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slava [35]
3 years ago
10

. An 1800-W toaster, a 1400-W electric frying pan, and a 75-W lamp are plugged into the same outlet in a 15-A, 120-V circuit. (T

he three devices are in parallel when plugged into the same socket.). (a) What current is drawn by each device? (b) Will this combination blow the 15-A fuse?
Physics
2 answers:
blagie [28]3 years ago
7 0

Answer:

Toaster: 15A

Electric Frying Pan: 11.7A

Light bulb: 0.625A

(B) Yes the combination will blow the fuse.

Explanation:

The foremula for calculating the power consumed by electrical appliances in a circuit is P = IV

The three appliances are connected in parallel which means that they will all have the same voltage but different current flowing through them.

The current I = P/V = power consumed by appliance / Voltage supplied to it.

The current drawn by:

the toaster = 1800/120 = 15A

the frying pan = 1400/120 = 11.7A

the light bulb = 75/120 = 0.625A

(B) To know if the combination of all three appliances will blow the fuse or not, we add all the current drawn by each of the electrical appliances. That ic the current drawn by the toaster, frying pan and the light bulb which is 15A + 11.7A + 0.625A = 27.3A. This current demand is greater than the fuse rating and as a result this combination of appliances will blow the fuse.

s2008m [1.1K]3 years ago
6 0

Answer:

Explanation:

The devices are connected in parallel, the same voltage will flow through them

1) I, current in the 1800 W toaster

P, power = I V

where p = 1800 W

1800 /120 = I

I drawn by the toaster = 15 A

2) I, current drawn by the 1400 W frying pan = 1400 / 120 = 11.667 A

3) I, current drawn by 75 W = 75 / 120 = 0.625 A

B) Total current drawn by the circuit = 15 A + 11.667 A + 0.625 A = 27.292 A which is greater than 15-A hence the combination will blow the fuse

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A watermelon is dropped off of a 50 ft bridge, and it explodes upon impact with the ground. How fast was it traveling in mph upo
Drupady [299]

Answer: 56.72 ft/s

Explanation:

Ok, initially we only have potential energy, that is equal to:

U =m*g*h

where g is the gravitational acceleration, m the mass and h the height.

h = 50ft and g = 32.17 ft/s^2

when the watermelon is near the ground, all the potential energy is transformed into kinetic energy, and the kinetic energy can be written as:

K = (1/2)*m*v^2

where v is the velocity.

Then we have:

K = U

m*g*h = (m/2)*v^2

we solve it for v.

v = √(2g*h) = √(2*32.17*50) ft/s = 56.72 ft/s

6 0
3 years ago
An 80-kilogram skier slides on waxed skis along a horizontal surface of snow at constant velocity while pushing with his poles.
ivann1987 [24]

Answer:

The force F is created by the reaction of the Earth to the thrust of the rods, whereby the thrust is created by a force of action and reaction.

Explanation:

To answer this question, let's write Newton's second law of the two axes

Y Axis  

        Fy + N - W = 0

        Fy + N = W

X axis

       Fx - fr = 0

      Fx = fr

The force F is created by the reaction of the Earth to the thrust of the rods, whereby the thrust is created by a force of action and reaction.

   The direction of this force is along the length of the rods that are in an Angle, where the x and y components of the force come from

In general this force is small because the rubbing of the skis is small

8 0
3 years ago
A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.
cupoosta [38]

Answer:

a) The x and y components of the momentum are 8.731\,\frac{kg\cdot m}{s} and -11.661\,\frac{kg\cdot m}{s}, respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

\vec p = m\cdot \vec v (1)

Where:

\vec p - Vector momentum, measured in kilogram-meters per second.

m - Mass of the particle, measured in kilograms.

\vec v - Vector velocity, measured in meters per second.

If we know that m = 2.93\,kg and \vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right], then the momentum is:

\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]

\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]

The x and y components of the momentum are 8.731\,\frac{kg\cdot m}{s} and -11.661\,\frac{kg\cdot m}{s}, respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}} (2)

\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right) (3)

Where:

\|\vec p\| - Magnitude of momentum, measured in kilogram-meters per second.

\theta - Direction of momentum, measured in sexagesimal degrees.

If we know that p_{x} = 8.731\,\frac{kg\cdot m}{s} and p_{y} = -11.661\,\frac{kg\cdot m}{s}, then the magnitude and direction of momentum are, respectively:

\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}

\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}

\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)

\theta \approx 306.823^{\circ}

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

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