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3 years ago

Miles (322,000 kilometers [km]) or

2 answers:

4 0

B because as long as you keep it good and take good care of it like it says you shouldn’t need a change unless soemthing happens

slega [8]3 years ago

3 0

The answer is B . have a good day

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1) Each of the following would be considered company-confidential except

AysviL [449]

Answer is your company’s address

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3 years ago

One of the best ways to increase engine power and control detonation and preignition is to?

Reika [66]

Answer:

use water injection.

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1 year ago

When should u check ur review mirror

vaieri [72.5K]

Answer:

All the time and to see if someone is behind you.

Explanation:

After you check your side mirrors you should check your review mirror to see who is behind you.

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3 years ago

The Ethernet (CSMA/CD) alternates between contention intervals and successful transmissions. Assume a 100 Mbps Ethernet over 1 k

Vesnalui [34]

<h3><u>CSMA/CD Protocol: </u></h3>

Carrier sensing can transmit the data at anytime only the condition is before sending the data sense carrier if the carrier is free then send the data.

But the problem is the standing at one end of channel, we can’t send the entire carrier. Because of this 2 stations can transmit the data (use the channel) at the same time resulting in collisions.

There are no acknowledgement to detect collisions, It's stations responsibility to detect whether its data is falling into collisions or not.

<u>Example: </u>

$T_{P}=1 H r$ , at time t = 10.00 AM, A starts, 10:59:59 AM B starts at time 11:00 AM collision starts.

12:00 AM A will see collisions

Pocket Size to detect the collision.

$\begin{aligned}&T_{t} \geq 2 T_{P}\\&\frac{L}{B} \geq 2 T_{P}\\&L \geq 2 \times T_{P} \times B\end{aligned}$

CSMA/CD is widely used in Ethernet.

<u>Efficiency of CSMA/CD:</u>

In the previous example we have seen that in worst case $2 T_{P}$ time require to detect a collision.

There could be many collisions may happen before a successful completion of transmission of a packet.

We are given number of collisions (contentions slots)=4.

$\text { Propagation day }=\frac{\text {distance}}{\text {speed}}$

Distance = 1km = 1000m

$\begin{aligned}&\text { Speed }=2 \times 10^{8} \mathrm{m} / \mathrm{sec}\\ &T_{P}=\frac{1000}{2 \times 10^{8}}=(0.5) \times 10^{-5}=5 \times 10^{-6}\\ &T_{t}=5 \mu \mathrm{sec}\end{aligned}$

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3 years ago

Suppose the loop is moving toward the solenoid (to the right). Will current flow through the loop down the front, up the front,

Tems11 [23]

Answer:

See explanation

Explanation:

The magnetic force is

F = qvB sin θ

We see that sin θ = 1, since the angle between the velocity and the direction of the field is 90º. Entering the other given quantities yields

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6 0

3 years ago

Read 2 more answers