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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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An electron of mass 9.11x10^-31 kg has an initial speed of 2.40x10^5 m/s. It travels in a straight line, and its speed increases
egoroff_w [7]

Answer: A) Force = 3.841*10^-18 N.

B) force (f) is 4.30* 10^12 times greater than the weight (Fg).

Explanation: mass of electronic charge = 9.11*10^-31kg

v = final velocity = 6.80*10^5 m/s

u = initial velocity = 2.40 * 10^5 m/s

S= distance covered = 4.8cm = 0.048m

a = acceleration

Since the acceleration of the electron is assumed to be constant, newton's laws of motion are valid.

Thus, recall that

v² = u² + 2aS

(6.80*10^5)² = ( 2.40*10^5)² + 2*a( 0.048)

46.24 * 10^10 = 5.76 * 10^10 + 0.096a

46.24 *10^10 - 5.76* 10^10 = 0.096a

40.48* 10^10 = 0.096a

a = 40.48 * 10^10/0.096

a = 4.2167*10^12m/s².

Force = mass * acceleration

Force = 9.11*10^-31 * 4.2167*10^12

Force = 3.841*10^-18 N.

Weight =Fg= mg where g = acceleration due gravity = 9.8m/s²

Fg= 9.11*10^-31 * 9.8

Fg = 8.9278* 10^-30 N

By comparing the force and the weight, we have that

F/Fg = 3.841 * 10^-18/8.9278 * 10^-30 = 4.30* 10^12.

This implies that the force (f) is 4.30* 10^12 times greater than the weight (Fg).

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3 years ago
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Two identical particles of charge 6 μμC and mass 3 μμg are initially at rest and held 3 cm apart. How fast will the particles mo
krek1111 [17]

Answer:

Explanation:

The charges will repel each other and go away with increasing velocity , their kinetic energy coming from their potential energy .

Their potential energy at distance d

= kq₁q₂ / d

= 9 x 10⁹ x 36 x 10⁻¹² / 2 x 10⁻² J

= 16.2 J

Their total kinetic energy will be equal to this potential energy.

2 x 1/2 x mv² = 16.2

= 3 x 10⁻⁶ v² = 16.2

v = 5.4 x 10⁶

v = 2.32 x 10³ m/s

When masses are different , total P.E, will be divided between them as follows

K E of 3 μ = (16.2 / 30+3) x 30

= 14.73 J

1/2 X 3 X 10⁻⁶ v₁² = 14.73

v₁ = 3.13 x 10³

K E of 30 μ = (16.2 / 30+3) x 3

= 1.47 J

1/2 x 30 x 10⁻⁶ x v₂² = 1.47

v₂ = .313 x 10³ m/s

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3 years ago
A 5-cm-high peg is placed in front of a concave mirror with a radius of curvature of 20 cm.
Andrei [34K]

Answer:

Explanation:

Using the magnification formula.

Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)

M = v/u = H1/H2

v/u = H1/H2...1

3) Given the radius of curvature of the concave lens R = 20cm

Focal length F = R/2

f = 20/2

f = 10cm

Object distance u = 5cm

Object height H2= 5cm

To get the image distance v, we will use the mirror formula

1/f = 1/u+1/v

1/v = 1/10-1/5

1/v = (1-2)/10

1/v =-1/10

v = -10cm

Using the magnification formula

(10)/5 = H1/5

10 = H1

H1 = 10cm

Image height of the peg is 10cm

4) If u = 15cm

1/v = 1/f-1/u

1/v = 1/10-1/15

1/v = 3-2/30

1/v = 1/30

v = 30cm

30/15 = H1/5

15H1 = 150

H1/= 10cm

5) if u = 20cm

1/v = 1/f-1/u

1/v = 1/10-1/20

1/v = 2-1/20

1/v = 1/20

v = 20cm

20/20 = H1/5

20H1 = 100

H1 = 5cm

6) If u = 30cm

1/v = 1/f-1/u

1/v = 1/10-1/30

1/v = 3-1/30

1/v = 2/30

v = 30/2 cm

v =>15cm

15/30 = Hi/5

30H1 = 75

H1 = 75/30

H1 = 2.5cm

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2 years ago
What day of the year is solar time the same as sidereal time?
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Answer:

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