Answer:
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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.
Answer:
The frequency of the oscillation is 2.45 Hz.
Explanation:
Given;
mass of the spring, m = 0.5 kg
total mechanical energy of the spring, E = 12 J
Determine the spring constant, k as follows;
E = ¹/₂kA²
kA² = 2E
k = (2E) / (A²)
k = (2 x 12) / (0.45²)
k = 118.519 N/m
Determine the angular frequency, ω;

Determine the frequency of the oscillation;
ω = 2πf
f = (ω) / (2π)
f = (15.396) / (2π)
f = 2.45 Hz
Therefore, the frequency of the oscillation is 2.45 Hz.
Answer:
6 cm long
Explanation:
F = 4110N
Vo(speed of sound) = 344m/s
Mass = 7.25g = 0.00725kg
L = 62.0cm = 0.62m
Speed of a wave in string is
V = √(F / μ)
V = speed of the wave
F = force of tension acting on the string
μ = mass per unit density
F(n) = n (v / 2L)
L = string length
μ = mass / length
μ = 0.00725 / 0.62
μ = 0.0116 ≅ 0.0117kg/m
V = √(F / μ)
V = √(4110 / 0.0117)
v = 592.69m/s
Second overtone n = 3 since it's the third harmonic
F(n) = n * (v / 2L)
F₃ = 3 * [592.69 / (2 * 0.62)
F₃ = 1778.07 / 1.24 = 1433.927Hz
The frequency for standing wave in a stopped pipe
f = n (v / 4L)
Since it's the first fundamental, n = 1
1433.93 = 344 / 4L
4L = 344 / 1433.93
4L = 0.2399
L = 0.0599
L = 0.06cm
L = 6cm
The pipe should be 6 cm long