Answer:
= 15.57 N
= 2.60 N
= 16.98 N
The mass of the bag is the same on the three planets. m=1.59 kg
Explanation:
The weight of the sugar bag on Earth is:
g=9.81 m/s²
m=3.50 lb=1.59 kg
=m·g=1.59 kg×9.81 m/s²= 15.57 N
The weight of the sugar bag on the Moon is:
g=9.81 m/s²÷6= 1.635 m/s²
=m·g=1.59 kg× 1.635 m/s²= 2.60 N
The weight of the sugar bag on the Uranus is:
g=9.81 m/s²×1.09=10.69 m/s²
=m·g=1.59 kg×10.69 m/s²= 16.98 N
The mass of the bag is the same on the three planets. m=1.59 kg
Answer:
See below explanation
Explanation:
The correspondent chemical reaction for copper carbonate decomposed by heat is:
CuCO₃ (s) → CuO (s) + CO₂ (g)
Considering all molar mass (MM) for each element ( we consider rounded numbers) :
MM CuCO₃ = 123 g/mol
MM CuO = 79 g/mol
MM CO₂ = 44 g/mol
Statement mentions that scientis heated 123.6 g of CuCO₃ (almost a MM), until a black residue is obtained, which weights 79.6 g : this solid residue is formed by CuO, and the remaining mass (approximatelly 44 g) belongs to teh second product, this is, CO₂; as it is a gas compund, it is not certainly included on the solid residue.
So, law of conservation mass is true for this case, since: 123.6 g = 79.6 g + 44 g. As explained, on the solid residue, we don not include the 44 g, which "escaped" from our system, since it is a gas compound (CO₂)
Explanation:
LD₁ = 10⁵ mm⁻²
LD₂ = 10⁴mm⁻²
V = 1000 mm³
Distance = (LD)(V)
Distance₁ = (10⁵mm⁻²)(1000mm³) = 10×10⁷mm = 10×10⁴m
Distance₂ = (10⁹mm⁻²)(1000mm³) = 1×10¹² mm = 1×10⁹ m
Conversion to miles:
Distance₁ = 10×10⁴ m / 1609m = 62 miles
Distance₂ = 10×10⁹m / 1609 m = 621,504 miles.
It’s hard to perfectly measure the distance something travels, as well as the exact time it takes, making the results have some variation.
Answer:
Fₓ = 0, 
= 0 and 
<em> = - 3.115 10⁻¹⁵ N</em>
Explanation:
The magnetic force given by the expression
F = q v xB
the bold are vectors, the easiest analytical way to determine this force in solving the determinant
![F = q \left[\begin{array}{ccc}i&j&k\\5.8&-6.7&0\\-0.81&0.6&0\end{array}\right] 10^{4}](https://tex.z-dn.net/?f=F%20%3D%20q%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C5.8%26-6.7%260%5C%5C-0.81%260.6%260%5Cend%7Barray%7D%5Cright%5D%20%2010%5E%7B4%7D)
F = 1.6 10⁻¹⁵ [ i( 0-0) + j (0-0) + k^( 5.8 0.60 - 0.81 67) ]
F =i^0 + j^0
- k^ 3.115 10⁻¹⁵ N
Fₓ = 0
= 0
<em> = - 3.115 10⁻¹⁵ N</em>
