Answer :
Let's go through each of these questions step by step.
### 60. NAND gate is a combination of an;
For this question, let’s review the options to determine the correct gate combination for a NAND gate.
A. OR gate & a NOT gate
B. AND gate & a NOT gate
C. NAND gate & a NOT gate
D. OR gate & AND gate
A NAND gate is properly defined as a combination of an AND gate and a NOT gate, where the output of the AND gate is negated by the NOT gate.
So, the correct answer is:
B. AND gate & a NOT gate
### 61. A grinding wheel, initially at rest, is rotated with constant angular acceleration 5 rad/s² for 8 seconds. The wheel is then brought to rest with uniform negative acceleration in 10 revolutions. Determine the negative acceleration required and the time needed to bring the wheel to rest.
Let’s break this problem into two parts and calculate the required quantities.
First part (rotating the wheel):
1. Initial angular velocity, ω₀ = 0 (initially at rest).
2. Angular acceleration, α = 5 rad/s².
3. Time, t = 8 seconds.
Using the kinematic equation for angular velocity:
[tex]\[ ω = ω₀ + αt \][/tex]
[tex]\[ ω = 0 + (5 \text{ rad/s}²)(8 \text{ s}) \][/tex]
[tex]\[ ω = 40 \text{ rad/s} \][/tex]
Using the kinematic equation for angular displacement (θ):
[tex]\[ θ = ω₀t + \frac{1}{2} αt² \][/tex]
[tex]\[ θ = 0 + \frac{1}{2} (5 \text{ rad/s}²)(8 \text{ s})² \][/tex]
[tex]\[ θ = \frac{1}{2} (5)(64) \][/tex]
[tex]\[ θ = 160 \text{ rad} \][/tex]
Second part (bringing the wheel to rest):
1. Final angular velocity is zero (wheel brought to rest).
2. Initial angular velocity, [tex]\( ω = 40 \text{ rad/s} \)[/tex].
Given the wheel stops in 10 revolutions:
[tex]\[ θ = 10 \times 2π \][/tex]
[tex]\[ θ = 20π \text{ rad} \][/tex]
Using the kinematic equation:
[tex]\[ ω_f^2 = ω_i^2 + 2αθ \][/tex]
[tex]\[ 0 = (40)^2 + 2α(20π) \][/tex]
[tex]\[ -1600 = 40π α \][/tex]
[tex]\[ α = \frac{-1600}{40π} \][/tex]
[tex]\[ α \approx -12.7 \text{ rad/s}² \][/tex]
Then, time to stop:
[tex]\[ ω_f = ω_i + αt \][/tex]
[tex]\[ 0 = 40 + (-12.7)t \][/tex]
[tex]\[ t = \frac{40}{12.7} \][/tex]
[tex]\[ t \approx 3.15 \text{ s} \][/tex]
So, the correct answer is:
D. -12.7 rad/s² & 3.15 sec
### 62. If a person has a mass of 50.0 kg, what would be the force of gravitational attraction on him at Earth's surface?
The gravitational force can be calculated using the formula:
[tex]\[ F = mg \][/tex]
Where:
1. [tex]\( m \)[/tex] is the mass (50.0 kg).
2. [tex]\( g \)[/tex] is the acceleration due to gravity (9.8 m/s²).
[tex]\[ F = 50.0 \text{ kg} \times 9.8 \text{ m/s}² \][/tex]
[tex]\[ F = 490 \text{ N} \][/tex]
So, the correct answer is not among the given options. The typical approach would yield 490 N, but if there's a rounding that leans toward closer values in real-world calculations or educator preferences:
Correct calculation yields 490 N, but answers are problematic if all options are incorrect.
### 63. How tall must a water-filled closed-end manometer be to measure blood pressures as high as 300 mmHg?
Using the conversion:
[tex]\[ 1 \text{ mmHg} = 13.6 \text{ mmH}_2\text{O} \][/tex]
Given 300 mmHg:
[tex]\[ Height = 300 \text{ mmHg} \times 13.6 \][/tex]
[tex]\[ Height = 4080 \text{ mmH}_2\text{O} \][/tex]
So, the correct answer is:
A. 4080 mm
I hope that helps clarify the problems and solutions!
