Is there a link between the number of bulbs and current drawn from the power pack?

spayn [35]

Threshhold frequency is defined as the minimum light frequency necessary to start photo-electric emission from the surface. This frequency is just effecient enough to emit electrons without any additional force or energy. this minimum frequency creates the necessary result that is sought by the person. The photo electric emission for certain metals can occir with threshhold frequency. The importance of this threshhold frequency lies in the fact that this has been used to create numerous instruments like amplifiers for use of people.

3 0

3 years ago

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A direct result of the Newlands reclamation Act was the

Lina20 [59]

I think it is building and managing of irrigation systems

3 0

3 years ago

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A woman takes her dog Rover for a walk on a leash. She pulls on the leash with a force of 30.0 N at an angle of 29° above the ho

ValentinkaMS [17]

Answer:

The force parallel to the horizontal is 26.24 N

Explanation:

She pulls on the leash with a force F = 30 N, this force, since its at an angle of 29° (i will cal this angle $\theta$ ), it has a force component on x (the horizontal, i will call this force $F_{x}$ ) and a force component on y (the vertical, i will call this $F_{y}$ ).

This can be seen in the attached picture.

Since we are asked about the force parallel to the horizontal, we need to find the component of the force $F_{x}$ , since $F_{x}$ is the adjacent angle, we need to use cosine:

$F_{x}=Fcos \theta$

since $F=30N$ and $\theta=29$

$F_{x}=(30N)cos(29)$

$F_{x}=(30N)(0.8746)$

$F_{x}=26.24N$

The force parallel to the horizontal is 26.24 N

7 0

3 years ago

All stars go through a lifecycle.

Komok [63]

Mira is much bigger than the Sun.

Only very massive stars will go through a supernova stage, causing the outer layer to explode away and the core to collapse in on itself, becoming very dense.

5 0

2 years ago

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An airplane flies at 150 km/hr. (a) The airplane is towing a banner that is b = 0.8 m tall and l = 25 m long. If the drag coef-

maw [93]

Answer:

Power requirement P for the banner is found to be 30.62 W
Power requirement P for the solid flat plate is found to be 653.225 W
Answer for part(c) is explained below in the explanation section and can be summarized as: The main difference between the drags and power requirements of the two objects of same size was due to their significantly different drag-coefficients. The Cd for banner was given, whereas the Cd for a flat plate is generally found to be around 1.28 which is the value we used in our calculations that resulted in a huge increase of power to tow the flat plate
Power requirement P for the smooth spherical balloon was found to be 40.08 W

Explanation:

First of all we will establish variables and equations known that are known to us to solve this question. Since we are given the velocity of the airplane:

v = velocity of airplane i.e. 150 km/hr. To convert it into m/s we will divide it by 3.6 which gives us 41.66 m/s
The density of air at s.t.p (standard temperature pressure) is given as d = 1.225 kg / m^3
The power can be determined this equation: P = F . v, where F represents the drag-force that we will need to determine and v represents the velocity of the airplane
The equation to determine drag-force is: $F = 1/2 * d * C_d * A$

In the drag-force equation Cd represents the co-efficient of drag and A represents the frontal area of the banner/plate/balloon (the object being towed)

Frontal area A of the banner is : 25 x 0.8 = 20 m^2

Part a) We will plug in in the values of Cd, d, A in the drag-force equation i.e. Fd = 1/2 * 0.06* 1.225 * 20 = 0.735 N. Now to find the power P we will use P = F . v i.e. 0.735 * 41.66 = 30.62 W

Part b) For this part the only thing that has fundamentally changed is the drag-coefficient Cd since it's now of a solid flat plate and not a banner. The drag-coefficient of a flat plate is approximately given as : Cd_fp = 1.28

Now we will plug-in our values into the same equations as above to determine drag-force and then power. i.e. Fd = 1/2 * 1.28 * 1.225 * 20 = 15.68 N. Using Fd to determine power, P = 15.68 * 41.66 = 653.225 W

Part c) The main reason for such a huge power difference between two objects of same size was due to their differing drag-coefficients, as drag-coefficients are generally large for objects that are not of a streamlined shape and leave a large wake (a zone of low air pressure behind them). The flat plate being solid had a large Cd where as the banner had a considerably low Cd and therefore a much lower power consumption

Part d) The power of a smooth sphere can be calculated in the same manner as the above two. We just have to look up the Cd of a smooth sphere which is found to be around 0.5 i.e. Cd_s = 0.5. Area of sphere A is given as : pi* r^2 (r = d / 2). Now using the same method as above:

Fd = 1/2 * 0.5 * 3.14 * 1.225 = 0.962 N

P = 0.962 * 41.66 = 40.08 W

4 0

3 years ago