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Sergeu [11.5K]
3 years ago
14

Internal Forces are :

Physics
1 answer:
Ugo [173]3 years ago
8 0






Hi Pupil Here's your answer ::






➡➡➡➡➡➡➡➡➡➡➡➡➡



They are always balanced forces since if they are applied on an object the resultant of tgese forces on the object is Zero

So, the option A is the correct. They are always balanced forces.
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A standard 1 kilogram weight is a cylinder 54.0 mm in height and 55.0 mm in diameter. what is the density of the material
denis-greek [22]

The radius of the cylinder is equal to half the diameter:

r=\frac{d}{2}=\frac{55.0 mm}{2}=27.5 mm

The volume of the cylinder is given by:

V=\pi r^2 h=\pi (27.5 mm)^2 (54.0 mm)=1.28 \cdot 10^5 mm^3

where h is the heigth of the cylinder. Converting into meters,

V=1.28 \cdot 10^{-4} m^3

And the density of the material will be given by the ratio between the mass and the volume:

d=\frac{m}{V}=\frac{1 kg}{1.28 \cdot 10^{-4} m^3}=7812.5 kg/m^3

5 0
3 years ago
8. Semiconductors
Andreas93 [3]

do not obey ohm's law so it's a I believe

5 0
3 years ago
A fixed mass of an ideal gas is heated from 50°C to 80°C (a) at constant volume and (b) at constant pressure. For which case do
soldi70 [24.7K]

Answer:

Specific heat at constant pressure is =  1.005 kJ/kg.K

Specific heat at constant volume is =  0.718 kJ/kg.K

Explanation:

given data

temperature T1 =  50°C

temperature T2 = 80°C

solution

we know energy require to heat the air is express as

for constant pressure and volume

Q  = m ×  c × ΔT     ........................1

here m is mass of the gas and c is specific heat of the gas and Δ T is change in temperature of the gas

here both Mass and temperature difference is equal and energy required is dependent on specific heat of air.

and here at constant pressure Specific heat  is greater than the specific heat at constant volume,

so the amount of heat required to raise the temperature of one unit mass by one degree at constant pressure is

Specific heat at constant pressure is =  1.005 kJ/kg.K

and

Specific heat at constant volume is =  0.718 kJ/kg.K

3 0
3 years ago
In the equation F = Kq1 q2/r2 solve for q2. Solve for r.
Tpy6a [65]

Answer:

r = √(k q₁ q₂ / F)

Explanation:

F = k q₁ q₂ / r²

Multiply both sides by r²:

F r² = k q₁ q₂

Divide both sides by F:

r² = k q₁ q₂ / F

Take the square root of both sides:

r = √(k q₁ q₂ / F)

3 0
2 years ago
what kind of weather is associated with high pressure?How does density,humidity, and air motion compare to that low pressure sys
frutty [35]
Usually nice weather , i dont know the answer to the second part                                                                                       
6 0
3 years ago
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