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2 years ago

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3) A 30 kg child slides freely across a "Slip and Slide" on LEVEL GROUND. While the child slides, the force applied to keep them

sliding is 0 N. The coefficient of kinetic friction is 0.02. What is the acceleration experienced by the child?

1 answer:

zavuch27 [327]2 years ago

4 0

Answer:

a = 0.1962 m/s^2

Explanation:

The magnitude of kinetic friction exerted is given by

$F_k=\mu_kN$

Where, μ_k= coefficient of kinetic friction= 0.02 and N = reaction force = mg

Where m= mass = 30 Kg and, g is acceleration due to gravity =9.81 m/s^2

F_k=0.02×30×9.81 =5.886 N

Now, since, there is no applied force this kinetic friction force will cause acceleration of the child

⇒ ma = F_k

here, a is the acceleration

⇒30a = 5.886

⇒ a = 0.1962 m/s^2

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Natali5045456 [20]

The question is incomplete. The complete question is :

A plate of uniform areal density $\rho = 2 \ kg/m^2$ is bounded by the four curves:

$y = -x^2+4x-5m$

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Solution :

Given :

$y = -x^2+4x-5$

$y = x^2+4x+6$

$x=1$

$x=2$

and $\rho = 2 \ kg/m^2$ , $P_x=1 \$ , $P_y=-2 \$ .

So,

$dI = dmr^2$

$dI = \rho \ dA \ r^2$ , $r=\sqrt{(x-1)^2+(y+2)^2}$

$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \ dy \ dx$

$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$

$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$

$I=\frac{32027}{21} \times 2$

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So the moment of inertia is $3050.19 \ kg \ m^2$ .

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2 years ago

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vagabundo [1.1K]

Explanation:

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S=ut+½gt² ("ut" term will cancel because u=0).

=> S= ½gt²

=>S = ½×8×10²

=>S = 4×100

=>S = 400m .

Therefore, the distance traveled by the body in 10s is 400m.

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1. An excited lithium atom emits a red light with wavelength a = 671nm. What is the corresponding photon energy? hc (6.63 x 10-3

Rudiy27

Answer:

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Explanation:

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