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andrey2020 [161]
1 year ago
10

A swimming pool heater has to be able to raise the temperatureof the 40,000 gallons of water in the pool by 10 degress Celsius.H

ow many kilowatt-hours of energy are required? (one gallon ofwater has a mass of approximately 3.8kg and the specific heat ofwater is 4186 J/kg C)
a.) 1960 kWh
b.) 1770 kWh
c.) 330 kWh
d.) 216 kWh
Physics
1 answer:
andrew11 [14]1 year ago
3 0

Answer:b)1770 kWh

Explanation:

Given

volume of water V=40,000 gallons

Temperature rise \Delta T=10^{\circ}C

c=4186 J/kg-^{\circ} C

also 1 kg mass is approximately is 1 gallon

therefore 40,000 gallon is equivalent to 3.8\times 40000 kg

heat Required to raise temperature is

Q=mc\Delta T

Q=3.8\times 40000\times 4186\times 10

Q=63,627.2\times 10^5 J

Q=63,672.2\times 10^2 KJ

Q=1767.42 kWh\approx 1770 kWh

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Two roads intersect at right angles, one going north-south, the other east-west. an observer stands on the road 60 meters south
Sloan [31]

observer is standing at distance d = 60 m south from the intersection

cyclist is travelling at speed v = 10 m/s

now after t = 8 s its displacement from intersection is given by

x = 10*8 = 80 m

so the position of cyclist makes an angle with the observer

\theta = tan^{-1}\frac{80}{60} = 53 degree

now the component of velocity of cyclist along the line joining its position with the observer is given as

v = v_o cos\phi

here

\phi = 90 -\theta

\phi = 90 - 53 = 37 degree

v = 10 cos37 = 8 m/s

so at this instant cyclist is moving away with speed 8 m/s

7 0
2 years ago
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10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of
Rina8888 [55]

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

W=Fd cos \theta

where :

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

\theta=22^{\circ} is the angle between the direction of the force and the displacement

Substituting, we find the work done

W=(87)(44)(cos 22^{\circ})=3549 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
2 years ago
A physician orders Humulin R 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many
jekas [21]

The question is incomplete, the concentration of qam and humulin is not given unless R is used concentration

Complete question:

A physician orders Humulin 50/50 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many total units of insulin are administered each morning?

Answer:

the total units of insulin admistered each morning

= 22 units of qam and humulin

Explanation:

given

44 units and Humnlin N

with concentration 50/100 = 1/2 = 0.5

∴ 44 × 0.5 ≈ 22 units in the morning

regular insulin administered each day

(22 + 35)units of qam and humulin

= 57units

5 0
1 year ago
If a 5.0 kg box is pulled simultaneously by a 10.0 N force and a 5.0 N force, then its acceleration must be?
kicyunya [14]

For this case we have that by definition:

F = ma

Where,

  • <em>m: mass of the object </em>
  • <em>a: acceleration of the object </em>
  • <em>F: summation of forces </em>

We have then:

F = 10 + 5\\F = 15 N

Then, by clearing the acceleration we have:

a = \frac {F} {m}

Substituting values we have:

a = \frac {15} {5}\\a = 3 \frac {m} {s ^ 2}

Answer:

The acceleration of the box is equal to:

a = 3 \frac {m} {s ^ 2}

6 0
1 year ago
In conventional television, signals are broadcast from towers to home receivers. Even when a receiver is not in direct view of a
fgiga [73]

(a) The diffraction decreases

The formula for the diffraction pattern from a single slit is given by:

sin \theta = \frac{n \lambda}{a}

where

\theta is the angle corresponding to nth-minimum in the diffraction pattern, measured from the centre of the pattern

n is the order of the minimum

\lambda is the wavelength

a is the width of the opening

As we see from the formula, the longer the wavelength, the larger the diffraction pattern (because \theta increases). In this problem, since the wavelength of the signal has been decreased from 54 cm to 13 mm, the diffraction of the signal has decreased.

(b) 10.8^{\circ}

The angular spread of the central diffraction maximum is equal to twice the distance between the centre of the pattern and the first minimum, with n=1. Therefore:

sin \theta = \frac{(1) \lambda}{a}

in this case we have

\lambda=54 cm = 0.54 m is the wavelength

a=5.7 m is the width of the opening

Solving the equation, we find

\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.54 m}{5.7 m})=5.4^{\circ}

So the angular spread of the central diffraction maximum is twice this angle:

\theta = 2 \cdot 5.4^{\circ}=10.8^{\circ}

(c) 0.26^{\circ}

Here we can apply the same formula used before, but this time the wavelength of the signal is

\lambda=13 mm=0.013 m

so the angle corresponding to the first minimum is

\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.013 m}{5.7 m})=0.13^{\circ}

So the angular spread of the central diffraction maximum is twice this angle:

\theta = 2 \cdot 0.13^{\circ}=0.26^{\circ}

5 0
1 year ago
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