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7 months ago

Imagine the arc of a football as it flies through the air. How does this motion illustrate classical mechanics?

Engineering

2 answers:

BigorU [14]7 months ago

7 0

Answer: Maybe A

Explanation:

crimeas [40]7 months ago

4 0

A is the correct answer

hope it helps u

branliest?

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

Vesnalui [34]

<h3>CSMA/CD Protocol: </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.

Example:

$T_{P}=1 H r$ , 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.

Efficiency of CSMA/CD:

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

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

7 0

2 years ago

A hydraulic cylinder is to be used to move a workpiece in a manufacturing operation through a distance of 50 mm in 10 s. A force

svetlana [45]

Answer:

The answer to this question is 1273885.3 ∅

Explanation:

The first step is to determine the required hydraulic flow rate liquid if working pressure and if a cylinder with a piston diameter of 100 mm is available.

Given that,

The distance = 50mm

The time t =10 seconds

The force F = 10kN

The piston diameter is = 100mm

The pressure = F/A

10 * 10^3/Δ/Δ

P = 1273885.3503 pa

Then

Power = work/time = Force * distance /time

= 10 * 1000 * 0.050/10

which is =50 watt

Power =∅ΔP

50 = 1273885.3 ∅

5 0

2 years ago

1. A _____ is applied to a wall or roof rafters to add strength and keep the building straight and plumb.

PSYCHO15rus [73]

Answer:

A force must s applied to a wall or roof rafters to add strength and keep the building straight and plumb

7 0

2 years ago

Engineers are designing a cylindrical air tank and are trying to determine the dimensions of the tank. The proposed material for

lana66690 [7]

Answer:

The length of tank is found to be 0.6 m or 600 mm

Explanation:

In order to determine the length, we need to find a volume for the tank.

For this purpose, we use ideal gas equation:

PV = nRT

n = no. of moles = m/M

Therefore,

PV = (m/M)(RT)

V = (mRT)/(MP)

where,

V = volume of air = volume of container

m = mass of air = 4.64 kg

R = General Gas Constant = 8.314 J/mol.k

T = temperature of air = 10°C + 273 = 283 K

M = molecular mass of air = 0.02897 kg/mol

P = Pressure of Air = 20 MPa = 20 x 10^6 N/m²

V = (4.64 kg)(8.314 J/mol.k)(283 k)/(0.02897 kg/mol)(20 x 10^6 N/m²)

V = 0.01884 m³

Now, the volume of cylindrical tank is given as:

V = 0.01884 m³ = π(Diameter/2)²(Length)

Length = (0.01884 m³)(4)/π(0.2 m)²

Length = 0.6 m = 600 mm

4 0

3 years ago

Please help this is due today!!!!!

White raven [17]

Answer:

1:c 2:False

Explanation:

7 0

2 years ago