Answer:
[tex]0.845\ \text{N}[/tex]
Explanation:
g = Acceleration due to gravity at sea level = [tex]9.81\ \text{m/s}^2[/tex]
R = Radius of Earth = 6371000 m
h = Altitude of observatory = 4205 m
Change in acceleration due to gravity due to change in altitude is given by
[tex]g_h=g(1+\dfrac{h}{R})^{-2}\\\Rightarrow g_h=9.81\times(1+\dfrac{4205}{6371000})^{-2}\\\Rightarrow g_h=9.797\ \text{m/s}^2[/tex]
Weight at sea level
[tex]W=mg\\\Rightarrow W=65\times 9.81\\\Rightarrow W=637.65\ \text{N}[/tex]
Weight at the given height
[tex]W_h=mg_h\\\Rightarrow W_h=65\times 9.797\\\Rightarrow W_h=636.805\ \text{N}[/tex]
Change in weight [tex]W_h-W=636.805-637.65=-0.845\ \text{N}[/tex]
Her weight reduces by [tex]0.845\ \text{N}[/tex].
How many flip-flop values are complemented in an 8-bit binary ripple counter to reach the next count value after: 0110111 and 01010110?
Answer:
- Four (4) flip-flop values will complemented
- one (1) flip-flop value will complemented
Explanation:
To find how many flip flop number of bits complemented, we just need to figure out what the next count in the sequence is and find how many bits have changed.
taking a look at the a) 00110111
we need to just 1 to the value,
so
00110111 + 0000001 = 00111000
So here, only the first four bits are complemented.
Therefore Four (4) flip-flop values will complemented
Next
b) 01010110
we also add 1 to the value
01010110 + 00000001 = 01010111
only the first bit is complemented.
Therefore one (1) flip-flop value will complemented
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.
a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8mm. the beam carries a uniform load of 6.5kN/m along the entire length and, in the same direction, a concentrated force of 4kN at the free end. (a) determine the max bending stress (b) determine the max transervse shear stress (c) determine the max shear stress in the beam
Answer:
a) 159.07 MPa
b) 10.45 MPa
c) 79.535 MPa
Explanation:
Given data :
length of cantilever beam = 1.5m
outer width and height = 100 mm
wall thickness = 8mm
uniform load carried by beam along entire length= 6.5 kN/m
concentrated force at free end = 4kN
first we determine these values :
Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m
Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N
A) determine max bending stress
б = [tex]\frac{MC}{I}[/tex] = [tex]\frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}[/tex] = 159.07 MPa
B) Determine max transverse shear stress
attached below
ζ = 10.45 MPa
C) Determine max shear stress in the beam
This occurs at the top of the beam or at the centroidal axis
hence max stress in the beam = 159.07 / 2 = 79.535 MPa
attached below is the remaining solution
Write 83,120 in expanded form using powers of 10.
Answer:
8*10000+3*1000+1*00+2*10+2
Explanation:
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
