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Orlov [11]
2 years ago
15

Susan works in the research and development department. She has recently purchased a large high-speed external drive and has att

ached the drive to her computer using a USB cable. Her drive requires a minimum bandwidth of 400 Mbps and at least 900 milliamps (mA) to function.
Although the correct drivers are installed, the drive is not functioning. To troubleshoot the problem, she has connected her drive to her coworkers' computer where the drive functions properly. No additional cables are required for this drive.
Which of the following is the MOST likely reason Susan's external hard drive is not working?
A. Susan has connected her drive to a USB 3.0 port, which does not have enough power for her drive.
B. Susan has connected her drive to a USB 2.0 port, which does not have enough power for her drive.
C. Susan has connected her drive to a USB 3.0 port, which does not support the maximum bandwidth required.
D. Susan has connected her drive to a USB 2.0 port, which does not support the maximum bandwidth required.
Physics
1 answer:
ivann1987 [24]2 years ago
5 0

     The reason why Susan's drive is not working is that she has connected the drive to a USB 2.0 port, which does not have enough power for her drive.

<h3>What is the USB?</h3>

USB is a short acronym for the original word called a Universal Serial Bus. USB is a connector interface for connecting a computer to peripherals and other devices.

USB comes in different specs such as the:

  • USB 2.0
  • USB 3.0

Some of the characteristics differences between USB 2.0 and 3.0 are:

A USB 2.0 requires 480 Mbps for data transfer speed while USB 3.0 requires 4800 Mbps.

A USB 2.0 has the ability to use a maximum power of 500 mA, while USB 3.0 can make use of up to 900 mA.

From the given information, It is stated that the drive requires a minimum of 400 Mbps and at least 900 mA;

Therefore, it implies that the drive is supposed to work with a power of at least 900 mA for it to work.

But since it is not working, we can conclude that the drive is connected to a USB 2.0 port (500 mA) which does not have enough power for the drive.

Learn more about the USB port here:

brainly.com/question/13714615

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Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an
joja [24]

Answer:

Approximately 9.7\; \rm m \cdot s^{-2}.

Explanation:

Assuming that there is no other force on this vehicle, the 16000\; \rm N force from the road would be the only force on this vehicle. The net force would then be equal to this 16000\; \rm N\! force. The size of the net force would be 16000\; \rm N\!\!.

Let m denote the mass of this vehicle and let \Sigma F denote the net force on this vehicle.  

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, a, would be:

\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}.

For this vehicle, \Sigma F = 16000\; \rm N whereas m = 1650\; \rm kg. The acceleration of this vehicle would be:

\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}.

8 0
3 years ago
At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.50 gg? Assume the spacesh
Svetradugi [14.3K]

Answer:

The time needed is T  = 16.8 s

Explanation:

From the question we are told that

      The magnitude of the stimulated acceleration due gravity is  a  =  0.5 g

        The diameter of the spaceship is  d =  35m

       

Generally the force acting on the spaceship is  

       F  =  ma

Given that the spaceship is rotating it implies that the force experienced by the occupant is a centripetal force so

      F  = \frac{mv^2}{r}

Thus  

       ma  =  \frac{mv^2}{r}

=>    \frac{v^2}{r}  =  a

      Generally the speed of this spaceship is mathematically represented as

      v =  \frac{2 \pi}{T}

=>    v^2  =   [\frac{2\pi}{T}] ^2

=>     \frac{\frac{4\pi^2 r^2}{T^2} }{r}  = 0.5g

=>       \frac{4 \pi^2 r }{T^2} =  0.5 g

=>         T  = \sqrt{ \frac{4\pi^2 r}{0.5g}}

substituting values

          T  = \sqrt{ \frac{4* (3.142)^2 *(35)}{0.5 * 9.8}}

         T  = 16.8 s

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