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Virty [35]
2 years ago
5

A 0.30-m-radius automobile tire accelerates from rest at a constant 2.0 rad/s2. What is the centripetal acceleration of a point

on the outer edge of the tire after 5.0 s
Physics
1 answer:
lianna [129]2 years ago
8 0

Answer:

The centripetal acceleration is 30 m/s²

Explanation:

Given that,

Radius = 0.30 m

Acceleration = 2.0 rad/s²

Time = 5.0

We need to calculate the centripetal acceleration

Using formula of angular velocity

\omega=a\times t

Where,

a = acceleration

t = time

Put the value into the formula

\omega=2.0\times5.0

\omega=10\ rad/s

Using formula of centripetal acceleration

a_{c}=r\omega^2

a_{c}=0.30\times10^2

a_{c}=30\ m/s^2

Hence, The centripetal acceleration is 30 m/s²

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yKpoI14uk [10]
The Milky Way galaxy is the one that the sun is a member of, and it contains
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The Milky Way is known to be bigger than your average galaxy, but it's
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6 0
3 years ago
found the potential spring energy. not sure how to find the height of the block by knowing only the mass. found the weight of it
evablogger [386]

(h + .16) m g = 1/2 k x^2   total PE of block relative to where it stops

(h + .16) .82 * 9.8 = .5 * 120 * .16^2    PE released = PE of  spring

8.04 h + 1.29 = 1.536

h = (1.536 - 1.29) / 8.04 = .031 m = 3.1 cm

7 0
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
Evgesh-ka [11]

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:

(Intercept)            x  

      1.10         0.98  

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

For this case we have the following data:

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:

> linearmodel

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept)            x  

      1.10         0.98  

Part b

For this case we have the following trend equation given:

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

7 0
3 years ago
A boy pushes a cart with a constant velocity of 0.5m/s by applying a force of 60 N. What is the total frictional force acting on
Nutka1998 [239]

Answer:

60N

Explanation:

in this case the minimum amount of force required must be equal to the friction Force. i.e <u>Newton</u><u>'s</u><u> </u><u>first</u><u> </u><u>law</u><u> of</u><u> </u><u>mot</u><u>ion</u><u>.</u>

therefore the maximum amount of frictional force is equal to the applied force which is 60N.

because of the net force acting on the object is zero the object is in constant motion . i.e equal and opposite force must be applied so that the object is in constant velocity therefore the total frictional force must be 60N

8 0
3 years ago
A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release
Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done  to compress the spring initially=\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J

b.

By conservation law of energy

Initial energy of spring=Kinetic energy  of object

37.8=\frac{1}{2}(3)v^2

v^2=\frac{37.8\times 2}{3}

v=\sqrt{\frac{37.8\times 2}{3}}

v=5.02 m/s

c.Work done by friction on the incline,w_{friction}=P.E-spring \;energy

W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J

8 0
3 years ago
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