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Katena32 [7]
3 years ago
15

A go kart is traveling at a constant 13.8 meters per second around a horizontal circular track with a radius of 42.0 meters. The

Combined mass of a go kart and the driver is 143 kg. What is the magnitude of the centripetal force acting on the combined mass
Physics
1 answer:
Ivan3 years ago
4 0

Answer:

648.40N

Explanation:

Centripetal acceleration is the acceleration of an object around a circular path. Centripetal for e acting on the combined mass is expressed as:

F = ma

F = mv²/r

m is the mass of the object = 143kg

v is the linear velocity = 13.8m/s

r is the radius = 42.0m

Substitute the given parameters into the formula;

F = 143*13.8²/42

F = 27,232.92/42

F = 648.40N

Hence the magnitude of the centripetal force acting on the combined mass is 648.40N

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Answer:

The answer is left

Explanation:

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A square wave has amplitude 0 V for the low voltage and 4 V for the high voltage. Calculate the average voltage by integrating o
Margarita [4]

Answer:

V_{average} = \frac{1}{2}  V_o  ,     V_{average} = 2 V

Explanation:

he average or effective voltage of a wave is the value of the wave in a period

            V_average = ∫ V dt

in this case the given volage is a square wave that can be described by the function

           V (t) = \left \{ {{V=V_o \ \ \  t<  \tau /2} \atop {V=0 \ \  \ \  t> \tau /2 }   } \right.

to substitute in the equation let us separate the into two pairs

             V_average = \int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt

             V_average = V_o \ \int\limits^{1/2}_0 {} \, dt

             V_{average} = \frac{1}{2}  V_o

we evaluate  V₀ = 4 V

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3 years ago
Using Excel, or some other graphing software, plot the values of y as a function of x. (You will not submit this spreadsheet. Ho
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Answer:

a) > x<-c(1,2,3,4,5)

> y<-c(1.9,3.5,3.7,5.1,6)

> linearmodel<-lm(y~x)

And the output is given by:

> linearmodel

Call:

lm(formula = y ~ x)

Coefficients:

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b) y = 0.98 x +1.10

And if we compare this with the general model y = mx +b

We see that the slope is m= 0.98 and the intercept b = 1.10

Explanation:

Part a

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x: 1,2,3,4,5

y: 1.9,3.5,3.7,5.1, 6

For this case we can use the following R code:

> x<-c(1,2,3,4,5)

> y<-c(1.9,3.5,3.7,5.1,6)

> linearmodel<-lm(y~x)

And the output is given by:

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Call:

lm(formula = y ~ x)

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Part b

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And if we compare this with the general model y = mx +b

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