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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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A force of 12 N changes the momentum of a toy car from 3kgm/s t0 10kgm/s. Calculate the time the force took to produce this chan

yan [13]

Answer:

Time = 0.58 seconds

Explanation:

Given the following data;

Initial momentum = 3 kgm/s

Final momentum = 10 kgm/s

Force = 12 N

To find the time required for the change in momentum;

First of all, we would determine the change in momentum.

$Change \; in \; momentum = final \; momentum - initial \; momentum$

$Change \; in \; momentum = 10 - 3$

Change in momentum = 7 kgm/s

Now, we can find the time required;

Note: the impulse of an object is equal to the change in momentum experienced by the object.

Mathematically, impulse (change in momentum) is given by the formula;

$Impulse = force * time$

Making "time" the subject of formula, we have;

$Time = \frac {impulse}{force}$

Substituting into the formula, we have;

$Time = \frac {7}{12}$

Time = 0.58 seconds

6 0

3 years ago

Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur

Alenkasestr [34]

Answer:

the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$

the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be $\phi$ and $\beta$ respectively.

The velocity of the block to be = v

The equivalent mass of the system = $m_e$

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

$K.E = \frac{1}{2} m_ev^2$ --------------- equation (1)

The angular velocity of the cylinder = $\omega$ : &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

$v = \omega R$

$\omega =\frac{v}{R}$

The kinetic energy of the system in terms of individual mass can be written as:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2$

By replacing $\omega$ with $\frac{v}{R}$ ; we have:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2$

$K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2$ ------------------ equation (2)

Equating both equation (1) and (2); we have:

$m_e = m_1+m_2+\frac{I}{R^2}$

Therefore, the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$ which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

$f_{block }=m_1gsin \beta$

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

$f_{cylinder} = m_2gsin \phi$

Equating the above both equations; we have the equation of motion of the equivalent system to be

$m_e \bar x = f_{cylinder}-f_{block}$

which can be written as follows from the previous derivations

$(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Finally; the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

8 0

3 years ago

~choices below~

mart [117]

A is the correct answer

5 0

3 years ago

Read 2 more answers

A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force

Vesnalui [34]

Ya know what the day you are talking to you about it and then send me the back link you got it and I don’t want

7 0

3 years ago

Sound travels through air at 343 m/s,

sasho [114]

The sound wave will have traveled 2565 m farther in water than in air.

Answer:

Explanation:

It is known that distance covered by any object is directly proportional to the velocity of the object and the time taken to cover that distance.

Distance = Velocity × Time.

So if time is kept constant, then the distance covered by a wave can vary depending on the velocity of the wave.

As we can see in the present case, the velocity of sound wave in air is 343 m/s. So in 2.25 s, the sound wave will be able to cover the distance as shown below.

Distance = 343 × 2.25 =771.75 m

And for the sound wave travelling in fresh water, the velocity is given as 1483 m/s. So in a time interval of 2.25 s, the distance can be determined as the product of velocity and time.

Distance = 1483×2.25=3337 m.

Since, the velocity of sound wave travelling in fresh water is greater than the sound wave travelling in air, the distance traveled by sound wave in fresh water will be greater.

Difference in distance covered in water and air = 3337-772 m = 2565 m

So the sound wave will have traveled 2565 m farther in water than in air.

5 0

3 years ago