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laila [671]
3 years ago
7

In science though,what is it mainly used for????

Physics
2 answers:
erik [133]3 years ago
6 0
It is mainly used to give facts about living organisms

Leona [35]3 years ago
5 0

This branch of philosophy is handily called the philosophy of science. ... What kind of data can be used to distinguish between real causes and accidental regularities? ... Though they might seem elementary, these questions are actually quite ... views of scientists themselves — who mainly spend their time doing science, .
You might be interested in
If there is a current of 15A in a circuit for 5 minutes, what quantity of electric charge flow through the circuit?
poizon [28]

charge = current * time

q = 15 * 300    [5 min = 300 sec ]

q = 4500 C

7 0
3 years ago
Assume a beam of light hits the boundary separating medium 1, with index of refraction n1 and medium 2, with index of refraction
Reptile [31]

Answer:

option C

Explanation:

Given,

Refractive index of medium 1 = n₁

Refractive index of medium 2 = n₂

For total internal reflection to take place light should move from denser medium to the rarer medium.

Here Total internal reflection take place at the boundary of medium 1 and medium 2 so, the refractive index of medium 1 is more than medium 2

 n₁ > n₂

The correct answer is option C

3 0
3 years ago
a 2.0-mole sample of an ideal gas is gently heated at constant temperature 330 k. it expands from initial volume 19 l to final v
shutvik [7]
Isothermal Work =  PVln(v₂/v₁)

PV = nRT =  2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J

Isothermal Work =  PVln(v₂/v₁)            v₂ = ? v₁ = 19L, 

1.7 kJ = (5487.24)In(v₂/19)

1700 = (5487.24)In(v₂/19)

In(v₂/19) = (1700/5487.24) = 0.3098

In(v₂/19) = 0.3098

(v₂/19) = e^{0.3098}


v₂  =  19* e^{0.3098}

v₂ = 25.8999

v₂ ≈ 26 L        Option b.
6 0
3 years ago
To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch.
ki77a [65]

Answer:

0.0133A

Explanation:

Since we have two sections, for the Inductor region there would be a current i_1. In the case of resistance 2, it will cross a current i_2

Defined this we proceed to obtain our equations,

For i_1,

\frac{di_1}{dt}+i_1R_1 = V

I_1 = \frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})

For i_2,

I_2R_2 =V

I_2 = \frac{V}{R_2}

The current in the entire battery is equivalent to,

i_t = I_1+I_2

i_t = \frac{V}{R_2}+\frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})

Our values are,

V=1V

R_1 = 95\Omega

L= 1.5*10^{-2}H

R_2 =360\Omega

Replacing in the current for t= 0.4m/s

i=\frac{1}{360}+\frac{1}{95}(1-e^{-\frac{95*0.4}{1.5*10^{-2}}})

i= 0.0133A

i_1 = 0.01052A

3 0
3 years ago
Ari, a young patient who is regularly treated by a psychiatrist, feels compelled to carry out repetitive tasks the same way very
blondinia [14]
OCD would probably be the answer.
5 0
3 years ago
Read 2 more answers
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