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3 years ago

15

Is jewelry gold a compound or mixture?

2 answers:

USPshnik [31]3 years ago

7 0

I think it’s a mixture.

loris [4]3 years ago

4 0

Answer:

jewelry gold is a mixture.

Explanation:

it contains more than one element and its composition varies.

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A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release

Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done to compress the spring initially= $\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J$

b.

By conservation law of energy

Initial energy of spring=Kinetic energy of object

$37.8=\frac{1}{2}(3)v^2$

$v^2=\frac{37.8\times 2}{3}$

$v=\sqrt{\frac{37.8\times 2}{3}}$

v=5.02 m/s

c.Work done by friction on the incline, $w_{friction}=P.E-spring \;energy$

$W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J$

8 0

3 years ago

zloy xaker [14]

Answer:

0.75Hz

Explanation:

Given parameters:

Speed of the wave = 3m/s

Wavelength = 4m

Unknown:

Frequency of the wave = ?

Solution:

The speed of a wave is given by the expression below:

Speed = frequency x wavelength

Frequency = $\frac{Speed }{Wavelength}$ = $\frac{3}{4}$ = 0.75Hz

5 0

3 years ago

How do you capture two and a half wavelengths on a sound wave?

Serjik [45]

With the flow of the water

8 0

3 years ago

prohojiy [21]

Answer:

a

Explanation:

because it taken the rock more time to hit the water

8 0

4 years ago

A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl

zysi [14]

Answer: $\alpha =9.302\ rad/s^2$

Explanation:

Given

mass of sphere $m=250\ gm$

diameter of sphere $d=4.30\ cm$

radius $r=\frac{4.30}{2}\ cm$

$f=0.0200\ N$

friction will provide resisting torque so

$f\times r=I\times \alpha$

where $I=\text{moment of Inertia}$

$f=\text{friction force}$

$\alpha =\text{angular acceleration}$

$I=\frac{2}{5}mr^2$

$0.02\times r=\frac{2}{5}mr^2\times \alpha$

$\alpha =\frac{5}{2r}\times f$

$\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02$

$\alpha =9.302\ rad/s^2$

(b)time taken to decrease its rotational speed by $21\ rad/s$

$t=\dfrac{\Delta \omega }{\alpha }$

$t=\dfrac{21}{9.302}$

$t=2.25\ s$

6 0

3 years ago