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3 years ago

8

An object moves along a circular track of radius 3.00 m. The object completes one loop around the track in 4.00 s. Determine the

centripetal acceleration of the object.

1 answer:

zaharov [31]3 years ago

7 0

Answer:

a = 7.4 m / s²

Explanation:

The centripetal acceleration of an object that has constant speed is

a = v² / r

linear and angular velocity are related

v = w r

let's substitute

a = w² r

the angular velocity is related to the period

w = 2π / T

we substitute

a = $\frac{4\pi ^2 \ r}{T^2}$

let's calculate

a = 4π² 3.00 /4²

a = 7.4 m / s²

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3 years ago

A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi

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Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block $m = 6.60$ kg

Initial speed of block $v _{i} = 1.56$ $\frac{m}{s}$

Final speed of block $v_{f} = 0$ $\frac{m}{s}$

Coefficient of kinetic friction $\mu _{k} = 0.62$

Ramp inclined at angle $\theta =$ 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

$\mu _{k} mgd \cos 28.4 = \Delta K - \Delta U$

Where $\Delta U = mg d\sin 28.4$

$\mu _{k} mgd \cos 28.4 = \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4$

$\mu _{k} mgd \cos 28.4 +mgd\sin 28.4 = \frac{1}{2}mv_{i} ^{2}$

$d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}$

$d = 0.122$ m

Therefore, the distance travel by block before coming to rest is 0.122 m

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4 years ago

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4 years ago

The angular acceleration of a wheel is given in rad/s2 by 32 t 3 - 11t 4, where t is in seconds. if the wheel starts from rest a

disa [49]

Answer:

$t=3.64s$

Step by step explanation

Step 1

In this step we use the definition of acceleration to determine the expression to integrate. Note that the acceleration is the derivative of velocity with respect to time.

$a=\frac{dv}{dt} \\\implies adt=dv\\\implies dv=(32t^3-11t^4)dt$

Step 2

Perform integration on the expression from step 1. This calculation is performed as shown below.

$v=\int(32t^3-11t^4)dt\\v=\frac{32}{4}t^4- \frac{11}{5}t^5 +c\\\8t^4- \frac{11}{5}t^5 +c$

Step 3

In this step we use the condition that at $t=0$ , the velocity $v=0$ to find the exact form of the velocity function. We substitute this point into the velocity function we found in step 2.

$0=8(0)^4-\frac{11}{5} (0)^5+c\\\implies c=0$

Step 4

We now use the function in step 3 to find out when the velocity is zero again.

$v=0=8t^5-\frac{11}{5}t^4\\\implies 0=8t^4-\frac{11}{5} t^5\\\implies t^4(8-\frac{11}{5} t)\\\implies t=0, 8-\frac{11}{5}t=0\\\\\implies t=0,t=\frac{40}{11} =3.64$

The velocity is zero again when $t=3.64$

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3 years ago