A bookshelf is 2m long, with supports at its end (p and q). A book weighing 10N is placed in the middle of the shelf. what are the upward forces acting on p and q. if the book is moved 50 cm from q, what are the forces at p and q now?

Answers

Answer 1

Answer: When the book is placed in the middle, the forces acting on p and q is 5N. When the book is moved 50 cm from q, the forces at p and q can be solved by doing a moment balance
With p as the pivot
Fq (2 m) = 10 N (0.5 m)
Fq = 2.5 N
and
Fp = 10 N - 2.5 N = 7.5 N

Answer 2

Answer:

Answer:

Fq = 2.5N

Fp = 7.5N

Explanation:

Hello! Let's solve this!

When the book is in the middle, the force in p (Fp) and the force in q (Fq) are equal, and half of the total force

Fp = Fq = 5N

When the book moves 0.5m from q we have a balance

Fq * 2m = 10N * 0.5m

We cleared Fq

Fq = (10N * 0.5m) / 2m

Fq = 2.5N

Fp = 10N-Fq = 10N-2.5N = 7.5N

Conclusion

Fq = 2.5N

Fp = 7.5N

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Explain how an EVEN PARITY Circuit would be created and show the EVEN PARITY TRUTH table. Please do this in multisim

Answers

Answer:

Explanation:

Creating an even parity circuit in Multisim involves designing a digital logic circuit that checks if the number of '1's in a binary input is even. If the input has an even number of '1's, the circuit should output '1' (indicating even parity); otherwise, it should output '0' (indicating odd parity).

Here's how to create an even parity circuit in Multisim, along with the truth table:

**Creating the Even Parity Circuit:**

1. Open Multisim and create a new blank schematic.

2. Add the following components to your schematic:

- Input pins (for binary input bits)

- XOR gates

- An AND gate

- An inverter (NOT gate)

- Output display (LED or probe)

3. Connect the input pins to the XOR gates. Each input pin corresponds to one bit of the binary input.

4. Connect the outputs of the XOR gates to the inputs of the AND gate.

5. Connect the output of the AND gate to the input of the inverter (NOT gate).

6. Connect the output of the inverter to the output display.

7. Label your input pins for clarity (e.g., A0, A1, A2, ...).

**Designing the Even Parity Truth Table:**

To create the truth table for even parity, you'll need to list all possible input combinations (binary numbers) along with the corresponding output (even or odd).

Assuming you have a 3-bit input (A2, A1, A0), here's the truth table:

| A2 | A1 | A0 | Output (Even Parity) |

|----|----|----|-----------------------|

| 0 | 0 | 0 | 1 (Even) |

| 0 | 0 | 1 | 0 (Odd) |

| 0 | 1 | 0 | 0 (Odd) |

| 0 | 1 | 1 | 1 (Even) |

| 1 | 0 | 0 | 0 (Odd) |

| 1 | 0 | 1 | 1 (Even) |

| 1 | 1 | 0 | 1 (Even) |

| 1 | 1 | 1 | 0 (Odd) |

Each row in the truth table represents a unique combination of input bits (A2, A1, A0) and specifies whether the output is '1' (Even) or '0' (Odd).

Once you have created the circuit in Multisim and designed the truth table, you can simulate the circuit to verify its functionality. Ensure that the circuit produces the expected output (even parity) based on the input values.

a car travels 50 kilometers West and 10 minutes. after reaching the destination the car troubles back to the starting point, again taking 5 minutes. What is the average velocity of the car

Answers

SPEED is

   (distance covered) / (time to cover the distance) .

The car's average SPEED is

      (100 km) / (15min)

=     6-2/3 km per minute.

VELOCITY is

   (distance and direction between start-point and end-point)
divided by
   (time to travel from start-point to end-point).

The car ended up at the starting point, so the distance between
stating-point and end-point is zero.

Velocity = (zero) / (time to travel from start-point to end-point)

    = Zero .

Fluids only exert pressure downward. true or false

Answers

False it goes all directions. Hope this helped.

Let's assume you use green light (λ = 550 nm) to look at an electron. What is the uncertainty in determining the electron's velocity? Express your answer rounded up to the nearest hundredth.

Answers

By uncertainty principle

λ = h / p.

