Answer:
The answer is below
Explanation:
Given that:
Volume of carbide ([tex]V_{C}[/tex]) = 76% = 0.76, Volume of Nickel ([tex]V_{M}[/tex]) = 100% - 76% = 24% = 0.24, thermal conductivities of carbide ([tex]E_{C}[/tex]) = 30 W/m-K and thermal conductivities of meta. ([tex]E_M[/tex]) = 67 W/m-K
a) The maximum thermal conductivity is given by:
Max = [tex]E_CV_c+E_mV_m=(0.76*30) + (0.24*67) = 38.88\ W/mK[/tex]
b) The minimum thermal conductivity is given by:
Min = [tex]\frac{E_ME_C}{E_MV_C+E_CV_M}=\frac{30*67}{(0.76*67)+(0.24*30)}=34.58\ W/mK[/tex]
Five identical keys are suspended from a balance, which is held horizontally as shown. The two keys on the left are attached to the balance 6 cm from the pivot and the three keys on the right are attached 5 cm from the pivot. What will happen when the person lets go of the balance beam?
Answer:
movement in clockwise direction.
Explanation:
The following parameters or information are given from the question above, they are:
[1]. There are two identical keys, [2]. two out of the five keys are attached to 6cm from the pivot, [3]. three keys out of the five keys on the right are attached 5 cm.
Therefore, considering the moment of force, the two keys on the left = 2 × 6 = 12.
Also, considering the moment of force, the 3 keys on the right = 3 × 5 = 15.
Therefore, we have more weight on the right keys. So, in order to balance the force there must be movement in clockwise direction.
Determine the size of memory needed for CD recording of a piece of music, which lasts for 26 minutes, is done with a 20-bit Analog-to-Digital Converter (ADC) in stereo (2 channels), at the rate of 44.1 kSa/s, with the compression factor 6 (allow 10% error margin).
Answer: the size of memory needed for the CD recording is 28.7 MB
Explanation:
so in the case of stereo, the bitrate is;
⇒ 26 × 60 × 44.1 × 10³ × 2
= 137592 × 10³
for 10 bit
⇒ 137592 × 10³ × 10
= 1375920 × 10³ bits
now divide by 8 (convert to bytes)
⇒ (1375920 × 10³) / 8
= 171,990,000 BYTE
divide by 1000 (convert to kilobytes)
= 171,990,000 / 1000
= 171,990 KILOBYTES
now Given that, the compression ratio is 6
so
171,990 / 6
= 28665 KB
we know that. 1 MB = 1000 KB
x MB = 28665 KB
x MB = 28665 / 1000
⇒ 28.665 MB ≈ 28.7 MB
Therefore the size of memory needed for the CD recording is 28.7 MB
