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lina2011 [118]
1 year ago
8

A large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is subme

rged in the graduated cylinder, the total volume increases to 61.2 ml. (a) determine the density of this piece of jewelry. (b) assuming that the jewelry is made from only one substance, what substance is it likely to be
Physics
1 answer:
Lerok [7]1 year ago
4 0

The density of the substance is<u> 10.5 g/cm³.</u>

The jewelry is made out of <u>Silver.</u>

Density ρ is defined as the ratio of mass <em>m</em> of the substance to its volume V<em>. </em> The cylinder contains a volume <em>V₁ of water</em> and when the jewelry is immersed in it, the total volume of water and the jewelry is found to be V₂.

The volume <em>V</em> of the jewelry is given by,

V=V_2 -V_1

Substitute 48.6 ml for <em>V₁ </em>and 61.2 ml for V₂.

V=V_2 -V_1\\ =61.2 ml -48.6 ml\\ =12.6 ml

calculate the density ρ of the jewelry using the expression,

\rho =\frac{m}{V}

Substitute 132.6 g for <em>m</em> and 12.6 ml for <em>V</em>.

\rho =\frac{m}{V}\\ =\frac{132.6 g}{12.6 ml} \\ =10.5 g/ml

Since 1 ml=1 cm^3,

The density of the jewelry is <u> 10.5 g/cm³.</u>

From standard tables, it can be seen that the substance used to make the jewelry is <u>silver</u><em><u>, </u></em>which has a density 10.5 g/cm³.



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Calculate the change in the kinetic energy (KE) of the bottle when the mass is increased. Use the formula KE = mv2, where m is t
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1 year ago
Read 3 more answers
For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo
pickupchik [31]
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height. 

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
4 0
1 year ago
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