Easy POINTS computer genius help me plz

Match each context to the type of the law that is most suitable for it.

Bas_tet [7]

Answer:

sorry i dont understand the answer

Explanation:

but i think its a xd jk psml lol

5 0

2 years ago

Explain how voltage level are interpreted by a digital circuit

lisabon 2012 [21]

Answer:

the state of the circuit is a function of the voltage level. The interpretation is up to the user.

Explanation:

A binary digital circuit adopts one of two states, depending on whether the voltage level is above or below some threshold that depends on the design of the circuit. Within each state, the voltage may have some typical range. When the voltage is near the threshold, the state of the circuit may actually be "indeterminate".

The internal/output voltage is a function of the state of the circuit. The interpretation of that voltage as a true/false or 1/0 or other meaning is up to the user of the circuit.

The circuit interprets a given input voltage as intending to convey a particular input signal state according to the circuit specifications. Input voltages near the threshold between states may cause unexpected or even destructive results.

__

In order to conserve space, some digital circuits use more than 2 different voltage levels to signify more than 2 different states.

5 0

2 years ago

Ok there........................................................................

Juliette [100K]

Answer:

ok THERE

Explanation:

4 0

2 years ago

A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.

Nataliya [291]

Answer:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

Explanation:

We can assume that the general formula for the drag force is given by:

$D= C_D \frac{\rho}{2}V^2 A$

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

$D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)$

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

$D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})$

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

$\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D1} \frac{\rho}{2}V^2 (wh) V$