Where λ = wavelength, h = Planck's constant = 6.63 * 10⁻³⁴ Js

λ = wavelength = 550 nm = 550 * 10⁻⁹ m, Mass of electron = 9.1 * 10 ⁻³¹ kg

p = Momentum = mv

λ = h / mv

v = h / mλ

v = 6.63 * 10⁻³⁴ / (9.1 * 10⁻³¹ * 550 * 10⁻⁹)

v = 1 324 675.325 m/s

v ≈ 1.325 *10⁶ m/s

1. Coherent beams with wavelengths of 400 nm intersect at points on the screen. What will be observed at these points interference maximum or minimum, if the optical path difference of the beam is in the 1st case A1=20 um and in the 2nd - A2=15 um? Find the order of minimum and maximum. 2. An installation for observing Newton's rings in reflected light is illuminated by monochromatic light incident normally to the plate surface. The distance between the second and fourth dark rings is 5 mm. Find the distance between the eighth and sixteenth light rings. 3. Find the angle between the main planes of the polarizer and the analyzer if the intensity of the natural light passing through the polarizer and the analyzer decreased by 5 times.

Answers

1. The observed interference at points will be maximum for $A_1 = 20 \, \mu m$ and minimum for $A_2 = 15 \, \mu m$ , with orders 12th maximum and 12th minimum.
2. The distance between the eighth and sixteenth light rings is approximately 2.583 mm.
3. The angle between the main planes of the polarizer and the analyzer is approximately $63.43^\circ$ when the intensity decreases by 5 times.

1. For coherentbeams with wavelengths of 400 nm, the interference observed at points on the screen will be maximum when the optical path difference (OPD) is a whole number of wavelengths ( $m \lambda$ ), and minimum when the OPD is a half-integer number of wavelengths ( $(m + 0.5) \lambda$ ).

Given

$A_1 = 20 \, \mu m$

$A_2 = 15 \, \mu m$ , the difference in optical path between the two cases is $$\Delta A = A_1 - A_2 = 5 \, \mu m$.$

The number of wavelengths difference can be calculated as $(\Delta A)/(\lambda)$ , which is approximately 12.5 wavelengths.

2. In Newton's ringspattern, the radius of the m dark ring is given by $r_m = √(m \lambda R)$ , where R is the radius of curvature of the lens.

Given the distance between the second and fourth darkrings ( $m = 2\) to \(m = 4$ ) as 5 mm, we can set up an equation:

$$√(4 \lambda R) - √(2 \lambda R) = 5 \, \text{mm}$.$

Solving for R gives $$R \approx 1.904 * 10^(-2) \, \text{m}$.$

Now, the radius of the 16th light ring can be calculated as $r_(16) = √(16 \lambda R)$ , and the radius of the 8th light ring is $r_(8) = √(8 \lambda R)$ . The distance between these two light rings is then given by:

$$d = r_(16) - r_(8) \n\n= √(16 \lambda R) - √(8 \lambda R)$.$

Plugging in the values, we get $d \approx 2.583 \, \text{mm}$ .

3. When the intensity decreases by a factor of 5, the transmission of the polarizer and analyzer combination becomes $(1)/(√(5))$ times the original transmission, as intensity is proportional to the square of the transmission.

The relationship between the intensity and the angle $\theta$ between the main planes of the polarizer and the analyzer is given by Malus's law: $I = I_0 \cos^2 \theta$ , where $I_0$ is the initial intensity.

Given that $I_0$ decreases by a factor of 5, we have:

$(I)/(I_0) = (1)/(√(5)) \n\n= \cos^2 \theta$

Solving for $\theta$ , we get:

$\theta = \arccos \left( (1)/(√(5)) \right)\n\n \approx 63.43^\circ$

Thus, the angle is $63.43^\circ$ .

For more details regarding wavelength, visit:

brainly.com/question/31143857

#SPJ4

Answer:

a=3....................................

The mean distance from Saturn to the sun is 9 times greater than the mean distance from Earth to the sun. How long is a Saturn year?

Answers

It probably also has a nineth of the speed, so it's 81 earth years. If they had the same speed, it would be 9 earth years. Probably the first one is more accurate :-??

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