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nikklg [1K]
3 years ago
11

Two gliders collide on an air track moving in from opposite directions, and they bounce back. The masses of the gliders are 0.20

kg and 0.30 kg. The magnitudes of their initial velocities are 0.50 m/s and 0.40 m/s, respectively. After the collision, the magnitudes of their velocities are 0.40 m/s and 0.20 m/s, respectively. Assume the first glider is initially heading in the positive direction.The total momentum of the gliders before collision is closest to:
Physics
1 answer:
timurjin [86]3 years ago
3 0

Answer:

so initial momentum is 0.22kgm/s

Explanation:

m1=0.20kg

m2=0.30kg

initial velocity of m1=u1=0.50m/s

initial velocity of m2=u2=0.40m/s

total momentum of the system before collision

Pi=m1u1+m2u2

Pi=0.20kg×0.50m/s+0.30kg×0.40m/s

Pi=0.1kgm/s+0.12kgm/s

Pi=0.22kgm/s

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A disk 8.00 cm in radiu: rotates at a constant rate of 1200 revinin about its central axis Determine (a) its angular speed in zadians per second, (b) the tangential speed at a point 3.00 cm from its center, (c) the radial aceleration of a point on the rim, and (d) the total distance a point our tke rim noves ign-2.00 s (E) The moment of inertia if it's mass is 2Kg? is the answer

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If the observed test value of a hypothesis test is outside of the established critical value(s), a researcher would __________.
sashaice [31]
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6 0
3 years ago
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Which of the following is not affect by muscles a blood temperature b digestion c hair growth d speech
Murrr4er [49]

Answer:

c

Explanation:

i would think c would be correct because

blood temperature IS affected by muscles

digestion IS affected by muscles

speech IS affected by muscles

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3 0
3 years ago
A hot-water stream at 80 ℃ enters a mixing chamber with a mass flow rate of 0.5 kg/s where it is mixed with a stream of cold wat
rewona [7]

Answer:

\dot{m_{2}}=0.865 kg/s

Explanation:

\dot{m_1}= 0.5kg/s

from steam tables , at 250 kPa, and at

T₁ = 80⁰C ⇒ h₁ = 335.02 kJ/kg

T₂ = 20⁰C⇒ h₂ = 83.915 kJ/kg

T₃ = 42⁰C ⇒ h₃ = 175.90 kJ/kg

we know

\dot{m_{in}}=\dot{m_{out}}

\dot{m_{1}}+\dot{m_{2}}=\dot{m_{3}}

according to energy balance equation

\dot{m_{in}}h_{in}=\dot{m_{out}}h_{out}

\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}

\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=(\dot{m_{1}}+\dot{m_{2}})h_{3}\\(0.5\times 335.02)+(\dot{m_{2}}\times 83.915)=(0.5+\dot{m_{2}})175.90\\\dot{m_{2}}=0.865 kg/s

4 0
3 years ago
Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate
masha68 [24]

Answer:

The heat loss per unit length is   \frac{Q}{L}   = 2981 W/m

Explanation:

From the question we are told that

     The outer diameter of the pipe is d = 104mm = \frac{104}{1000} = 0.104 m

     The thickness is  D = 2mm = \frac{2}{1000} = 0.002m  

      The temperature  of water is  T = 90^oC = 90 + 273 = 363K  

      The outside air temperature is T_a = -10^oC = -10 +273 = 263K

        The water side heat transfer coefficient is z_1 = 300 W/ m^2 \cdot K

       The  heat transfer coefficient is  z_2 = 20 W/m^2 \cdot K

The heat lost per unit length is mathematically represented as

           \frac{Q}{L}   = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1}  +  \frac{ln [\frac{d}{D} ]}{z_2}}

Substituting values

         \frac{Q}{L}   = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300}  +  \frac{ln [\frac{0.104}{0.002} ]}{20}}

           \frac{Q}{L}   = \frac{628}{0.2107}

           \frac{Q}{L}   = 2981 W/m

6 0
3 years ago
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