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2 years ago

The work done by static friction can be : a. Zero

Physics

1 answer:

mario62 [17]2 years ago

5 0

Answer:

A. Zero

Explanation:

The definition of work is related with the energy that a force 'F' implies in a displacement 'Δx'; it means that it is necessary that the body which is over the force comes under a displacement to develop work. The static friction is a force which is only present when there is no movement; when a force overcome the static friction force, the dynamic friction force appears and it is said that the static one disappear. In that sense, the static friction won't ever be a force that can induce a displacement, so the work done by this force will always be zero.

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In a physics lab, Asha is given a 10.7 kg uniform rectangular plate with edge lengths 67.3 cm by 53.5 cm . Her lab instructor re

Leya [2.2K]

Answer:

$I=2.6363\ kg.m^2$

Explanation:

Given:

dimension of uniform plate, $(0.673\times 0.535)\ m^2$

mass of plate, $m=10.7\ kg$

Now we find the moment of inertia about the center of mass of the rectangular plate is given as:

$I_{cm}=\frac{1}{12} \times m(L^2+B^2)$

where:

$L=$ length of the plate

$B=$ breadth of the plate

$I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)$

$I_{cm}=0.6591\ kg.m^2$

We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .

Now we find the distance between the center of mass and the corner:

$s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}$

$s=0.4299\ m$

Now using parallel axis theorem:

$I=I_{cm}+m.s^2$

$I=0.6591+10.7\times 0.4299^2$

$I=2.6363\ kg.m^2$

6 0

3 years ago

PEDALING A BIKE : ACCELERATION:: PULLING A DOGS LEASH: _____.

Alex_Xolod [135]

Pulling a dogs leash: inertia

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3 years ago

Which form of the law of conservation of energy describes the motion of the block as it slides on the floor from the bottom of t

aivan3 [116]

Answer:

Explanation:

Let assume begins movement at zero point, that is, height is equal to zero. The block has an initial linear kinetic energy and no gravitational potential energy and end with no linear kinetic energy, some gravitational potential energy and work losses due to slide friction. In mathematical terms, this system can be model as follows:

$K_{1} = U_{2} + W_{loss, 1 \longrightarrow 2}$