </em><em>that</em><em> </em><em>led</em><em> </em>

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A heat pump is to be used for heating a house in winter. The house is to be maintained at 70°F at all times. When the temperatur
Anna35 [415]

Answer:

\dot{W_{H} } = 4244.48 Btu/h

Explanation:

Temperature of the house, T_{H} = 70^{0} F

Convert to rankine, T_{H} = 70^{0}+ 460 = 530 R

Heat is extracted at 40°F i.e T_{L} = 40^{0}F  = 40 + 460 = 500 R

Calculate the coefficient of performance of the heat pump, COP

COP = \frac{T_{H} }{T_{H} - T_{L}  } \\COP = \frac{530 }{530 - 500  }\\ COP = \frac{530}{30} \\COP = 17.67

The minimum power required to run the heat pump is given by the formula:

\dot{W_{H} } = \frac{\dot{Q_{H} }}{COP} \\...............(*)

Where the heat losses from the house, \dot{Q_{H} } = 75,000 Btu/h

Substituting these values into * above

\dot{W_{H} } = \frac{75000}{17.67} \\ \dot{W_{H} } = 4244.48 Btu/h

3 0
3 years ago
The escape speed from an object is v2 = 2GM/R, where M is the mass of the object, R is the object's starting radius, and G is th
Rom4ik [11]

Answer:

Approximate escape speed = 45.3 km/s

Explanation:

Escape speed

        v=\sqrt{\frac{2GM}{R}}

Here we have

   Gravitational constant = G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²

   R = 1 AU = 1.496 × 10¹¹ m

   M = 2.3 × 10³⁰ kg

Substituting

    v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 2.3\times 10^{30}}{1.496\times 10^{11}}}=4.53\times 10^4m/s=45.3km/s

Approximate escape speed = 45.3 km/s

6 0
3 years ago
Why do we use atomic models?
vlabodo [156]
Models help us to understand systems and their properties
3 0
2 years ago
Two tuning forks produce sounds of wavelengths of 3.4 meters and 3.3 meters.
KengaRu [80]

Answer:

Explanation:

wavelength, λ = 3.4 m

wavelength, λ' = 3.3 m

Speed, v = 340 m/s

f = v / λ = 340 / 3.4 = 100 Hz

f' = v / λ' = 340 / 3.3 = 103.03 Hz

Frequency of beat, n = f' - f = 103.03 - 100 = 3.03 Hz

5 0
3 years ago
A coin is placed on a large disk which rotates uniformly at a rate of 1 rot/s. The coefficient of friction between the coin and
s2008m [1.1K]

Answer:

r = 0.02 m

Explanation:

from the question we have :

speed = 1 rps = 1x 60 = 60 rpm

coefficient of friction (μ) = 0.1

acceleration due to gravity (g) = 9.8 m/s^{2}

maximum distance without falling off (r) = ?

to get how far from the center of the disk the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force

mv^2 / r =  m x g x μ

v^2 / r =  g x μ  .......equation 1

where

velocity (v) = angular speed (rads/seconds) x radius

angular speed (rads/seconds) = (\frac{2π}{60} ) x rpm

angular speed (rads/seconds) = (\frac{2 x π}{60} ) x 60 = 6.28 rads/ seconds

now

velocity = 6.28 x r = 6.28 r

now substituting the value of velocity into equation 1

v^2 / r =  g x μ

(6.28r)^2 / r = 9.8 x 0.1

39.5 x r = 0.98

r = 0.02 m

6 0
3 years ago
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