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butalik [34]
3 years ago
15

How to get on your screen on 2k20 in every mode

Engineering
2 answers:
VashaNatasha [74]3 years ago
5 0
D pad or rb or lb hop this helps
nadezda [96]3 years ago
4 0

Answer:

down on the d pad

Explanation:

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A rigid bar ABCD is pinned at A and supported by two steel rods connected at B and C, as shown. There is no strain in the vertic
mylen [45]

Answer:

See attached picture.

Explanation:

4 0
3 years ago
At what depth in water is the increased pressure five times greater than atmospheric pressure (101 kPa)?​
umka21 [38]

Explanation:

40.4m

Explanation:

Pressure at depth is given as

P = P, + pgh

Final pressure at depth h= 5 Po

5Po= Po + pgh

pgh = 4Po = 4 x 1.01 x 10^5

h = (4.04×10^5)/ (1000x10)

h=40.4m

8 0
3 years ago
Steel riverts in aluminium drain gutters leak after two years. is it galvanic corrosion? ​
Andre45 [30]

Answer:

<h2>A good way to reduce corrosion is to use an isolating coating or paint on the aluminum and the steel to isolate them electrically. Insulating washers are also effective in isolating the two dissimilar materials and creating a relatively safe surface area</h2>

8 0
3 years ago
Suppose there are 76 packets entering a queue at the same time. Each packet is of size 5 MiB. The link transmission rate is 2.1
tia_tia [17]

Answer:

938.7 milliseconds

Explanation:

Since the transmission rate is in bits, we will need to convert the packet size to Bits.

1 bytes = 8 bits

1 MiB = 2^20 bytes = 8 × 2^20 bits

5 MiB = 5 × 8 × 2^20 bits.

The formula for queueing delay of <em>n-th</em> packet is :  (n - 1) × L/R

where L :  packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate =  2.1 Gbps = 2.1 × 10^9 bits per second.

Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9

queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9

queueing delay for 48th packet = 0.938725181 seconds

queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds

4 0
3 years ago
Carbon dioxide enters an adiabatic compressor at 100 kPa and 300 K at a rate of 1.8 kg/s and exits at 600 kPa and 450 K. Neglect
Veronika [31]

Answer:

isentropic efficiency of the compressor (nc)=

89.4%

Explanation:

answer attached

6 0
3 years ago
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