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son4ous
20 days ago
10

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

away. The tractor beam exerts a constant force of 4.0 ✕104 N on the shuttlecraft. Both spacecraft are initially at rest. How far does the starship move as it pulls the shuttlecraft aboard?

Physics
1 answer:
Keith_Richards [2.2K]20 days ago
3 0
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)

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kicyunya [2264]

Answer:

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Explanation:

We start with the given parameters:

The electric field impacting the circular surface is E=(4000j+3000k)\ N/C

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\phi=E{\cdot}A

\phi=(4000j+3000k){\cdot}Ak

Applying properties of the dot product, we calculate the electric flux as:

\phi=3000\times Ak

\phi=3000\times \pi (1.83)^2

\phi=31562.63\ Nm^2/C

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4 0
22 days ago
Can pockets of vacuum persist in an ideal gas? Assume that a room is filled with air at 20∘C and that somehow a small spherical
kicyunya [2264]

Answer:

The time required is 20 μs

Explanation:

Here is the data provided:

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radius = 1 cm

atomic mass of air = 29 u

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solution:

We calculate the root mean square velocity of air particles. This can be expressed as:

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v = \sqrt{\frac{3RT}{M} }............................1

v = \sqrt{\frac{3(8.314)293}{29*10{-3}kg} }

Resulting in v = 501.99 m/s.

<pNow, to cover the distance of 1 cm,<pThe duration needed for air is calculated as:

time taken = \frac{r}{v}

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3 0
1 month ago
Calculate the minimum average power output necessary for a person to run up a 12.0 m long hillside, which is inclined at 25.0° a
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Answer:

Power output, P = 924.15 watts

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