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3 years ago

. An electric geyser was used 2 hours a day once a week. The power rating

Physics

1 answer:

iragen [17]3 years ago

3 0

Answer:

Annual energy = 1460000 Joules

Explanation:

Given the following data;

Time = 2 hours

Power = 2000 Watts

To find the energy consumption of the geyser;

Energy = power * time

Energy = 2000 * 2

Energy = 4000 Joules

Next, the annual energy consumption is;

Annual energy = 4000 * 365

Annual energy = 1460000 Joules

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The object is not accelerating

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3 years ago

The body weighing 2 kg moves through the horizontal surface and crosses the path x = 75 cm The coefficient of friction of the bo

Taya2010 [7]

The kinetic energy of the body in definitive position is 4.24 J.

Explanation:

As per the work energy theorem, the work done on any system or object to move it from one position to another is equal to the change in kinetic energy of the object. In this case, the body weighing 2 kg is moved over an horizontal surface for a distance of 75 cm. As there will be frictional force acting on the body while moving over the surface. This frictional force multiplied by the distance the object is moved will give the work done on the body.

Frictional force = Coeffficent of friction × Normal force.

As the weight of the body is 2 kg, the normal force acting on it will be mass multiplied with acceleration due to gravity.

Frictional force = - 0.8×9.8 × 2 =-15.68 N

So the work done will be the product of frictional force with the displacement of 75 cm or 0.75 m.

Work done = Frictional force × Displacement

Work done = -15.68×0.75 = -11.76 J.

So the work is done by the object.

If the kinetic energy of the body at starting is 16 J, then the kinetic energy of the body at definitive position will be obtained as below.

Work done = change in kinetic energy

-11.76 J = Final kinetic energy-16 J

Final Kinetic energy = - 11.76+16

Final kinetic energy = 4.24 J

Thus, the kinetic energy of the body in definitive position is 4.24 J.

3 0

3 years ago

A power plant to represent the need for and use for energy; label it energy production.

meriva

I’m confused about this question

6 0

3 years ago

How efficient is a pulley system if it enables you to lift a 600.0 Newton engine 0.600 meters if you exerted 35.7 Newtons of for

muminat

Answer:

η = 0.882 = 88.2 %

Explanation:

The efficiency of the pulley system can be given as follows:

$\eta = \frac{W_{out}}{W_{In}}\\\\$

where,

η = efficiency of pulley system = ?

W_out = Output Work = (600 N)(0.6 m) = 360 J

W_in = Input Work = (35.7 N)(11.43 m) = 408.051 J

Therefore,

$\eta = \frac{360\ J}{408.051\ J}$

4 0

3 years ago

A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

6 0

3 years ago

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