Answer:
83%
Explanation:
On the surface, the weight is:
W = GMm / R²
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.
In orbit, the weight is:
w = GMm / (R+h)²
where h is the height of the shuttle above the surface of the Earth.
The ratio is:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
The shuttle in orbit retains 83% of its weight on Earth.
<span>Px = 0
Py = 2mV
second, Px = mVcosφ
Py = –mVsinφ
add the components
Rx = mVcosφ
Ry = 2mV – mVsinφ
Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²)
and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
simplifying
Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²)
Vf = (V/3)âš((cosφ)² + (2 – sinφ)²)
Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ))
Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ))
using the identity sin²(Ď)+cos²(Ď) = 1
Vf = (V/3)âš1 + 4 – 2sinφ)
Vf = (V/3)âš(5 – 2sinφ)</span>
Hello
1) First of all, since we know the radius of the wire (

), we can calculate its cross-sectional area

2) Then, we can calculate the current density J inside the wire. Since we know the current,

, and the area calculated at the previous step, we have

3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity

of the aluminium, the electric field is given by
Answer:
Mass released = 8.6 g
Given data:
Initial number of moles nitrogen= 0.950 mol
Initial volume = 25.5 L
Final mass of nitrogen released = ?
Final volume = 17.3 L
Formula:
V₁/n₁ = V₂/n₂
25.5 L / 0.950 mol = 17.3 L/n₂
n₂ = 17.3 L× 0.950 mol/25.5 L
n₂ = 16.435 L.mol /25.5 L
n₂ = 0.644 mol
Initial mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.950 mol × 28 g/mol
Mass = 26.6 g
Final mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.644 mol × 28 g/mol
Mass = 18.0 g
Mass released = initial mass - final mass
Mass released = 26.6 g - 18.0 g
Mass released = 8.6 g
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Answer:
the greater the height of an object the *greater* its gravitational potential energy
greater