Ask question

Ask question

All categories

3 years ago

11

A computer software update involved updating the flash memory of a hardware component. The update failed. A phone technician sai

d the component required replacement, but based on knowledge of how flash memory worked, the user suggested manually downloading the software, which worked.
A. True
B. False

1 answer:

Alexandra [31]3 years ago

3 0

Answer:

Yes, it's true. Computers do work that way. It's experienced by one of the authors of the book how computers work.

Explanation:

You might be interested in

Friction is generated when?

valkas [14]

Things are rubbed against each other

8 0

3 years ago

A student sees a newspaper ad for an apartment that has 2030 square feet (ft2) of floor space. How many square meters of area ar

Vilka [71]

Square feet or square meters?

8 0

3 years ago

Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is

marshall27 [118]

Answer:

$x=0.0049\ m= 4.9\ mm$

$d=0.01153\ m=11.53\ mm$

Explanation:

Given:

mass of the object, $m=0.75\ kg$

elastic constant of the connected spring, $k=150\ N.m^{-1}$

coefficient of static friction between the object and the surface, $\mu_s=0.1$

(a)

Let x be the maximum distance of stretch without moving the mass.

<em>The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.</em>

$f=k.x$

$\mu_s.N=k.x$

where:

N = m.g = the normal reaction force acting on the body under steady state.

$0.1\times (9.8\times 0.75)=150\times x$

$x=0.0049\ m= 4.9\ mm$

(b)

Now, according to the question:

Amplitude of oscillation, $A= 0.0098\ m$

coefficient of kinetic friction between the object and the surface, $\mu_k=0.085$

Let d be the total distance the object travels before stopping.

<em>Now, the energy stored in the spring due to vibration of amplitude:</em>

$U=\frac{1}{2} k.A^2$

<u><em>This energy will be equal to the work done by the kinetic friction to stop it.</em></u>

$U=F_k.d$

$\frac{1}{2} k.A^2=\mu_k.N.d$

$0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d$

$d=0.01153\ m=11.53\ mm$

<em>is the total distance does it travel before stopping.</em>

7 0

3 years ago

A worker is holding a filled gas cylinder still. Which two sentences are true about the energy of the filled gas cylinder?

MA_775_DIABLO [31]

The answer is B and C

3 0

3 years ago

Read 2 more answers

William was swimming at a speed of 3 m/s when he started to slow down. He slowed to a stop in 2 seconds. What is his acceleratio

schepotkina [342]

Answer:

$-1.5\;\rm m \cdot s^{-2}$ on average.

Explanation:

The acceleration of an object is the rate of change in its velocity. The average acceleration of an object is equal to the change in velocity over the time required for that change to take place.

For the William in this question:

The initial speed of William is $3\; \rm m \cdot s^{-1}$ .
Because William came to a complete stop after $2\; \rm s$ , his speed at that time would be $0\; \rm m \cdot s^{-1}$ .

Had William not came to a complete stop, it would be necessary to consider whether there is a change to his orientation. However, because in this question William came to a stop, the change in his velocity would be:

$\begin{aligned} &\text{Change in velocity} \\&= \text{Final velocity} - \text{Initial velocity}\\ &= 0\; \rm m \cdot s^{-1} - 3\; \rm m \cdot s^{-1} = -3\; \rm m \cdot s^{-1}\end{aligned}$ .

That change in velocity happened over $2\; \rm s$ . Therefore, the average acceleration would be:

$\begin{aligned}&\text{Average Acceleration} = \frac{\text{Change in velocity}}{\text{Time taken}} = \frac{-3\; \rm m \cdot s^{-1}}{2\; \rm s} \approx -1.5\; \rm m \cdot s^{-1}\end{aligned}$ .

3 0

3 years ago