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Mrrafil [7]
3 years ago
12

In an area in which electricity costs 8 cents/kilowatt-hour, a 5 kW clothes dryer runs for 90 minutes to dry a load of laundry.

How much did it cost to dry these clothes?
These are the choices
$0.20
$0.30
$0.45
$0.60
Please help!!!!!!!!!!!!
Physics
2 answers:
aliya0001 [1]3 years ago
6 0
Firstly find kilowatt*hour
so you have 5*1.5(90min=1.5hours)
7.5*8 =60
so the cost is 60 cent
8cent/kilowatthour*7.5kilowatt hour
Over [174]3 years ago
5 0
$0.60
90 mins is 1.5 hrs
5 kW × 1.5 = 7.5 kWh
7.5 kWh × $0.08 = $0.60
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1000 millicoulombs of charge passes through a point. The amount of current passing through the point is
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The amount of current passing through the point is 1 A

The amount of current passing through the point can be calculated using the formula below.

⇒ Formula:

  • Q = i/t......................... Equation 1

⇒ Where:

  • Q = Charge
  • i = current
  • t = time.

⇒ Make "i" the subject of the equation.

  • i = Qt....................... Equation 2

From the question,

⇒ Given:

  • Q = 1000 millicoulombs = 1 coulombs
  • t = 1 seconds. (Assuming the time is 1 seconds)

⇒ Substitute these values into equation 2

  • i = 1/1
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Hence, The amount of current passing through the point is 1 A.

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Answer:

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D physiological condition

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A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.
kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

W=Fd cos \theta

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

where

\Delta K is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

\Delta K=69.3 J

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s

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3 years ago
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