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3 years ago

What is the lithosphere? Select all that apply.

Physics

2 answers:

alekssr [168]3 years ago

8 0

The lithosphere is the solid, outer part of the Earth, including the brittle upper portion of the mantle and the crust.

jeka943 years ago

8 0

Answer:

i think that the option A is right

Explanation:

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Belle is walking south at an average velocity of 6 km/h. How many kilometers can she walk in three hours?

omeli [17]

Answer:

I think the answer is 18 km/h

Explanation:

6 0

3 years ago

A certain material has a mass of 565 g while occupying 50 cm3 of space. What is this material?

EastWind [94]

<u>Answer:</u>

Lead

<u>Explanation:</u>

To get the density of the material, the formula would be:

mass divided by volume which is given by $d = \frac { m } { v }$ .

Here in this problem, we are given a mass of $565 g$ which occupies a volume of $50 cm^3$ .

So plugging the data in the above formula to find the density:

Density = $\frac { 565 } { 50 } = 11.3$

From the table, we can see that the material is Lead which has a density of 11.3c/cm^3.

8 0

3 years ago

What is the general pattern of the planetary distances from the sun

eimsori [14]

Answer:

The Titius–Bode law (sometimes termed just Bode's law) is a hypothesis that the bodies in some orbital systems, including the Sun's, orbit at semi-major axes in a function of planetary sequence. The formula suggests that, extending outward, each planet would be approximately twice as far from the Sun as the one before.

Explanation:

8 0

4 years ago

Calculate the power produced by a 12 V battery charger if it delivers:A. 10 A in the fast charge modeB. 2 A in trickle charge mo

Novay_Z [31]

The electric power (P) produced by an electric current (I) passing through a voltage (V) is given by the equation:

$P=VI$

If the battery charger works with a voltage of 12V, replace V = 12V into the equation, as well as the corresponding values for the electric current to find the power produced in each case.

A) I = 10A

Replace V=12V and I=10A to find the power produced in the fast charge mode:

$P=(12V)(10A)=120W$

B) Replace V=12V and I=2A to find the power produced in the trickle charge mode:

$P=(12V)(2A)=24W$

Therefore, the power produced by a 12V battery charger is equal to 120 Watts when it delivers 10A in fast charge mode, and 24 Watts when it delivers 2A in trickle charge mode.

3 0

1 year ago

A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca

svetoff [14.1K]

Answer:

$q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Explanation:

Given that $L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V$ , we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

$E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000 \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}$

#To find the particular solution:

$Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Hence the charge at any time, t is $q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

6 0

4 years ago