Who wants to do my missing physics work

JulsSmile [24]

<u>Communications:</u> Laura develops programs that flag suspicious network activity.

<h3>What is communication?</h3>

Communication involves the transfer of information from one person (sender) to another (recipient), especially through the use of semiotics, symbols, signs and network devices.

<h3>How to match the scenarios?</h3>

<u>Communications:</u> Laura develops programs that flag suspicious network activity.

<u>Healthcare:</u> Greg calibrates electronic scales.

<u>Manufacturing:</u> Tim ensures the output of a process remains in a defined range.

Read more on communication here: brainly.com/question/26152499

#SPJ1

8 0

1 year ago

A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sph

arsen [322]

Answer:

option E

Explanation:

given,

I is moment of inertia about an axis tangent to its surface.

moment of inertia about the center of mass

$I_{CM} = \dfrac{2}{5}mR^2$ .....(1)

now, moment of inertia about tangent

$I= \dfrac{2}{5}mR^2 + mR^2$

$I= \dfrac{7}{5}mR^2$ ...........(2)

dividing equation (1)/(2)

$\dfrac{I_{CM}}{I}= \dfrac{\dfrac{2}{5}mR^2}{\dfrac{7}{5}mR^2}$

$\dfrac{I_{CM}}{I}=\dfrac{2}{7}$

$I_{CM}=\dfrac{2}{7}I$

the correct answer is option E

4 0

2 years ago

Would you have been a Federalist or an Anti-Federalist? Why?

Readme [11.4K]

Answer:

Explanation:

I would have been an anti-federalist. They favored a weak central government because they related it to the British tyranny. Federalists believed in a strong central government, almost controlling. I would not have respected this type of government.

6 0

2 years ago

A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.0

ivolga24 [154]

Since you solved a and b I will start with part c.
Part C
To answer this question we need to find zeros of a velocity function:
$v(t)=0.04t^3-0.06t^2$
We can factor this polynomial:
$v(t)=0.04t^3-0.06t^2=t^2(0.04t-0.06)$
Now it's pretty easy to find zeros. This function will be equal to zero when any of the factors are equal zero.
$t^2=0;\\ 0.04t-0.06=0$
We solve these two equations and we get our zeros:
$t_1=0; t_2=\frac{3}{2}$
The particle is at rest at t=0 and t=3/2.
Part D
To solve this we need to determine when our velocity function is greater than zero. We will use factored form.
We determine when each factor is greater than zero and with that information, we build the following table:
$\centering \label{my-label} \begin{tabular}{lllll} Range & -\infty & 0 & 3/2 & +\infty \\ t^2 & - & + & + & + \\ 0.04t-0.06 & - & - & + & + \\ t^2 (0.04t-0.06) & + & - & + & + \end{tabular}$
We can see, from the table, that our function is positive when $- \infty < t$ and t>3/2.
That is the range in which particle is moving in positive direction.
Part E
We know that distance traveled is given with:
$s(t)=0.01t^4 - 0.02t^3$
We simply plug in t=12 to find total distance traveled:
$s(12)=0.01(12)^4 - 0.02(12)^3=172.80 ft$
Part F
We know that acceleration is defined as a rate of change of velocity.
We find acceleration by taking the first derivative of velocity with respect to time.
$a(t)=\frac{dv}{dt}=(0.04t^3-0.06t^2)'=0.12t^2-0.12t$
To find acceleration after 1 second we simply plug in t=1s in above equation:
$a(1)=0.12-0.12=0$

7 0

3 years ago

The formula shown below is used to calculate the energy released when a specific quantity of fuel is burned. Calculate the energ

mafiozo [28]

Answer: 8400 J

Explanation:

The formula referenced in the question is:

$Q=m. c. \Delta T$