### 60. NAND gate is a combination of an;
For this question, let’s review the options to determine the correct gate combination for a NAND gate.
A. OR gate & a NOT gate
B. AND gate & a NOT gate
C. NAND gate & a NOT gate
D. OR gate & AND gate
A NAND gate is properly defined as a combination of an AND gate and a NOT gate, where the output of the AND gate is negated by the NOT gate.
So, the correct answer is:
B. AND gate & a NOT gate
### 61. A grinding wheel, initially at rest, is rotated with constant angular acceleration 5 rad/s² for 8 seconds. The wheel is then brought to rest with uniform negative acceleration in 10 revolutions. Determine the negative acceleration required and the time needed to bring the wheel to rest.
Let’s break this problem into two parts and calculate the required quantities.
First part (rotating the wheel):
1. Initial angular velocity, ω₀ = 0 (initially at rest).
2. Angular acceleration, α = 5 rad/s².
3. Time, t = 8 seconds.
Using the kinematic equation for angular velocity:
[tex]\[ ω = ω₀ + αt \][/tex]
[tex]\[ ω = 0 + (5 \text{ rad/s}²)(8 \text{ s}) \][/tex]
[tex]\[ ω = 40 \text{ rad/s} \][/tex]
Using the kinematic equation for angular displacement (θ):
[tex]\[ θ = ω₀t + \frac{1}{2} αt² \][/tex]
[tex]\[ θ = 0 + \frac{1}{2} (5 \text{ rad/s}²)(8 \text{ s})² \][/tex]
[tex]\[ θ = \frac{1}{2} (5)(64) \][/tex]
[tex]\[ θ = 160 \text{ rad} \][/tex]
Second part (bringing the wheel to rest):
1. Final angular velocity is zero (wheel brought to rest).
2. Initial angular velocity, [tex]\( ω = 40 \text{ rad/s} \)[/tex].
Given the wheel stops in 10 revolutions:
[tex]\[ θ = 10 \times 2π \][/tex]
[tex]\[ θ = 20π \text{ rad} \][/tex]
Using the kinematic equation:
[tex]\[ ω_f^2 = ω_i^2 + 2αθ \][/tex]
[tex]\[ 0 = (40)^2 + 2α(20π) \][/tex]
[tex]\[ -1600 = 40π α \][/tex]
[tex]\[ α = \frac{-1600}{40π} \][/tex]
[tex]\[ α \approx -12.7 \text{ rad/s}² \][/tex]
Then, time to stop:
[tex]\[ ω_f = ω_i + αt \][/tex]
[tex]\[ 0 = 40 + (-12.7)t \][/tex]
[tex]\[ t = \frac{40}{12.7} \][/tex]
[tex]\[ t \approx 3.15 \text{ s} \][/tex]
So, the correct answer is:
D. -12.7 rad/s² & 3.15 sec
### 62. If a person has a mass of 50.0 kg, what would be the force of gravitational attraction on him at Earth's surface?
The gravitational force can be calculated using the formula:
[tex]\[ F = mg \][/tex]
Where:
1. [tex]\( m \)[/tex] is the mass (50.0 kg).
2. [tex]\( g \)[/tex] is the acceleration due to gravity (9.8 m/s²).
[tex]\[ F = 50.0 \text{ kg} \times 9.8 \text{ m/s}² \][/tex]
[tex]\[ F = 490 \text{ N} \][/tex]
So, the correct answer is not among the given options. The typical approach would yield 490 N, but if there's a rounding that leans toward closer values in real-world calculations or educator preferences:
Correct calculation yields 490 N, but answers are problematic if all options are incorrect.
### 63. How tall must a water-filled closed-end manometer be to measure blood pressures as high as 300 mmHg?
Using the conversion:
[tex]\[ 1 \text{ mmHg} = 13.6 \text{ mmH}_2\text{O} \][/tex]
Given 300 mmHg:
[tex]\[ Height = 300 \text{ mmHg} \times 13.6 \][/tex]
[tex]\[ Height = 4080 \text{ mmH}_2\text{O} \][/tex]
So, the correct answer is:
A. 4080 mm
I hope that helps clarify the problems and solutions!
