Compare the time period of two simple pendulums of length 4m and 16m at a place.

Physics

1 answer:

Ostrovityanka [3.2K]1 month ago

5 0

Answer:

The period of the pendulum measuring 16 m is double that of the 4 m pendulum.

Explanation:

Recall that the period (T) of a pendulum with length (L) is defined by:

$T=2\,\pi\,\sqrt{ \frac{L}{g} }$

where "g" denotes the local gravitational acceleration.

Since both pendulums are positioned at the same location, the value of "g" will be consistent for both, and when we compare the periods, we find:

$T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2$

Thus, the duration of the 16 m pendulum is two times that of the 4 m one.

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A Federation starship (8.5 ✕ 106 kg) uses its tractor beam to pull a shuttlecraft (1.0 ✕ 104 kg) aboard from a distance of 14 km

Keith_Richards [3271]

Consider the diagram below.

m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ = acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted

Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t² (1)
x = 0.5a₂t² (2)

F = m₁a₁ = m₂a₂ (3)
Additionally,
x + y = 14000 m (4)

From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²

From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s

Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m

The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.

Result: The starship shifts by 16.5 m (to the nearest tenth)