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2 years ago

Many battery-powered devices come with cords that allow them to be

Physics

1 answer:

lara [203]2 years ago

4 0

Answer:

The change that has to take place inside the power cord in order for the device to function properly include;

The changing of the alternating current (AC) from the electrical outlet to the direct current having a specific voltage and current value with which the device can be powered

Explanation:

A battery powered device makes use of direct current (DC) electric power from a battery, while the power normally given out at an electrical outlet comes as an alternating current (AC) electric power source.

The power cord for battery-powered device, also known as an AC/DC adapter, that allows them to be plugged into electrical outlets converts the AC electric current it obtains from the electrical outlets to DC electrical current of the appropriate voltage and amperage that the device can make use of for electric power to function and for charging the battery which is the power source for the device

Therefore, the change that has to take place in the power cord is the conversion of the electric outlet alternating current (AC) voltage and amperage values, into direct current (DC) of the required voltage and current for the device

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If vector A =i+2j-k and vec A cross vec B =3i-j+5k. find vec B

postnew [5]

Let B = a i + b j + c k. Then the cross product of A = i + 2j - k with B is

A × B = ( i + 2j - k ) × ( a i + b j + c k )

A × B = a ( i × i ) + 2a ( j × i ) - a ( k × i )

… … … + b ( i × j ) + 2b ( j × j ) - b ( k × j )

… … … + c ( i × k ) + 2c ( j × k ) - c ( k × k )

A × B = 0 - 2a k - a j

… … … + b k + 0 + b i

… … … - c j + 2c i - 0

A × B = (b + 2c) i + (-a - c) j + (b - 2a) k

So we have

3 i - j + 5 k = (b + 2c) i + (c - a) j + (b - 2a) k

which gives us the system of equations,

{ b + 2c = 3

{ -a - c = -1

{ -2a + b = 5

Solve for a, b, and c.

• Eliminate c from the first two equations:

(b + 2c) + 2 (-a - c) = 3 + 2 (-1)

-2a + b = 1

But -2a + b = 5, and 5 ≠ 1, so there is no such vector B that satisfies the cross product!

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2 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

nignag [31]

The position of the particle is given by:

x(t) = t³ - 12t² + 21t - 9

Differentiate x(t) with respect to t to find the velocity x'(t):

x'(t) = 3t² - 24t + 21

Differentiate x'(t) with respect to t to find the acceleration x''(t):

x''(t) = 6t - 24

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2 years ago

How might observations of the sun's outermost layers reveal what's happening in the interior? and how could this information be

LuckyWell [14K]

A raging activity can be found in the Sun's interior, with pressure waves being produced and travelling back and forth, from the core to the surface and back to the core. By looking closely at the 'surface' we can see these "ripples". It gives us an idea of how dense the material was that the waves passed through. In a way, this can help to predict solar storms in the future.

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3 years ago

Read 2 more answers

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

Differences between Light year and Cosmic year in two points

rusak2 [61]

Answer:

A cosmic year is 365.25 days, some times called a side real year and is just the time it takes for us to go round the sun once.

A light year is the distance light travels in a year. Now light travels at about 186,262 miles a Second! Which is not slow by any ones book.

An experiment was conducted just after Christmas a few years ago. Two girls were selected from the audience and went into two phone boxes a few feet apart. They could only hear each other via the phones. The phone call went to a ground station about 200 miles away, then up to a geostationary coms satellite, back to a ground station 1/3 of the way around the world, then repeated, with a third satellite before being sent from another ground station back to London and the other phone box. We the audience could hear both sides of the conversation from both boxes. And could hear the delay between sending and receiving. So even at the speed of light, there was about 1.5 seconds of delay. So because distances in space are so vast that saying a star is x millions of miles away causes problems, you run out of zero’s! So our nearest other star is about 4.5 light years away. Our sun (our nearest start) is about 8 light minuets away. Varies slightly as our orbit is not 100% cirular.

I HOPE THIS IS HELPFUL.

3 0

2 years ago