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Umnica [9.8K]
1 year ago
5

Glycerin at a temperature of 30 degrees celcius flows at a rate of 8×10−6m3/sthrougha horizontal tube with a 30mmdiameter. what

is the pressure drop in pascals per100m?
Physics
1 answer:
marusya05 [52]1 year ago
3 0

The pressure drop in pascal is 3.824*10^4 Pascals.

To find the answer, we need to know about the Poiseuille's formula.

<h3>How to find the pressure drop in pascal?</h3>
  • We have the Poiseuille's formula,

                     Q=\frac{\pi r^4P}{8\beta l}

  • where, Q is the rate of flow, P is the pressure drop, r is the radius of the pipe, is the coefficient of viscosity (0.95Pas-s for Glycerin) and l being the length of the tube.
  • By substituting values and rearranging we will get the pressure drop as,

                  P=3.284Pascals

Thus, we can conclude that, the pressure drop in pascal is 3.824*10^4.

Learn more about the Poiseuille's formula here:

brainly.com/question/13180459

#SPJ4

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In a ballistics test, a 52g bullet hits a sand bag and stops after moving 1.34 m. If the initial bullat
olya-2409 [2.1K]

Answer:

Friction force on the bullet is 58.7 N opposite to its velocity

Explanation:

As we know that initial speed of the bullet is 55 m/s

after travelling into the sand bag by distance d = 1.34 m it comes to rest

so final speed

v_f = 0

now we can use kinematics top find the acceleration of the bullet

v_f^2 - v_i^2 = 2 a d

so we have

0 - 55^2 = 2(a)(1.34)

a = -1128.7 m/s^2

now by Newton's II law we know that

F = ma

so we have

F = (0.052)(-1128.7)

F = -58.7 N

8 0
3 years ago
In vacuum , the shorter the wavelength of an electromagnetic wave is , the:
skad [1K]

Answer:

Higher its Energy

Explanation:

7 0
2 years ago
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Firlakuza [10]
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3 0
3 years ago
"Determine the magnitude of the net force of gravity acting on the Moon during an eclipse when it is directly between Earth and
spayn [35]

Answer:

Net force = 2.3686 × 10^(20) N

Explanation:

To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.

Now, from standards;

Mass of earth;M_e = 5.98 × 10^(24) kg

Mass of moon;M_m = 7.36 × 10^(22) kg

Mass of sun;M_s = 1.99 × 10^(30) kg

Distance between the sun and earth;d_se = 1.5 × 10^(11) m

Distance between moon and earth;d_em = 3.84 × 10^(8) m

Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m

Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²

Now formula for gravitational force between the earth and the moon is;

F_em = (G × M_e × M_m)/(d_em)²

Plugging in relevant values, we have;

F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²

F_em = 1.9909 × 10^(20) N

Similarly, formula for gravitational force between the sun and moon is;

F_sm = (G × M_s × M_m)/(d_sm)²

Plugging in relevant values, we have;

F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×

7.36 × 10^(22))/(1496.96 × 10^(8))²

F_se = 4.3595 × 10^(20) N

Thus, net force = F_se - F_em

Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N

8 0
3 years ago
A 34.1-mL sample of benzene at 20.8°C was cooled to its melting point, 5.5 °C, and then frozen at 5.5 °C. Calculate the quantity
posledela

Answer:

4541.8 J

Explanation:

First we find the mass of benzene available

mass = density x volume

         = 0.867 x 34.1

         = 29.5647 g

Then we find the amount of heat transferred by two processes:

heat tranferred = heat lost during temp drop + heat lost during freezing

                         = mcΔT + mL

                         = 29.5647 x 1.74 x (20.8 - 5.5) + 29.5647 x 127

                         = 4541.7883434 J

                         = 4541.8 J

                         

3 0
3 years ago
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