Answer:
Limitation in the level of possible accuracy of the that can be obtained to mainly whole degrees
Explanation:
The angle measurement values are located around the circularly shaped bevel protractor to which a Vernier scale can be attached to increase the accuracy of the angle measurement reading
The accuracy of the bevel protractor is up to 5 arc minutes or 1/12° with the space on the Vernier scale having graduations of 1/12°, such that two spaces on the main scale is 5 arc minutes more than a space on the Vernier scale and the dimensions of the coincidence of the two scale will have an accuracy of up to 1/12°
Therefore, whereby the Vernier scale is damaged, the accuracy will be limited to the accuracy main scale reading in whole degrees.
5/2055 classes displayed there’s Nooooob changes
Answer:
Goal address: Based on the above Ethernet outline design, we get that the goal address begins from the first hex worth and is of size 12 hex values (got from question 1). In light of the bundle given, we get that the goal address is 6c40 0889 c448.
Source address: Similarly, the source address begins after the goal address and is of size 12 hex values (got from question 1). Thus, the appropriate response is f832 e4a7 fb38.
Type/Length: The sort/length begins after the source address and is of size 4 hex values (got from question 1). Thus, the appropriate response is 0806.
Data (Payload): The data(payload) begins after the sort/length and finishes not long before FCS(Checksum) which is of 4 bytes for example 4 * 2 = 8 hex qualities. So the information comprises of everything between the 'type/length' and 'CRC'. We persuade the CRC to be c0a8 01f2. Henceforth, the appropriate response is 0001 0800 0604 0002 f832 e4a7 bf38 c0a8 0101 6c40 0889 c448.
Question 4:
The Ethernet parcel type characterizes the convention utilized for sending the bundle information. We realize that type 0x0800 demonstrates the IPv4 convention, 0x0806 shows an ARP convention, and 0x86DD demonstrates an IPv6 convention. In light of the appropriate response we got in Question 3, we realize that the sort/length esteem for the given parcel is 0806, which implies that convention utilized for sending the information bundle was ARP convention.
Answer: At time 18.33 seconds it will have moved 500 meters.
Explanation:
Since the acceleration of the car is a linear function of time it can be written as a function of time as
![a(t)=5(1-\frac{t}{15})](https://tex.z-dn.net/?f=a%28t%29%3D5%281-%5Cfrac%7Bt%7D%7B15%7D%29)
![a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bd%5E%7B2%7Dx%7D%7Bdt%5E%7B2%7D%7D%5C%5C%5C%5C%5Ctherefore%20%5Cfrac%7Bd%5E%7B2%7Dx%7D%7Bdt%5E%7B2%7D%7D%3D5%281-%5Cfrac%7Bt%7D%7B15%7D%29)
Integrating both sides we get
![\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7Bd%5E%7B2%7Dx%7D%7Bdt%5E%7B2%7D%7Ddt%3D%5Cint%205%281-%5Cfrac%7Bt%7D%7B15%7D%29dt%5C%5C%5C%5C%5Cfrac%7Bdx%7D%7Bdt%7D%3Dv%3D5t-%5Cfrac%7B5t%5E%7B2%7D%7D%7B30%7D%2Bc)
Now since car starts from rest thus at time t = 0 ; v=0 thus c=0
again integrating with respect to time we get
![\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7Bdx%7D%7Bdt%7Ddt%3D%5Cint%20%285t-%5Cfrac%7B5t%5E%7B2%7D%7D%7B30%7D%29dt%5C%5C%5C%5Cx%28t%29%3D%5Cfrac%7B5t%5E%7B2%7D%7D%7B2%7D-%5Cfrac%7B5t%5E%7B3%7D%7D%7B90%7D%2BD)
Now let us assume that car starts from origin thus D=0
thus in the first 15 seconds it covers a distance of
![x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m](https://tex.z-dn.net/?f=x%2815%29%3D2.5%5Ctimes%2015%5E%7B2%7D-%5Cfarc%7B15%5E%7B3%7D%7D%7B18%7D%3D375m)
Thus the remaining 125 meters will be covered with a constant speed of
![v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s](https://tex.z-dn.net/?f=v%2815%29%3D5%5Ctimes%2015-%5Cfrac%7B15%5E%7B2%7D%7D%7B6%7D%3D37.5m%2Fs)
in time equalling ![t_{2}=\frac{125}{37.5}=3.33seconds](https://tex.z-dn.net/?f=t_%7B2%7D%3D%5Cfrac%7B125%7D%7B37.5%7D%3D3.33seconds)
Thus the total time it requires equals 15+3.33 seconds
t=18.33 seconds
Answer:
The percent elongation in the length of the specimen is 42%
Explanation:
Given that:
The gage length of the original test specimen
= 50 mm
The final gage length
= 71 mm
The area = 206 mm²
maximum load = 162,699 N
To determine the percent elongation in %, we use the formula:
![\%EL = \dfrac{L_f-L_o}{L_o}\times 100](https://tex.z-dn.net/?f=%5C%25EL%20%3D%20%5Cdfrac%7BL_f-L_o%7D%7BL_o%7D%5Ctimes%20100)
![\%EL = \dfrac{71 \ mm-50 \ mm}{50 \ mm}\times 100](https://tex.z-dn.net/?f=%5C%25EL%20%3D%20%5Cdfrac%7B71%20%5C%20mm-50%20%5C%20mm%7D%7B50%20%5C%20mm%7D%5Ctimes%20100)
![\%EL = \dfrac{21 mm}{50 \ mm}\times 100](https://tex.z-dn.net/?f=%5C%25EL%20%3D%20%5Cdfrac%7B21%20mm%7D%7B50%20%5C%20mm%7D%5Ctimes%20100)
![\%EL = 0.42 \times 100](https://tex.z-dn.net/?f=%5C%25EL%20%3D%200.42%20%5Ctimes%20100)
![\mathbf{\%EL = 42 \%}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5C%25EL%20%3D%2042%20%5C%25%7D)
The percent elongation in the length of the specimen is 42%