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SCORPION-xisa [38]

3 years ago

5

What is one limitation of using the jawbreaker as a model for Earth's layers? A. The layers are all made out of the same materia

l. B. The outside layer is the thinnest layer. C. The jawbreaker has four different layers. D. The jawbreaker can be split to show the inside.

2 answers:

schepotkina [342]3 years ago

7 0

The layers of a jawbreaker is all made of the same material. The layers of the Earth are not, it is made of different kinds of rock, magma, and metal is inside of the core

Margarita [4]3 years ago

4 0

The correct answe is A

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The formula K2S indicates that

jekas [21]

There are two atoms of potassium bonded to one atom of sulfur.

4 0

3 years ago

1. What is meant by half-life?

Bezzdna [24]

1. The time for a radioactive sample to reduce to half of its original mass.
4. 10% of 232 is 23.2 times 6 equals 139.2
68.9 x6 equals 413.4
5. 0.37kg
6. 2000 years
7. 6.84 seconds

8 0

4 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

<h3>a)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

<h3>b)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

7 0

3 years ago

Two identical silver spheres of mass m and radius r are placed a distance R (sphere 1) and 2R (sphere 2) from the Sun, respectiv

lys-0071 [83]

Answer:

The ratio of T2 to T1 is 1.0

Explanation:

The gravitational force exerted on each sphere by the sun is inversely proporational to the square of the distance between the sun and each of the spheres.

Provided that the two spheres have the same radius r, the pressure of solar radiation too, is inversely proportional to the square of the distance of each sphere from the sun.

Let F₁ and F₂ = gravitational force of the sun on the first and second sphere respectively

P₁ and P₂ = Pressure of solar radiation on the first and second sphere respectively

M = mass of the Sun

m = mass of the spheres, equal masses.

For the first sphere that is distance R from the sun.

F₁ = (GmM/R²)

P₁ = (k/R²)

T₁ = (F₁/P₁) = (GmM/k)

For the second sphere that is at a distance 2R from the sun

F₂ = [GmM/(2R)²] = (GmM/4R²)

P₂ = [k/(2R)²] = (k/4R²)

T₂ = (F₂/P₂) = (GmM/k)

(T₁/T₂) = (GmM/k) ÷ (GmM/k) = 1.0

Hope this Helps!!!

3 0

3 years ago

A girl throws a rock horizontally at 10 m/s from the top of a building, 22 m above street level. Assuming free fall conditions a

MAXImum [283]

Answer:21.18 m

Explanation:

Given

initial speed u=10 m/s

height of building h=22 m

time taken to complete 22 m

$h=ut+\frac{1}{2}at^2$

initial vertical velocity =0

$22=\frac{1}{2}gt^2$

$t=\sqrt{\frac{22\times 2}{g}}$

$t=2.11 s$

Horizontal Distance moved

$R=u_x\times t$

$R=10\times 2.11$

$R=21.18 m$

6 0

3 years ago