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3 years ago

Help me with number 28.

Physics

1 answer:

Ray Of Light [21]3 years ago

3 0

What you wrote in the space under the question is correct, but
you can go a lot farther than that.

The question gives you the mass of the object, and enough information
to calculate its volume.

So go on from where you left off ... take the formula you wrote (correctly),
plug in the numbers for mass and volume, and find the density. You can do it !

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A 0.16 kg billiard ball moving at 1.20 m/s collides elastically with one of the same mass at rest. It continues at a speed of 0.

Rainbow [258]

Use energy conservation to calculate the speed!
Total kinetic energy before collision = total kinetic energy after the collision.

5 0

3 years ago

I shared a picture of the problem. It’s a basic Physics question and an Algebra question.

julia-pushkina [17]

Hence the expression of ω in terms of m and k is

$\omega = \sqrt{\frac{k}{m}$

Given the expressions;

$T_s = 2 \pi \sqrt{\frac{m}{k} } \ and \ T_s = \frac{2 \pi}{\omega}$

Equating both expressions we will have;

$2 \pi \sqrt{\frac{m}{k} } = \frac{2 \pi}{\omega}$

Divide both equations by 2π

$\frac{2 \pi\sqrt{\frac{m}{2 \pi} } }{2 \pi}=\frac{\frac{2 \pi}{\omega} }{2\pi}\\\sqrt{\frac{m}{2 \pi} } = \frac{1}{\omega}\\$

Square both sides

$(\sqrt{\frac{m}{k} } )^2 = (\frac{1}{\omega} )^2\\\frac{m}{k} = \frac{1}{\omega ^2} \\\omega ^2 = \frac{k}{m}$

Take the square root of both sides

$\sqrt{\omega ^2} =\sqrt{\frac{k}{m} } \\\omega = \sqrt{\frac{k}{m}$

Hence the expression of ω in terms of m and k is

$\omega = \sqrt{\frac{k}{m}$

3 0

3 years ago

A 14000N car traveling at 25m/s rounds a curve of radius 200m. Find the following: a. The centripetal acceleration of the car.

tamaranim1 [39]

Answer:

Explanation:

Given

Weight of car $W=14,000\ N$

mass of car $m=\frac{14,000}{9.8}=1428.57\ N$

velocity of car $v=25\ m/s$

radius $r=200\ m$

(a)Centripetal acceleration is given by

$a_c=\frac{v^2}{r}$

$a_c=\frac{25^2}{200}$

$a_c=3.125\ m$

(b)Force that provide centripetal acceleration

$F=F_c=\frac{mv^2}{r}$

$F=\frac{1428.57\times 25^2}{200}$

$F=4464.285\ N$

(c)Friction force between car and tires is given by

$=\mu N$

where $\mu$ =coefficient of static friction

N=normal reaction

Centripetal force will balance the friction force

$F_c=F_r$

$4464.285=\mu \times 1428.57\times 9.8$

$\mu =0.318$

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3 years ago

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