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3 years ago

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What is horoscope? what is its uses

2 answers:

Lemur [1.5K]3 years ago

8 0

What is horoscope?
A forecast of a person's future, typically including delineation of character and circumstances, based on the relative positions of the stars and planets at the time of that person's birth.

*A short forecast for people born under a particular sign, especially as published in a newspaper or magazine.

*A birth chart.

What is its uses?
It can also be calculated for an event, a question, and even a country. Symbols are used to represent planets, signs, and geometric connections called aspects. In most cases, the horoscope in Western astrology is drawn on a circular wheel.

Eduardwww [97]3 years ago

6 0

A horoscope is an astrological chart that is calculated based upon the date,time and place of birth.
I think , may be it tells our future ....

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16x^2y^2-25a^2b^2<br>factorize the expression

SIZIF [17.4K]

Answer:

(4xy+5ab)(4xy-5ab)

Explanation:

16 $x^{2}$ $y^{2}$ -25 $a^{2}$ $b^{2}$

4^2 is 16 and 5^2 is 25,

Also, (x-a)(x+a) = x^2-a^2

So, this factorized is:

(4xy+5ab)(4xy-5ab)

Hope this helps!

8 0

2 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

2 years ago

Heat flows into a gas in a piston and work is performed on the gas by its surroundings. The amount of work done is equal to the

inna [77]

Answer:

The Internal energy of the gas did not change

Explanation:

In this situation the Internal energy of the gas did not change and this is because according the the first law of thermodynamics

Δ U = Q - W ------ ( 1 )

Δ U = change in internal energy

Q = heat added

W = work done

since Q = W. the value of ΔU will be = zero i.e. No change

4 0

2 years ago

Mechanical energy is the sum of ________ energy and potential energy.apex

crimeas [40]

Mechanical energy is the sum of kinetic energy and potential energy

6 0

3 years ago

True or false please help me now.

jolli1 [7]

Answer:

false

Explanation:

8 0

2 years ago