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3 years ago

9

A box of weight 50 N is at rest on a floor where the coefficient of static friction is 0.20. A rope is attached to the box and p

ulled horizontally to the right. What is the friction force that exerts by the floor on the box and which direction

1 answer:

Bas_tet [7]3 years ago

5 0

Answer:

The friction force that exerts by the floor on the box is 10 N.

Explanation:

Given that,

Weight of the box, W = 50 N

The coefficient of static friction is 0.2

A rope is attached to the box and pulled horizontally to the right. We need to find the friction force that exerts by the floor on the box. The direction of frictional force is in left side as we know that the frictional force acts in the opposite direction of motion. The force of friction is given by :

$F=\mu N$

N is normal reaction, N = W = 50 N

$F=0.2\times 50\\\\F=10\ N$

So, the friction force that exerts by the floor on the box is 10 N.

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Answer:

A. Its velocity is decreasing

Explanation:

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3 years ago

Edit question A 37-cm-long wire of linear density 18 g/m vibrating at its second mode, excites the third vibrational mode of a t

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Answer:

176.9N

Explanation:

The following data were given

wire length,L=37cm=0.37m

linear density=18g/m

tube length,=192cm=1.92m,

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Since it is an open-closed tube, the second harmonic frequency is expressed as

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The relationship between the tension, linear density and second harmonic frequency is expressed as

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3 years ago

Two spherical asteroids have the same radius R. Asteroid 1 has mass M and asteroid 2 has mass 1.97·M. The two asteroids are rele

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Answer:

$0.536\sqrt{\frac{GM}{R}}$

Explanation:

We are given that

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Mass of asteroid 2, $m_2=1.97 M$

Initial distance between their centers,d=13.63 R

Radius of each asteroid=R

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Initial velocity of both asteroids

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We have to find the speed of second asteroid just before they collide.

According to law of conservation of momentum

$(m_1+m_2)u=m_1v_1+m_2v_2$

$(M+1.97 M)\times 0=Mv_1+1.97Mv_2$

$Mv_1=-1.97 Mv_2$

$v_1=-1.97v_2$

According to law of conservation of energy

$Gm_1m_2(\frac{1}{d'}-\frac{1}{d})=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2$

$GM(1.97M)(\frac{1}{2R}-\frac{1}{13.63R})=\frac{1}{2}M(-1.97v_2)^2+\frac{1}{2}(1.97M)v^2_2$

$1.97M^2G(\frac{13.63-2}{27.26R})=\frac{1}{2}Mv^2_2(3.8809+1.97)$

$1.97MG(\frac{11.63}{27.26 R})=\frac{1}{2}(5.8509)v^2_2$

$v^2_2=\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}$

$v_2=\sqrt{\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}}$

$v_2=0.536\sqrt{\frac{GM}{R}}$

Hence, the speed of second asteroid = $0.536\sqrt{\frac{GM}{R}}$

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3 years ago

Calculate kp at 298.15 k for the reactions (a), (b), and (c) using δg°f values.

vivado [14]

(a) 2NO(g) + O₂(g) ⇄2NO₂(g)kp
(b) 2N₂O(g)⇄2NO(g) + N₂(g) kp
(c) N₂(g) + O₂(g)⇄ 2NO(g) kp
Now A is
2NO +O₂⇄2NO₂
ΔG° =ΔG° products - ΔG reactants
=2× 51.3-(256.6)
-70.6kJ/mol.
ΔG° = -RT Inkp
-70.6 = -8.314 ×10⁻³ ˣ 298.15 ˣInkJ
InkJ = 28.48
kp=2.34 ˣ 10¹²

B is
ΔG° = 2× 86.6 - 2 × 104.2 = -35.2
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kp = 1.47ˣ 10⁶

C is
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3 years ago

If a child starts from rest at point A and lands in the water at point B, a horizontal distance L = 2.52 m from the base of the

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Answer:

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Explanation:

Given that,

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$s=ut+\dfrac{1}{2}gt^2$

Put the value in the equation

$1.80=0+\dfrac{1}{2}\times9.8\times t^2$

$t^2=\dfrac{1.80\times2}{9.8}$

$t=\sqrt{\dfrac{1.80\times2}{9.8}}$

$t=0.606\ sec$

We need to calculate the velocity

Using formula of velocity

$v = \dfrac{d}{t}$

Put the value into the formula

$v=\dfrac{2.52}{0.606}$

$v=4.15\ m/s$

We need to calculate height

Using conservation of energy

$\dfrac{1}{2}mv^2=mgh$

$h=\dfrac{v^2}{2g}$

Put the value into the formula

$h=\dfrac{4.15^2}{2\times9.8}$

$h=0.878\ m$

Hence, The height of the water slide is 0.878 m.

4 0

3 years ago