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3 years ago

If the period of a clock signal is 500 ps, the frequency is:_____

Physics

1 answer:

NNADVOKAT [17]3 years ago

4 0

Answer:

The frequency of the signal is 2 GHz

Explanation:

Given;

period of the clock signal, T = 500 ps = 500 x 10⁻¹² s

the frequency of the signal is given by;

$F= \frac{1}{T}\\\\F = \frac{1}{500*10^{-12}}\\\\F = 2*10^{9} \ Hz$

F = 2 GHz

Therefore, the frequency of the signal is 2 x 10⁹ Hz or 2 GHz

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A demented scientist creates a new temperature scale, the "Z scale." He decides to call the boiling point of nitrogen 0°Z and th

slavikrds [6]

Answer:

A) $Z = 0.577 C +112.931$

$Z = 0.577*(100) +112.931=170.631 Z$

B) $C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C$

C) $K = C +273.15$

$K = -22.41 +273.15 =250.739 K$

Explanation:

For this case we want to create a function like this:

$Z = a C + b$

Where Z represent the degrees for the Z scale C the Celsius grades and tha valus a and b parameters for the model.

The boiling point of nitrogen is -195,8 °C

The melting point of iron is 1538 °C

We know the following equivalences:

-195.8 °C = 0 °Z

1538 °C = 1000 °Z

Let's say that one point its (1538C, 1000 Z) and other one is (-195.8 C, 0Z)

So then we can calculate the slope for the linear model like this:

$a = \frac{z_2 -z_1}{c_2 -c_1}= \frac{1000 Z- 0Z}{1538C -(-195.8 C)}=\frac{1000 Z}{1733.8 C}=0.577 \frac{Z}{C}$

And now for the slope we can use one point let's use for example (-195.8C, 0Z), and we have this:

$0 = 0.577 (-195.8) + b$

And if we solve for b we got:

$b = 0.577*195.8 =112.931 Z$

So then our lineal model would be:

$Z = 0.577 C +112.931$

Part A

The boiling point of water is 100C so we just need to replace in the model and see what we got:

$Z = 0.577*(100) +112.931=170.631 Z$

Part B

For this case we have Z =100 and we want to solve for C, so we can do this:

$Z-112.931 = 0.577 C$

$C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C$

Part C

For this case we know that $K = C +273.15$

And we can use the result from part B to solve for K like this:

$K = -22.41 +273.15 =250.739 K$

6 0

3 years ago

Choose all of the true statements regarding the relationship between voltage, resistance, and current. Current is measured in am

AveGali [126]

-- Current is measured in amps. (You can use any symbol you want to represent current, but the most common one is " I ", not "Δ".)

-- The relationship between current, voltage, and resistance is mathematically defined by Ohm's Law.

-- Current is the flow of electrons through a circuit.

-- (Ohm's Law is NOT mathematically represented by the equation V=I/R.) It should be V = I · R .

(When solving for Resistance in a circuit and both voltage and current are known values, the equation I =V*R is not true, and not the way to solve it.) If the resistance is what you're looking for, then the equation to use is R = V / I .

-- If the voltage in a circuit is increased, the current will also increase.

6 0

3 years ago

Read 2 more answers

An office window has dimensions 3.1 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.954 a

Virty [35]

Answer:

Net forces which pushes the window is 30342.78 N.

Explanation:

Given:

Dimension of the office window.

Length of the window = $3.1$ m

Width of the window = $2.1$ m

Area of the window = $(3.1\times 2.1) = 6.51\ m^2$

Difference in air pressure = Inside pressure - Outside pressure

= $(1.0-0.954)$ atm = $0.046$ atm

Conversion of the pressure in its SI unit.

⇒ $1$ atm = $101325$ Pa

⇒ $0.046$ atm = $0.046\times 101325 =4660.95$ Pa

We have to find the net force.

We know,

⇒ Pressure = Force/Area

⇒ $Pressure=\frac{Force }{Area}$

⇒ $Force =Pressure\times Area$

⇒ Plugging the values.

⇒ $Force =4660.95\times 6.51$

⇒ $Force=30342.78$ Newton (N)

So,

The net forces which pushes the window is 30342.78 N.

3 0

3 years ago

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

3 years ago

Do you attract the earth or the earth attract you ?Which one is attracting with a large force ?You or the earth

Katen [24]

The earth

The earths mass is what generates the force to draw you in. The deeper you go, sorry the more above you the less the pull will be

6 0

2 years ago