Based on the passage, how do the boys resolve their conflict?

ivanzaharov [21]

C. cooked noodles and water
because noodles are long and water has no shape or size.
if you have any problems with this answer,
comment and I will fix it.
Thank s!

6 0

3 years ago

Two common terms for a decrease in velocity are

Colt1911 [192]

deceleration or rėtardation i’m pretty sure (it won’t let me say the second word but it’s correct)

6 0

3 years ago

Read 2 more answers

From mechanics, you may recall that when the acceleration of an object is proportional to its coordinate, d2xdt2=−kmx=−ω2x , suc

marysya [2.9K]

Answer:

q = q₀ sin (wt)

Explanation:

In your statement it is not clear the type of circuit you are referring to, there are two possibilities.

1) The circuit of this problem is a system formed by an Ac voltage source and a capacitor, in this case all the voltage of the source is equal to the voltage at the terminals of the capacitor

ΔV = Δ $V_{C}$

we assume that the source has a voltage of the form

ΔV = ΔV₀o sin wt

The capacitance of a capacitor is

C = q / ΔV

q = C ΔV sin wt

the current in the circuit is

i = dq / dt

i = c ΔV₀ w cos wt

if we use

cos wt = sin (wt + π / 2)

we make this change by being a resonant oscillation

we substitute

i = w C ΔV₀ sin (wt + π/2)

With this answer we see that the current in capacitor has a phase factor of π/2 with respect to the current

2) Another possible circuit is an LC circuit.

In this case the voltage alternates between the inductor and the capacitor

V_{L} + V_{C} = 0

L di / dt + q / C = 0

the current is

i = dq / dt

they ask us for a solution so that

L d²q / dt² + 1 / C q = 0

d²q / dt² + 1 / LC q = 0

this is a quadratic differential equation with solution of the form

q = A sin (wt + Ф)

to find the constant we derive the proposed solution and enter it into the equation

di / dt = Aw cos (wt + Ф)

d²i / dt²= - A w² sin (wt + Ф)

- A w² + 1 /LC A = 0

w = √ (1 / LC)

To find the phase factor, for this we use the initial conditions for t = 0

in the case of condensate for t = or the charge is zero

0 = A sin Ф

Ф = 0

q = q₀ sin (wt)

6 0

3 years ago

A stationary 15 kg object is located in a table near the surface of the earth. The coefficient of static friction between the su

madreJ [45]

maximum static friction acting on the object will be

$F_s = \mu_s mg$

plug in all values

$F_s = 0.40 \times 15 \times 9.8 = 58.8 N$

So here it means that if applied force is less than or equal to 58.8 N then the object will remain stationary as friction can balance the external force upto this limit of external force

So here it is given that applied force is 20 N

so here object will not move due to this force and it will remain at rest always

due to this applied force

6 0

3 years ago

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,