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.
A = h / n => h = a*n
a = 0.290 hit / time
n = 300 times
=> h = 0.290 hit / time * 300 time = 87 hits
Answer: 87 hits
Answer:
R₂ / R₁ = D / L
Explanation:
The resistance of a metal is
R = ρ L / A
Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section
We apply this formal to both configurations
Small face measurements (W W)
The length is
L = W
Area
A = W W = W²
R₁ = ρ W / W² = ρ / W
Large face measurements (D L)
Length L = D= 2W
Area A = W L
R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L
The relationship is
R₂ / R₁ = 2W²/L
Answer:
B. i=2.79A
C. F=0.066N
Explanation:
A) By the right hand rule we have that
F=iL x B
F=iLBsin(α)
If the wire jump toward the observer the top pole face is the magnetic southpole.
B) The diameter of the pole face is 15cm. We can take this value as L (the length in which the wire perceives the magnetic field). Hence, we have
C) Now the length of the wire that feels B is
and the force will be (by taking the degrees between the magnetic field vector and current vector as 80°)
I hope this is useful for you
regards
