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3 years ago

I NEED HELP RIGHT NOW PLEASE HELP WITH question 2-13 !!!!! 19pts to anyone who helps me !!!!!!!

1 answer:

5 0

True
True
False*
False*
True
True
False
A,B,AB,O
10.)?
11.)water
carbon dioxide
12.)geocentric
13.)Juptier

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Which most likely has the greatest conductivity?

Rasek [7]

A warm copper wire will have more conductivity

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3 years ago

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The power radiated by the sun is 3.90 1026 W. The earth orbits the sun in a nearly circular orbit of radius 1.50 1011 m. The ear

WITCHER [35]

Answer:

3234.2 W

Explanation:

Since intensity I = Power/Area. The intensity of the light from the sun, I = power radiated by sun/area of sphere of radius, r = 1.5 × 10¹¹ m.

So, I = 3.9 10²⁶W/4π(1.5 × 10¹¹ m)² = 2.069 × 10³ W/m².

Now, the power radiated on the patch of area 0.570 m² at the equator is

P = Icos27/A = 2.069 × 10³ W/m² cos27/0.570 m² = 1843.49/0.570 = 3234.2 W

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2 years ago

RIGHT ANSWER GETS BRAINLIEST+FIVE STARS!

Reptile [31]

C is not mechanical weathering. It is chemical weathering.

4 0

3 years ago

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A 12kg cheetah accelerates 24 m/s". What is the force the cheetah needed to run?

Kobotan [32]

Answer:

288N

Explanation:

Given parameters:

Mass of Cheetah = 12kg

Acceleration = 24m/s²

Unknown:

Force needed by the cheetah to run = ?

Solution:

The force needed by the Cheetah to run is the net force.

According to Newton's law;

Force = mass x acceleration

Insert the given parameters and solve;

Force = 12 x 24 = 288N

7 0

3 years ago

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

4 0

3 years ago