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2 years ago

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A 24.0 cm long screwdriver is used as a lever to open a can of paint. If the fulcrum is 0.500 cm away from the end of the screwd

river, what is the screwdriver's ideal mechanical advantage?

1 answer:

ELEN [110]2 years ago

7 0

The ideal mechanical advantage of the screwdriver is 47

Explanation:

In this problem, the screwdriver acts as a lever.

The Ideal Mechanical Advantage (IMA) of a lever is given by:

$IMA=\frac{d_i}{d_o}$

where:

$d_i$ is the distance of the point of application of the input force from the fulcrum

$d_o$ is the distance of the output force from the fulcrum

In this problem, we have:

$d_o = 0.500 cm$ , since the fulcrum is 0.500 cm from the end

$d_i = 24.0 cm - 0.500 cm = 23.5 cm$ (the distance between the fulcrum and the point where are we holding the screwdriver)

Substutiting,

$IMA=\frac{23.50 cm}{0.50 cm}=47$

Learn more about levers here:

brainly.com/question/5352966

#LearnwithBrainly

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In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of

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Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

(c) the angular speed of the merry-go-round and child after the child has jumped on.

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

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2 years ago

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A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m

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Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, $I_1=900\ kg.m^2$

Initial angular velocity of the platform, $\omega=0.95\ rad/s$

Part A,

Let $\omega_2$ is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

$I_1\omega_1=I_2\omega_2$

Here, $I_2=I_1+mr^2$

$I_1\omega_1=(I_1+mr^2)\omega_2$

$900\times 0.95=(900+70\times (2.9)^2)\omega_2$

Solving the above equation, we get the value as :

$\omega_2=0.574\ rad/s$

Part B,

The initial rotational kinetic energy is given by :

$k_i=\dfrac{1}{2}I_1\omega_1^2$

$k_i=\dfrac{1}{2}\times 900\times (0.95)^2$

$k_i=406.12\ rad/s$

The final rotational kinetic energy is given by :

$k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2$

$k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2$

$k_f=245.24\ rad/s$

Hence, this is the required solution.

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2 years ago