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qaws [65]
3 years ago
15

According to Bernoulli's fluid formula a An increase in the speed will lower the internal pressure b An increase in the speed wi

ll increase the internal pressure c The internal pressure doesn't change with the speed
Physics
1 answer:
lorasvet [3.4K]3 years ago
5 0

Answer:

a An increase in the speed will lower the internal pressure

Explanation:

Bernoulli's fluid formula

P_1+\frac{1}{2}\rho v_1^2+\rho gh_1=P_1+\frac{1}{2}\rho v_2^2+\rho gh_2

where

P = Pressure

ρ = Density of fluid

g = Acceleration due to gravity

h = Height

v = Velocity of fluid

If there is no change in height then we get

P_1+\frac{1}{2}\rho v_1^2=P_1+\frac{1}{2}\rho v_2^2\\\Rightarrow P+\frac{1}{2}\rho v^2=constant

According to the Bernoulli's principle when the speed of the fluid is larger in a region of streamline flow the pressure is smaller in that region. From the above equation it can be seen that increase in speed should simultaneously reduce pressure in order for their sum to be constant.

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A 50 cm^3 block of iron is removed from an 800 degrees Celsius furnance and immediately dropped into 200 mL of 20 degrees Celsiu
NNADVOKAT [17]

Answer:

 % of water boils away= 12.64 %

Explanation:

given,

volume of block  = 50 cm³ removed from temperature of furnace = 800°C

mass of water = 200 mL = 200 g

temperature of water  = 20° C

the density of iron = 7.874 g/cm³ ,

so the mass of iron(m₁)  = density × volume = 7.874 × 50 g = 393.7 g

the specific heat of iron C₁ = 0.450 J/g⁰C

the specific heat of water Cw= 4.18 J/g⁰C

latent heat of vaporization of water is L_v = 2260 k J/kg = 2260 J/g

loss of heat from iron is equal to the gain of heat for the water

m_1\times C_1\times \Delta T = M\times C_w\times \Delta T + m_2\times L_v

393.7\times 0.45\times (800-100) = 200\times 4.18\times(100-20) + m_2\times 2260

m₂ = 25.28 g

25.28 water will be vaporized

% of water boils away =\dfrac{25.28}{200}\times 100

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3 years ago
Why do solar eclipses move from west to east?
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4 years ago
you will For this problem, you will need to look up physical parameters for objects in space want to keep about 4 significant fi
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Answer:

a)  r₁ = 3.836 10⁷ m,  b)   F = - 3,390 10⁸ N , c)  R = 120.3 m

Explanation:

a) This is a problem of equilibrium where the force is gravitational, we call F1 the force of the Moon and F2 the force from the earth.

       F₁ -F₂ = 0

       F₁ = F₂

       G m M_{m} / r₁² = G m M_{e} / r₂²

Let's look for the measured distance from a Coordinate System located on the Moon,

         r₂ = D - r₁

Where D is the average distance from Terra to the Moon 3.84 10⁸ m

        M_{m} / r₁² = M_{e} / (D - r₁)²

       (D² - 2 D r₁ + r₁²) = M_{e} /M_{m} r₁²

       (1 - M_{e} / M_{m})r₁² - 2D r₁ + D² = 0

Let's replace and solve the second degree equation

       (1 - 5.98 10²⁴ / 7.36 10²²) r₁² - 2 3.84 10⁸ r₁ + (3.84 10⁸)² = 0

       -80.25 r₁² - 7.68 10⁸ r₁ + 14.75 10¹⁶ = 0

        r₁² + 9.57 10⁶ r₁ - 1.838 10¹⁵ = 0

        r₁ = [-9.57 10⁶ ±√ (91.58 10¹² + 7.352 10¹⁵)] / 2

        r₁ = [-9.57 10⁶ + - 86.28 10⁶] / 2

The results are:

       r₁’= 38.355 10 6 m

       r₁ ’’ = -47.915 10 6 m

We take the positive result that a distance between the moon and the Earth, the equalization point is    r₁ = 3.836 10⁷ m

b) The force at point R = 2 r₁

        R = 2 3.8355 10⁷ = 7.671 10⁷ m

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        F = G m  M_{m} / R² - G m  M_{e} / (D- R)²

        F = G m ( M_{m} / R² -  M_{e} / (D-R)²)

   F = m 6.67 10⁻¹¹ (7.36 10²² / (7.671 10⁷)² - 5.98 10²⁴ / (3.84 10⁸ - 0.7671 10⁸)²

        F = m 6.67 10⁻¹¹ (0.125076 10⁸ - 0.63329 10⁸)

        F = m (-3.3897 10⁸) N

The mass m of the rocket must be known, suppose it is worth 1 kg (m = 1 kg)

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c) let's use gravitational force from the moon

        F = G m  M_{m} / R²

        R =√ G m  M_{m} / F

        R = √ (6.67 10⁻¹¹ 1 7.36 10²² / 3.3897 10⁸)

        R = √ (1.4482 10⁴)

        R = 1.2034 10² m

        R = 120.3 m

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