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Dovator [93]
3 years ago
15

A child on a bridge throws a rock straight down to the water below. The point where the child released the rock is 82 m above th

e water and it took 1.3 s for the rock to reach the water. Determine the rock's velocity (magnitude
Physics
1 answer:
eimsori [14]3 years ago
6 0

Answer:

40 m/s

Explanation:

given,

height of the fall, h = 82 m

time taken to fall, t = 1.3 s

rock velocity, v = ?

acceleration due to gravity, g = 9.8 m/s²

rock is released initial velocity, u = 0 m/s

using equation of motion

v² = u² + 2 a s

v² = 0 + 2 x 9.8 x 82

v² = 1607.2

v = 40 m/s

hence, rock's velocity is equal to 40 m/s

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4 0
4 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
There is a 50 g sample of Ra-229. It has a half-life of 4 minutes.
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Via half-life equation we have:


A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h} }


Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes.  By plugging those values in we get:

A_{final}=50(\frac{1}{2})^\frac{12}{4}=50(\frac{1}{2})^{3}=50(\frac{1}{8})=6.25g


There is 6.25 grams left of Ra-229 after 12 minutes.

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4 years ago
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