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3 years ago

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Which type of matter is gold an element used in making jewelry A.) A heterogeneous mixture B.) A pure substance C.) A crystallin

e compound D.) A homogeneous mixture

2 answers:

Mnenie [13.5K]3 years ago

7 0

Answer:

A

Explanation:

<em>The gold used in the making of jewelry is usually not pure but a heterogeneous mixture of metals. Pure gold is quite soft and even though it may look better in appearance compared to those made using heterogeneous mixtures, it usually bends easily. Hence, in order to make the jewelry more durable, gold is usually mixed with other metals to form a heterogeneous mixture. </em>

The correct option is A.

DaniilM [7]3 years ago

4 0

Answer:

BBBBBBBBB

Explanation:

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Answer:

$v = 3869 m/s$

Explanation:

As we know that the orbital speed of the satellite is given as

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also we know that

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now from above equation we know that

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so we will have

$v = (\frac{2\pi GM}{T})^{1/3}$

now plug in all data in this equation

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How large a net force is required to accelerate a 1600-kg SUV from rest to a speed of 25 m/s in a distance of 200 m

Mars2501 [29]

Answer:

F=2496 N

Explanation:

Given that,

Mass of SUV, m = 1600 kg

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We need to find the net force. Firstly, let's find acceleration using equation of motion.

$v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(25)^2-(0)^2}{2\times 200}\\\\a=1.56\ m/s^2$

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3 years ago

Captain Hook is ghting Peter Pan, and they are about to step onto a tightrope strung horizontally between two maststhat are 16 m

andreyandreev [35.5K]

Complete Question

Captain Hook is fighting Peter Pan, and they are about to step onto a tightrope strung horizontally between two masts that are 16 m apart. When Pan and Hook are standing exactly halfway between the masts, the rope makes a 3 anglewith the horizontal. The rope has a diameter of 0.02 m and a Youngs modulus of 35 GPa.

What is the combined mass,in kilograms, of Peter Pan and Captain Hook?

Answer:

Their combined mass is $m= 161.2kg$

Explanation:

A sketch that describes the question is shown on the first uploaded image

From the question we are told that

The distance apart is $d_A = 16m$

The angle the rope makes is $\theta = 3^o$

The diameter of the rope is $d = 0.02m$

The Young modulus is $Y = 35Pa$

From the diagram we see that the elongation of the rope can be mathematically evaluated as

$\Delta L = 2x - 16$

And applying SOHCATOH rule $x = \frac{8}{cos \theta}$

Substituting values

$x = \frac{8}{cos (3)}$

$= 8.01m$

And $\Delta L = \frac{16}{cos 3} -16$

$\Delta L = 0.02196m$

The Tension on the rope can be mathematically represented as

$T = Y A * \frac{\Delta L}{L}$

Where A is the area and is mathematically represented as

$A = \frac{\pi}{4} d^2$

Substituting values

$A = \frac{\pi}{4} (0.02)^2$

Now Substituting values into the formula for the tension on the rope

$T = (35*10^9) * \frac{\pi}{4} (0.02)^2 * \frac{(0.02196)}{16}$

$=15093.4 N$

From the diagram we can mathematically evaluate the the weight of peter and hook as

$W = 2T sin \theta$

Where $W = mg$

Now substituting this into the equation and making m the subject

$m = \frac{2Tsin \theta}{g}$

Substituting values

$m = \frac{2* 15093.4 sin(3)}{9.8}$

$m= 161.2kg$

Note SOHCATOH is

$Sin \theta = \frac{opposite}{hypotenuse}\\ Cos \theta = \frac{adjacent }{hypotenuse} \\Tan \theta = \frac{opposite}{adjacent}$

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