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Vedmedyk [2.9K]
2 years ago
15

How to get to physics questions on this site?thanks

Physics
2 answers:
tester [92]2 years ago
6 0
Select physics on the ask question drop down menu.
NeX [460]2 years ago
3 0
-- If you want to ASK a Physics question . . .
Click on the orange block over there  =====>
that says " + Ask a question".
In the new window, there is a drop-down menu where you can
select the category you want your question to go into.

-- If you want to SEE all the Physics questions that other people
have already asked . . .
. . . Click on the orange funnel up at the top, next to the "brainly.com" .
. . . This will open a screen full of all the questions that have been asked
in EVERY category ... all the questions on the site.
. . . At the top of the blue part of the screen, there's a strip that says "All Subjects".
When you click in there, you get a list of categories.  You can click on the one
you're interested in, and the screen changes to only show questions in the
subject you picked.
You might be interested in
A feather, a tennis ball, and a bowling ball are dropped from a high tower on
Vlada [557]

Answer:

In the absence of air resistance. I think no. D ) The bowling ball.

<em><u>Hope</u></em><em><u> this</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

7 0
2 years ago
En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a
Maru [420]

Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

Q=mc(T_2-T_1)

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

Q_1=-Q_2

for water we have to

c = 4.18J / g ° C

replacing we have

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

Q=mc(T_2-T_1)

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

Q_1=-Q_2

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

7 0
2 years ago
Question 9
Anastasy [175]

nAnswer:

uyui

Explanation:

4 0
2 years ago
You have about 10 quarts of blood in your body. At REST your heart pumps about 5 quarts each minutes. That is half of your blood
Zina [86]

Answer:

8 times

Explanation:

Given that You have about 10 quarts of blood in your body. At REST your heart pumps about 5 quarts each minutes.

That means the heart will pump 10 quarts in 2 minutes.

That is half of your blood volume per minute.

If during exercise it can pump 40 quarts per minute, that is, 80 quarts in 2 minutes.

To know how many times does all of your blood complete the cycle around your body during exercise, you must divide 80 quarts by 10 quarts. That is,

80 / 10 = 8

Therefore, your blood complete the cycle around your body 8 times during the exercise.

3 0
3 years ago
One litre of crude oil weighs 9.6N. Calculate its specific weight, density and specific gravity.​
Zepler [3.9K]

Answer:

The answer is "\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}"

Explanation:

Given:

\to v=1\ liter= 10^{-3} \ m^3\\\\\to  w= 9.6 \ N\\

calculation:

Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978

4 0
2 years ago
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