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faltersainse [42]
3 years ago
7

Who’s of the following is classified as a salt: ammonia or sodium chloride?

Physics
1 answer:
Temka [501]3 years ago
8 0

Sodium Chloride could be classified as a salt.

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Explain how thermal conductivity occurs in poor conductor materials​
DIA [1.3K]

Answer:

• As heat is applied to one end of this material, atoms in the hotter region gain vibràtory energy at a maximum amplitude. They transfer it on to their neighboring atoms and heat is transfered along this material in form of vibràtory energy.

Explanation:

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3 0
3 years ago
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10. when creating a cardiovascular fitness routine which program would be appropriate?
sasho [114]

Answer:

10. B

13. It's actually 30 seconds but 20 comes closest so B

14. B, any type of stretching is good before you perform any type of physical activity because it loosens up the muscles

15. D, all of the above

Explanation:

I have done many sports in my life and am still very active, I grew to know most of these things through the years.

5 0
3 years ago
Physics questions please help me
Ahat [919]
Hi,Find  answers from Task 5

1.(X+4)+(X)+(X+4)+(X)=50cm

4x+8=50cm

4x=42

X=10.5cm

Length=10.5+4=14.5cm

Width=10.5cm

Area= length × width=(10.5/100) × (14.5/100) =0.0152m2

2. Volume of a sphere= 4/3 ×π×r³

4/3 ×π×r³=3.2×10^-6 m³

r³=3.2×10^-6 m³/1.33×π

r³=7.64134761e-7

r=0.00914m

Surface area of the blood drop= 4πr²

=4×3.142×0.00914×0.00914=0.00105m²

3.

Equation of an ideal gas     = PV =n RT  

Equation for pressure, = P= n RT/V

Equation for the volume of an ideal gas= V= n RT/P

If the volume of gas doubles ,V(new)=  2n RT/P

Equation for temperature of an ideal gas, T = PV/n R

If temperature of gas triples, T (new)= 3PV/n R

New Equation for Pressure, = n× R× (3PV/n R)/(2n RT/P)

Pressure factor increase= P(new)/P(old)  ={ n× R× (3PV/n R)/(2n RT/P)}/{ n RT/V}

=3PV²/2n RT
5 0
3 years ago
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Sever21 [200]

Answer:2.89\approx 2.9^{\circ}C/s

Explanation:

Given

Power\left ( P\right )=150 MW

mass of core\left ( m\right )=1.60\times 10^5 kg

Average specific heat \left ( C\right )=0.3349 KJ/kg^{\circ}C

And rate of increase of temperature =\frac{\mathrm{d}T}{\mathrm{d} t}

Now

P=mc\frac{\mathrm{d}T}{\mathrm{d} t}

150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}

Thus \frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}

\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s

6 0
4 years ago
The place where things are pushed together is.. <br><br> (Plz help me, I’m in a quiz)
kogti [31]
Can I see the book you read for I can answer and it more info because your question doesn’t make sense
7 0
3 years ago
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