Two stones resembling diamonds are suspected of being fakes. To determine if the stones might be real, the mass and volume of ea
1 answer:
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 of stone A
density of stone B.
Hence, the density of the stone A is the equal to Diamond then stone A is diamond.
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Answer:
a) E = 8.63 10³ N /C, E = 7.49 10³ N/C
b) E= 0 N/C, E = 7.49 10³ N/C
Explanation:
a) For this exercise we can use Gauss's law
Ф = ∫ E. dA = /ε₀
We must take a Gaussian surface in a spherical shape. In this way the line of the electric field and the radi of the sphere are parallel by which the scalar product is reduced to the algebraic product
The area of a sphere is
A = 4π r²
if we use the concept of density
ρ = q_{int} / V
q_{int} = ρ V
the volume of the sphere is
V = 4/3 π r³
we substitute
E 4π r² = ρ (4/3 π r³) /ε₀
E = ρ r / 3ε₀
the density is
ρ = Q / V
V = 4/3 π a³
E = Q 3 / (4π a³) r / 3ε₀
k = 1 / 4π ε₀
E = k Q r / a³
let's calculate
for r = 4.00cm = 0.04m
E = 8.99 10⁹ 3.00 10⁻⁹ 0.04 / 0.05³
E = 8.63 10³ N / c
for r = 6.00 cm
in this case the gaussine surface is outside the sphere, so all the charge is inside
E (4π r²) = Q /ε₀
E = k q / r²
let's calculate
E = 8.99 10⁹ 3 10⁻⁹ / 0.06²
E = 7.49 10³ N/C
b) We repeat in calculation for a conducting sphere.
For r = 4 cm
In this case, all the charge eta on the surface of the sphere, due to the mutual repulsion between the mobile charges, so since there is no charge inside the Gaussian surface, therefore the field is zero.
E = 0
In the case of r = 0.06 m, in this case, all the load is inside the Gaussian surface, therefore the field is
E = k q / r²
E = 7.49 10³ N / C
Answer:
6.18 m/s
Explanation:
Roller skate collision
The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;
x-axis component form (+x east);
+
+
=
+
Ф
60 ·8 + 0 = (60 + 80)Ф
480 = 140Ф................. (I)
y-axis component form (+y north);
+
+
=
+
Ф
0 + 80.9 = (60 + 80)Ф
720=
140Ф
140Vf=Ф......................................(2)
Substituting (2) into (1) to give the angle;
480 = 720tan Ф
Ф = arctan(0.67) =33.69°.......................(3)
Evaluating (1) with (3) gives the velocity magnitude
480 = 140Vfsin 33.69°
Vf=6.18 m/s
note 1:
This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.