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vivado [14]
3 years ago
5

What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 11.0 m and the downward accel

eration of the car is 1.50 g?
Physics
1 answer:
8090 [49]3 years ago
3 0
<span>When an object moves in a circle, the acceleration points toward the center of the circle. This acceleration is called centripetal acceleration. We can use a simple equation to find centripetal acceleration. a = v^2 / r We can use this same equation to find the speed of the car. v^2 = a * r v = sqrt { a * r } v = sqrt{ (1.50)(9.80 m/s^2)(11.0 m) } v = 12.7 m/s The speed of the roller coaster is 12.7 m/s</span>
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in the primitive yo-yo apparatus (figure 1), you replace the solid cylinder with a hollow cylinder of mass m , outer radius r ,
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The magnitude of the downward acceleration of the hollow cylinder is 6m/s^2.

Z = I α

T.R =1/2 M ( R^{2} + (R/2)^{2} )α

T.R = 1/2M 5R^{2}/4 α

T = 5Ma/8

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4 0
1 year ago
Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is
Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of c_p is determined via the expression:

c_p = \frac{KR}{K_1}

where ;

R = universal gas constant = \frac{8.314 \ J}{28.97 \ kg.K}

k = constant = 1.4

c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}

c_p= 1.004 \ kJ/kg.K

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}     ------ equation(1)

we can rewrite the above equation as :

0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}

T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}

where:

T_1  = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K

T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}

T_2 = 402.36 \ K

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation:\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}

P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}

P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}

P_2 = 17.79 \ bar

Therefore, the exit pressure is 17.79 bar

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