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3 years ago

Suppose the price of electricity is 12.75 cents per kWºh. What

Physics

1 answer:

adell [148]3 years ago

5 0

Answer:

The difference between the cost of operating LED and incandescent bulb is $5.1

Explanation:

We are given the cost of electricity that is 12.75 cents per kWh. We want to find out the difference in the operating cost of an incandescent and LED bulb for a time period of 2,000 hours.

Since we are not given the rating of the incandescent bulb and LED bulb, we will assume their ratings.

For a light intensity of 250 Lumens;

The average rating of an LED bulb is approximately 5 Watts.

The average rating of an incandescent bulb is approximately 25 Watts.

Now lets find out the kWh of each bulb.

Energy = Power×Time

For LED bulb:

E = 5×2,000 = 10,000 Wh

Divide by 1000 to convert into kWh

E = 10,000/1000 = 10 kWh

Cost = 12.75×10 = 127.5 cents

Cost = $1.27

For Incandescent bulb:

E = 25×2,000 = 50,000 Wh

Divide by 1000 to convert into kWh

E = 50,000/1000 = 50 kWh

Cost = 12.75×50 = 637.5 cents

Cost = $6.37

Difference in Cost:

Difference = $6.37 - $1.27 = $5.1

Therefore, the difference between the cost of operating LED and incandescent bulb is $5.1.

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How does the observed pitch of the buzzer change as it moves

andriy [413]

The answer is

Pitch of the buzzer increased (higher tone) as it moves towards the observer

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2 years ago

Holding onto a tow rope moving parallel to a frictionless ski slope, a 68.7 kg skier is pulled up the slope, which is at an angl

Furkat [3]

Answer:

a) $F = 78.606\,N$ , b) $F = 88.911\,N$

Explanation:

a) Let consider two equations of equilibrium, the first parallel to ski slope and the second perpendicular to that. The equations are, respectively:

$\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = 0\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0$

The force on the skier is:

$F = m \cdot g \cdot \sin \theta$

$F = (68.7\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 6.7^{\textdegree}$

$F = 78.606\,N$

b) The equations of equilibrium are the following:

$\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = m\cdot a\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0$

The force on the skier is:

$F = m\cdot (a + g \cdot \sin \theta)$

$F = (68.7\,kg)\cdot (0.150\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}}\cdot \sin 6.7^{\textdegree})$

$F = 88.911\,N$

3 0

3 years ago

A cohesive force between the liquids molecules is responsible for the fluids is called

kiruha [24]

Answer:

static force

Explanation:

mark me brainliest

5 0

3 years ago

Read 2 more answers

A car speedometer has a 4% uncertainty. What is the range of possible speeds (in km/h) when it reads 110 km/h?

Kruka [31]

4% of 110 is 4.4. So the possible range of speeds is the interval from 110-4.4 till 110+4.4.
105.6 till 114.4

4 0

3 years ago

A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turn

kumpel [21]

Answer:

a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

Explanation:

Statement is incomplete. The complete description is now described below:

A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off.

(a) What is the magnitude of the satellite's velocity when the thruster turns off

(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis

Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite ( $\vec{v}_{S}$ ), measured in meters per second, is:

$\vec{v}_{S} = (v_{o,x}+a_{x}\cdot t)\,\hat{i}+v_{y}\,\hat{j}$

Where:

$v_{o,x}$ - Initial velocity in +x direction, measured in meters per second.

$a_{x}$ - Acceleration in +x direction, measured in meter per square second.

$t$ - Time, measured in seconds.

$v_{y}$ - Velocity in +y direction, measured in meters per second.

If we know that $v_{o,x} = 0\,\frac{m}{s}$ , $a_{x} = 0.250\,\frac{m}{s^{2}}$ , $t = 45\,s$ and $v_{y} = 21.4\,\frac{m}{s}$ , the final velocity of the satellite is:

$\vec{v}_{S} = \left[0\,\frac{m}{s}+\left(0.250\,\frac{m}{s^{2}} \right)\cdot (45\,s) \right]\,\hat{i}+\left(21.4\,\frac{m}{s} \right)\,\hat{j}$

$\vec{v_{S}} = 11.25\,\hat{i}+21.4\,\hat{j}\,\,\left[\frac{m}{s} \right]$

a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:

$\|\vec {v}_{S}\| = \sqrt{\left(11.25\,\frac{m}{s} \right)^{2}+\left(21.4\,\frac{m}{s} \right)^{2}}$

$\|\vec{v}_{S}\| \approx 24.177\,\frac{m}{s}$

The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:

$\tan \alpha = \frac{v_{y}}{v_{x}} = \frac{21.4\,\frac{m}{s} }{11.25\,\frac{m}{s} }$

$\tan \alpha = 1.902$

$\alpha = \tan^{-1}1.902$

$\alpha \approx 62.266^{\circ}$

The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

4 0

3 years ago