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Colt1911 [192]
3 years ago
12

300 ml of a gas at 27°C is Cooled at -3°c at Constant pressure the final volume is plzz answer fast i will mark brainliest ​

Physics
1 answer:
Scorpion4ik [409]3 years ago
4 0

Answer:V₁=300ml

T₁=27°C

V₂=?

T₂= -3°C

as we know

V₁T₁=V₂T₂

By putting values in formula

300ml×27°C=V₂×(-3°C)

300ml×27°C/-3°C=V₂

8100ml/-3=V₂

-2700ml=V₂

or V₂=  -2700ml

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A compound has a melting point of 812°C. what type of bonding is in the compound?
AURORKA [14]

Answer:

Ionic compound.

Explanation:

812° C is a very high melting point. Such high melting points are generally ionic compound. Ionic compounds are have very strong bond between the elements ( electrostatic bond). In order to break this bond, large amount of heat energies are needed. So, they have high melting point. Also, Ionic compound are very good conductors of electricity.

7 0
3 years ago
What is happening to the ATOMS INSIDE of a magnet that gives the
Sliva [168]

Answer:

The atoms are aligned in a particular direction

Explanation:

The atoms become aligned in a particular direction in regions called domains, thus resulting in an overall resultant magnetism due to the spin of the electrons.

7 0
3 years ago
You are trying to overhear a most interesting conversation, but from your distance of 10.0 m , it sounds like only an average wh
Alexandra [31]

Answer:

r₂=0.1 m

Explanation:

Given that

r₁= 10 m  , β₁ = 20 dB

At r₂ ,β₂= 60 dB

As we know that intensity level of sound given as

\beta =10\ log\dfrac{I}{10^{-12}}

\beta _1=10\ log\dfrac{I_1}{10^{-12}}

20=10\ log\dfrac{I_1}{10^{-12}}

10² x 10⁻¹² = I₁

I₁=10⁻¹⁰ W/m²

\beta _2=10\ log\dfrac{I_2}{10^{-12}}

60=10\ log\dfrac{I_1}{10^{-12}}

10⁶ x  10⁻¹² = I₂

I₂ = 10⁻⁶ W/m²

I₁=10⁻¹⁰ W/m²

P = I A

P=Power ,I =Intensity  ,A=Area

\dfrac{I_1}{I_2}=\dfrac{r^2_2}{r^2_1}

\dfrac{10^{-10}}{10^{-6}}=\dfrac{r^2_2}{10^2}

r₂=0.1 m

4 0
3 years ago
Could you help me with this question?
Scorpion4ik [409]

Answer:

Magnitude of static friction force is 70 sin40° = 44.99 N.  

No, it is not necessary that it is maximum static friction.

Normal force is equal to 70 cos40° = 53.62 N.

Explanation:

We apply newton law of moton equation along the plane and perpendicular to plane;

Along the plane,

70 sin 40° = f_{r}  ---------------(1)

70 cos 40° = N --------------(2)

f_{r}_{max} = μN -----------------(3)

So, it depends on the value of μ that the friction is maximum or not .

7 0
3 years ago
In reaching her destination, a backpacker walks with an average velocity of 1.41 m/s, due west. This average velocity results, b
V125BC [204]
The total average velocity v=+1.41 m/s (I assume west as positive direction) is given by the total displacement, S, divided by the total time taken, t:
v= \frac{S}{t}= \frac{S_1+S_2}{t_1+t_2}
where:
-The total displacement S is the algebraic sum of the displacement in the first part of the motion (S_1=6.30 km=6300 m, due west) and of the displacement in the second part of the motion (S_2, due east).
-The total time taken t is the time taken for the first part of the motion, t_1, and the time taken for the second part of the motion, t_2. t_1 can be found by using the average velocity and the displacement of the first part:
t_1= \frac{S_1}{v_1}= \frac{6300 m}{2.49 m/s}=2530 s

t_2, instead, can be written as \frac{S_2}{v_2}, where v_2=-0.630 m/s is the average velocity of the second part of the motion (with a negative sign, since it is due east). 

Therefore, we can rewrite the initial equation as:
v=1.41 = \frac{6300+S_2}{2530- \frac{S_2}{0.630} }
And by solving it, we find the displacement in the second part of the motion (i.e. how far did the backpacker move east):
S_2=-844 m=-0.844 km

4 0
3 years ago
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