Answer:
(a) L = 128.75 m
(b) L = 182.56 m
(c) L = 114.28 m
(d) Mass of Gold = 7.68 kg = 7680 gram
(e) Cost of Gold Wire = $ 307040
Explanation:
The resistance of the wire is given as:
R = ρL/A
where,
R = Resistance
ρ = resistivity
L = Length
A = cross-sectional area
(a)
For Gold Wire:
ρ = 2.44 x 10⁻⁸ Ω.m
A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²
R = 1 Ω
Therefore,
1 Ω = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)
L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)
<u>L = 128.75 m</u>
<u></u>
(b)
For Copper Wire:
ρ = 1.72 x 10⁻⁸ Ω.m
A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²
R = 1 Ω
Therefore,
1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)
L = (1 Ω)(3.14 x 10⁻⁶ m²)/(1.72 x 10⁻⁸ Ω.m)
<u>L = 182.56 m</u>
<u></u>
(c)
For Aluminum Wire:
ρ = 2.75 x 10⁻⁸ Ω.m
A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²
R = 1 Ω
Therefore,
1 Ω = (2.75 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)
L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.75 x 10⁻⁸ Ω.m)
<u>L = 114.28 m</u>
<u></u>
(d)
Density = Mass/Volume
Mass = (Density)(Volume)
Volume of Gold = AL = (3.14 x 10⁻⁶ m²)(128.75 m) = 4.04 x 10⁻⁴ m³
Therefore,
Mass of Gold = (1.9 x 10⁴ kg/m³)(4.04 x 10⁻⁴ m³)
<u>Mass of Gold = 7.68 kg = 7680 gram</u>
<u></u>
(e)
Cost of Gold Wire = (Unit Price of Gold)(Mass of Gold)
Cost of Gold Wire = ($ 40/gram)(7680 grams)
<u>Cost of Gold Wire = $ 307040</u>
The electrical force between these two charges remains the
same. In coulomb’s law, it states that the magnitude of two charges (product of
two charges) is inversely proportional to the square of the distance. Since both
the magnitude and the distance are halved, therefore, the change in both quantities
will have no effect in the value of electrical force.
Answer:
e) Be four times greater
Explanation:
Here we have to use Newton's gravitational law that relates the gravitational force between two objects with their masses (
& 
) and the distance between them (
) in the next way:
(2)
Now if distance between asteroids is halved:
Note that 
because (1) is F so:
It's four times greater!
Since this is a projectile motion problem, break down each of the five kinematic quantities into x and y components. To find the range, we need to identify the x component of the displacement of the ball.
Let's break them down into components.
X Y
v₁ 32 cos50 m/s 32 sin50 m/s
v₂ 32 cos50 m/s ?
Δd ? 0
Δt ? ?
a 0 -9.8 m/s²
Let's use the following equation of uniform motion for the Y components to solve for time, which we can then use for the X components to find the range.
Δdy = v₁yΔt + 0.5ay(Δt)²
0 = v₁yΔt + 0.5ay(Δt)²
0 = Δt(v₁ + 0.5ayΔt), Δt ≠ 0
0 = v₁ + 0.5ayΔt
0 = 32sin50m/s + 0.5(-9.8m/s²)Δt
0 = 2<u>4</u>.513 m/s - 4.9m/s²Δt
-2<u>4</u>.513m/s = -4.9m/s²Δt
-2<u>4</u>.513m/s ÷ 4.9m/s² = Δt
<u>5</u>.00s = Δt
Now lets put our known values into the same kinematic equation, but this time for the x components to solve for range.
Δdₓ = v₁ₓΔt + 0.5(a)(Δt)²
Δdₓ = 32cos50m/s(<u>5</u>.00s) + 0.5(0)(<u>5</u>.00)²
Δdₓ = 32cos50m/s(<u>5</u>.00s)
Δdₓ = 10<u>2</u>.846
Therefore, the answer is A, 102.9m. According to significant digit rules, neither would be correct, but 103m is the closest to 102.9m so I guess that is what it is.
<span>When a meteoroid passes through Earth's atmosphere it's traveling very fast. The friction of the air makes gets the surface so hot it begins to burn or glow red</span>
