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Solnce55 [7]
4 years ago
12

What is the average power consumption in watts of an appliance that uses 5.00 kWh of energy per day? How many joules of energy d

oes this appliance consume in a year?
Physics
1 answer:
denpristay [2]4 years ago
5 0

Answer:

(A)  power  = 0.208 kW = 208 watts

(B)  energy = 6.6 x 10^{9} joules

Explanation:

energy consumed per day = 5 kWh

(a) find the power consumed in a day

         1 day = 24 hours

        power = \frac{energy}{time}

        power = \frac{5}{24}

          power  = 0.208 kW = 208 watts

         

(b) find the energy consumed in a year

    assuming it is not a leap year and number of days = 365 days

     1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds

            energy = power x time

            energy = 208 x 31,536,000

            energy = 6.6 x 10^{9} joules

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Name TWO WEAKNESSES of the model pictured below
Nikolay [14]

Answer:

Here are a few:

1) The orbital radius of these planets is ridiculously small an in no way representative of their actual radii.

2) The planets will only line up like that once every 5200 years, making this very unrepresentative of their usual relations - although this does make their order in distance from the sun.

3) The nebulae, comet, lens flare,  and other junk in the background is incorrect.

4) If this is meant as a representation of the planets, then Pluto should not be there as it is now considered a planetoid.

5) The planets are incorrectly scaled both to each other and to the sun.

7 0
3 years ago
Which calculates the intensity of an electric field at a point where a 0.50 C charge experiences a force of 20. N?
tester [92]

Answer: 40\ N/C

Explanation:

Given

Magnitude of charge is q=0.5\ C

Force experienced is F=20\ N

Electric field intensity is the electrostatic force per unit charge

\therefore E=\dfrac{F}{q}\\\\\Rightarrow E=\dfrac{20}{0.50}\\\\\Rightarrow E=40\ N/C

Thus, the electric field intensity is 40\ N/C

6 0
3 years ago
An airplane flies eastward and always accelerates at a constant rate. At one position along its path it has a velocity of 34.3 m
Tomtit [17]

Explanation:

We'll need two equations.

v² = v₀² + 2a(x - x₀)

where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.

x = x₀ + ½ (v + v₀)t

where t is time.

Given:

v = 47.5 m/s

v₀ = 34.3 m/s

x - x₀ = 40100 m

Find: a and t

(47.5)² = (34.3)² + 2a(40100)

a = 0.0135 m/s²

40100 = ½ (47.5 + 34.3)t

t = 980 s

7 0
4 years ago
A ball is dropped from a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the
AlekseyPX

Answer:

73.5 m/s

Explanation:

The position of the first ball is:

y = y₀ + v₀ t + ½ at²

y = h + (0)(18) + ½ (-9.8)(18)²

y = h − 1587.6

The position of the second ball is:

y = y₀ + v₀ t + ½ at²

y = h + (-v) (18−6) + ½ (-9.8)(18−6)²

y = h − 12v − 705.6

Setting the positions equal:

h − 1587.6 = h − 12v − 705.6

-1587.6 = -12v − 705.6

1587.6 = 12v + 705.6

882 = 12v

v = 73.5

The second ball is thrown downwards with a speed of 73.5 m/s

8 0
3 years ago
An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the wa
Lyrx [107]

Answer:

A) 17.7 m/s

B) 15.98 m

C) Zero

E) 9.8 m/s²

Explanation:

given information

distance, h = - 34 m

time, t = 5 s

A) What is the initial speed of the egg?

h - h₀ = v₀t - \frac{1}{2} gt², h₀ = 0

- 34 = v₀ 5 - \frac{1}{2} 9.8 5²

- 34 = 5 v₀ - 122.5

v₀ = 122.5 - 34/5

    = 17.7 m/s

B) How high does it rise above its starting point?

v² = v₀² - 2gh

v = 0 (highest point)

2gh = v₀²

h = v₀²/2g

  = 17.7²/2 (9.8)

  = 15.98 m

C) What is the magnitude of its velocity at the highest point?

v = 0 (at highest point)

E) What are the magnitude and direction of its acceleration at the highest point?

g= 9.8 m/s², since the egg is moved vertically, the acceleration is the same as the gravitational acceleration.

7 0
4 years ago
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