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Assoli18 [71]
3 years ago
15

The diagram shows the top view of a 65-kg student at point A on an amusement park ride. The ride spins the student in a horizont

al circle with a radius of 2.5 m, at a constant speed of 8.6 m/s. The floor is lowered and the student remains against the wall without falling to the floor.
The magnitude of the centripetal force acting on the student at point A is approximately

Question 4 options:

1398 N

47 N

223 N

1923 N

Physics
1 answer:
Elza [17]3 years ago
3 0

Answer:

1923 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 65 Kg

Radius (r) = 2.5 m

Velocity (v) = 8.6 m/s

Centripetal force (F) =?

The centripetal force, F, can be obtained by using the following formula:

F = mv²/r

F = 65 × 8.6² / 2.5

F = 65 × 73.96 / 2.5

F = 4807.4 / 2.5

F = 1922.96 ≈ 1923 N

Thus, the magnitude of the centripetal's force acting on the student is approximately 1923 N

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A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the m
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Answer

given,                                                

Tension of string is F                                                  

velocity is increased and the radius is not changed.      

the string makes two complete revolutions every second

consider the centrifugal force acting on the stone          

  = \dfrac{mv^2}{r}                          

now centrifugal force is balanced by tension

T =\dfrac{mv^2}{r}                                

From the above expression we can clearly see that tension is directly proportional to velocity and inversely proportional to radius.

When radius is not changing velocity is increasing means tension will also increase in the string.

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One of the most efficient heat engines ever built is a coal-fired steam turbine in the Ohio River valley, operating between 1 87
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Answer:The answer

Explanation:

5 0
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What is the internal energy of 2.00 mol of diatomic hydrogen gas (H2) at 35°C?
djyliett [7]
As you mentioned, we will use <span>Equipartition Theorem.
</span><span>H2 has 5 degrees of freedom; 3 translations and 2 rotation
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4 0
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Which phase of matter does not conduct heat well?
inessss [21]
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4 0
2 years ago
Read 2 more answers
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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