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chubhunter [2.5K]
3 years ago
13

At what minimum speed must a roller coaster be traveling when upside down at the top of a circle (fig. 5-34) so that the passeng

ers will not fall out? assume a radius of curvature of 9.6 m.
Physics
1 answer:
almond37 [142]3 years ago
4 0
At top, 2 forces act on passengers: normal force, force of gravity. The two act downward in the direction of the center of the circle. 
To answer the problem we would be using this formula which is apply N2L: 
Fnet = ma 
mg + N = mv^2/r 
minimum speed occurs when N = 0 
mg = mv^2/r 
v = sqrt (rg) 
= √94.09
v = 9.7 m/s
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A trunk rotation is a common dyamic flexibility assessment ture or false
Ilia_Sergeevich [38]

True. It would be false if the statement was "trunk rotation is the most common <em>static</em> flexibility assessment."


So, you're answer should be "true". Hope that helped!

6 0
3 years ago
A box full of charged plastic balls sits on a table. The electric force exerted on a ball near one upper corner of the box has c
tatuchka [14]

We have that the values for F north, F east, F up are

  • F_N=1.09090909*10^{-5}
  • F_E=5.18181818*10^{-6}
  • F_E=2*10^{-6}

From the Question we are told that

electric force F_1 = 1.2 x 10^{-3} N(N)

electric force , F_2=5.7 x 10^{-4} N(E)

electric force , F_3=2.2 x 10^{-4} N (U)

charge on this ball one q_1= 110 nC.

charge on this ball two q_2= -50 nC.

Generally the equation for the F north  is mathematically given as

F_N=\frac{F_1}{q_1}\\\\F_N=\frac{ 1.2 * 10^{-3} )}{110}

F_N=1.09090909*10^{-5}

For F East

F_E=\frac{F_2}{q_1}\\\\F_E=\frac{5.7 x 10^-4 }{110}

F_E=5.18181818*10^{-6}

For F UP

F_U=\frac{F_3}{q_1}\\\\F_U=\frac{2.2 x 10^-4 }{110}

F_E=2*10^{-6}

For more information on this visit

brainly.com/question/21811998

5 0
3 years ago
What is the gravitational potential energy of a 15.0kg object that is 5.00m above the ground relative to a point 8.00m above the
Licemer1 [7]

Explanation:

an object's gravitational potential energy Eg is m×g×h where:

m=mass

g=9.8m/s²

h=height relative to the closest object below it (because it cannot potentially fall through it

so Eg = 15×9.8×5=735J

4 0
2 years ago
If the air pressure is doubled, the speed of sound
Tom [10]

it remains unchanged

6 0
3 years ago
A thermometer initially reading 212F is placed in a room where the temperature is 70F. After 2 minutes the thermometer reads 125
frez [133]

Answer:

91.3°F

Explanation:

Let T be the temperature of the thermometer at any time

T∞ be the temperature of the room = 70°F

T₀ be the initial temperature of the thermometer = 212°F

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the cake = Rate of Heat gain by the environment

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 70) = (212 - 70)e⁻ᵏᵗ

(T - 70) = 142 e⁻ᵏᵗ

At t = 2 minute, T = 125°F

125 - 70 = 142 e⁻ᵏᵗ

55/142 = e⁻ᵏᵗ

- kt = In (55/142) = In (0.3873)

- k(2) = - 0.9485

k = 0.4742 /min

At time t = 4 mins

kt = 0.4742 × 4 = 1.897

(T - 70) = 142 e⁻ᵏᵗ

e^(-1.897) = 0.15

T - 70 = 142 × 0.15 = 21.3

T = 91.3°F

7 0
3 years ago
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