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Bad White [126]
3 years ago
6

Cold syrup flows sluggishly because of??​

Physics
2 answers:
Eva8 [605]3 years ago
5 0

Answer:

Water is very different from honey, syrup, glycerine, or oil. It pours easily and is not thick and sticky like the others. The property that determines how easily a liquid pours is called VISCOSITY. Water has a low viscosity; syrup has a high viscosity. Liquids with a high viscosity are said to be viscous.

8_murik_8 [283]3 years ago
4 0
Water has a low viscosity and syrup has a high
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Several types of radiation may be emitted during radioactive decay. The equation represents
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3 years ago
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How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0
3 years ago
In an LC circuit containing a 40-mH ideal inductor and a 1.2-mF capacitor, the maximum charge on the capacitor is 45mC during os
GalinKa [24]

Answer:

E) 6.5 A

Explanation:

Given that

L = 40 m H

C= 1.2 m F

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The maximum current I given as

I = Q.ω

ω =angular frequency

\omega=\dfrac{1}{\sqrt{CL}}

By putting the values

\omega=\dfrac{1}{\sqrt{CL}}

\omega=\dfrac{1}{\sqrt{40\times 10^{-3}\times 1.2 \times 10^{-3}}}

ω  = 144.33 rad⁻¹

Maximum current

I = 45 x 10⁻³ x 144.33  A

I= 6.49 A

I = 6.5 A

E) 6.5 A

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Answer:

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Baseball player swings and hits a pop fly straight up in the air to the catcher. the height of the baseball in meters t seconds
andre [41]
The height at time t is given by
h(t) = -4.91t² + 34.3t + 1

When the ball reaches maximum height, its derivative, h'(t) = 0.
That is,
-2(4.91)t+34.3 = 0
-9.82t + 34.3 = 0
t = 3.4929 s

Note that h''(t) = -9.82 (negative) which confirms that h will be maximum.

The maximum height is
hmax = -4.91(3.4929)² + 34.3(3.4929) + 1
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Answer:
The ball attains maximum height in 3.5 s (nearest tenth).
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4 0
3 years ago
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