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Karo-lina-s [1.5K]
1 year ago
5

Two stones resembling diamonds are suspected of being fakes. To determine if the stones might be real, the mass and volume of ea

ch are measured. Both stones have the same volume, 0.15 cm3. However, stone A has a mass of 0.52 g, and stone B has a mass of 0.42 g. If diamond has a density of 3.5 g/cm3, could either of the stones be real diamonds?
Physics
1 answer:
dimaraw [331]1 year ago
6 0

Answer:

stone A is diamond.

Explanation:

given,

Volume of the two stone =  0.15 cm³

Mass of stone A = 0.52 g

Mass of stone B = 0.42 g

Density of the diamond =  3.5 g/cm³

So, to find which stone is gold we have to calculate the density of both the stone.

We know,

density\density = \dfrac{mass}{volume}

density of stone A

\rho_A = \dfrac{0.52}{0.15}

\rho_A = 3.467\ g/cm^3

density of stone B.

\rho_B = \dfrac{0.42}{0.15}

\rho_B = 2.8\ g/cm^3

Hence, the density of the stone A is the equal to Diamond then stone A is diamond.

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Robin Hood wishes to split an arrow already in the bull's-eye of a target 40 m away.
tamaranim1 [39]

Answer:

5.843 m

Explanation:

suppose that the arrow leave the bow with a horizontal speed , towards he bull's eye.

lets consider that horizontal motion

distance = speed * time

time = 40/ 37 = 1.081 s

arrow doesnot have a initial vertical velocity component. but it has a vertical motion due to gravity , which may cause a miss of the target.

applying motion equation

(assume g = 10 m/s²)

s=ut+\frac{1}{2}*gt^{2}  \\= 0+\frac{1}{2}*10*1.081^{2}\\= 5.843 m

Arrow misses the target by 5.843m ig the arrow us split horizontally

4 0
1 year ago
A 58.7-g sample of nickel (s = 0.443 J/(g ∙ °C)), initially at 147.4°C, is placed in an insulated vessel containing 184.0 g of w
kari74 [83]

Answer:

Explanation:

Let t be the final temp

temp. drop of nickel=147.4-t

temp. gain by water=t-29.9

in equilibrium

Heat lost=Heat gained

58.7×0.443×(147.4-t)=184×4.18×(t-29.9)

58.7×0.443×147.4-58.7×0.443×t=184×4.18×t-184×4.18×29.9

     3833.00434 -26.0041 t=769.12t-22996.688

(769.12 +26.0041)t= ?

find t

3 0
2 years ago
An ideal gas at temperature t0 is slowly compressed at constant pressure of 2 atm from a volume of 10 liters to a volume of 2 li
Sati [7]
<span>Answer: 1600 J

Explanation:

1) Data:

a) ideal gas: ⇒ pV = nRT and work = ∫ pdV
b) slowly compressed ⇒ constant temperature and not heat exchange
c) pressure: p =  2 atm
d) intitial volume: Vi = 10 liters
e) final volumen: Vf = 2 liters.
f) then the volume of the gas is held constant ⇒ not work in this stage.
g) calculate the work done on the gas: W = ?

2) Equation

W = ∫pdV

3) Solution:

Since p = constant,  W = p ∫dV = p ΔV = p (Vf - Vi)

p = 2 atm × 1.0 ×10⁵ Pa / atm = 200.000 Pa

Vi = 10 liter ×  0.001 m³ ./ liter = 0.01 m³

Vf = 2 liter × 0.001 m³ / liter = 0.002 m³

W = 200.000 Pa × (0.002 m³ - 0.01m³) = - 1.600 J.

The negative sign means the work is done over the system.

That is all the work in the system because at the second stage the volume is held constant.
</span>
8 0
2 years ago
A 4.00-kg box sits atop a 10.0-kg box on a horizontal table. The coefficient of kinetic friction between the two boxes and betwe
natta225 [31]
First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.


Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.
4 0
2 years ago
A 20-kilogram box is sitting on an inclined plane with a 30-degree slope. If the box is at rest, what is the force
ad-work [718]

Answer:

Explanation:

The box is at rest because , the component of weight along the surface downwards is equal to frictional force .

frictional force = mgsinθ

= 20 x 9.8 sin 30

= 98 N .

7 0
2 years ago
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