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RSB [31]
3 years ago
12

According to the Law of Conservation of Energy, why does the first hill on a roller coaster always have to be the tallest of all

the other hills?
Physics
1 answer:
enyata [817]3 years ago
5 0
Because it generates enough momentum to keep the train going with out really having to speed up
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On a smooth horizontal floor an object
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what do you mean by this?

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What is the total mass of the object if the scale is balanced? (The black triangles show the location of the riders.) Do NOT inc
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If I'm understanding the picture, it looks like the riders add up to <em>117.8</em> .

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We are trying to reduce the current time of 15s by 30% but this leads to an increase of 18% in the CPI. For each processor, what
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Answer:

Explanation:

For reduction

Execution time = (instructions * CPI)/(Clock rate)

Since CPI is increased by 18%

18/100 +1 = 1.18

Therefore

P1 CPI = CPI x 1.18

P2 CPI = CPIx 1.18

P3 CPI = CPI x 1.18

Since clockrate reduces by 30% we have it to be a 0.7

P1: (Instructions x P1CPI) / 0.7

P2: (Instructionsx P2CPI) / 0.7

P3: (Instructions x P2CPI) /0 .7

5 0
3 years ago
A compact disc rotates from rest up to an angular speed of 31.4 rad/s in a time of 0.892 s.
Serggg [28]

Answer:

(a)  α = 35.20 rad/s^2

(b)  θ = 802°

(c)   v = 139.73 cm/s

(d)   a = 156.64 cm/s^2

Explanation:

(a) To find the angular acceleration of the disc you use the following formula:

\alpha=\frac{\omega-\omega_o}{t}         (1)

w: angular speed of the disc = 31.4 rad/s

wo: initial angular speed = 0 rad/s

t: time = 0.892s

You replace the values of the parameters in the equation (1):

\alpha=\frac{31.4rad/s-0rad/s}{0.892s}=35.20\frac{rad}{s^2}

The angular acceleration of the disc, for the given time, is 35.20rad/s^2

(b) To calculate the angle describe by the disc in such a time you use:

\theta=\frac{1}{2}\alpha t^2         (2)

\theta=\frac{1}{2}(35.20rad/s^2)(0.892s)^2=14.00rad

In degrees you have:

\theta=14.00rad*\frac{180\°}{\pi \ rad}=802\°

The angle described by the disc is 802°

(c) To calculate the tangential speed of the microbe for t=0.892s, you use the following formula:

v=\omega r         (3)

w: angular speed for t = 0.892s = 31.4rad/s

r: radius of the disc = 4.45cm

v=(31.4rad/s)(4.45cm)=139.73\frac{cm}{s}

The tangential speed is 139.73 cm/s

(d) The tangential acceleration is calculated by using the following formula:

a=\alpha r

α: angular acceleration for t=0.892s

a=(35.20rad/s^2)(4.45cm)=156.64\frac{cm}{s^2}

The tangential acceleration is 156.64cm/s^2

7 0
3 years ago
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