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Dvinal
1 month ago
13

In a novel from 1866 the author describes a spaceship that is blasted out of a cannon with a speed of about 11.000 m/s. The spac

eship is approximately 274 m long but part of it is packed with gunpowder, so it accelerates over a distance of only 213 m. What was the acceleration (in m/s2) experienced by the occupants of the spaceship during the launch?
Physics
1 answer:
Sav [3.1K]1 month ago
5 0

Answer:

a=0.284\ m/s^2

Explanation:

We start with the fact that

Initially, the spacecraft was at rest, u = 0

The final velocity of the rocket is given as v = 11 m/s

The distance that the rocket covers during acceleration is given as d = 213 m

We seek to determine the acceleration that the occupants of the spacecraft experience during launch, which is derived from the principles of kinematics. By applying the

third motion equation we can find the acceleration.

v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(11)^2-(0)^2}{2\times 213}\\\\a=0.284\ m/s^2

Thus, the acceleration felt by those inside the spacecraft is 0.284\ m/s^2.

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4. A 505-turn circular-loop coil with a diameter of 15.5 cm is initially aligned so that
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The intensity of the magnetic field is 4.8\cdot 10^{-5} T

Explanation:

According to Faraday's Law, the magnitude of the induced electromotive force (emf) in the coil corresponds to the rate of flux change that links through the coil:

\epsilon = \frac{N\Delta \Phi}{\Delta t} (1)

where

N = 505 represents the number of turns in the coil

\Delta \Phi is the variation in magnetic flux through the coil

\Delta t = 2.77 ms = 2.77\cdot 10^{-3} s indicates the time interval

\epsilon = 0.166 V

Rotating the coil from perpendicular to parallel alignment with the Earth's magnetic field results in the final flux equaling zero, making the magnitude of flux change simply the initial flux:

\Delta \Phi = B A cos \theta

where

B denotes the intensity of the magnetic field

A signifies the area of the coil

\theta=0^{\circ} represents the angle between the normal to the coil and the field

The area of the coil can be expressed as

A=\pi r^2

where

r=\frac{15.5 cm}{2}=7.75 cm = 7.75\cdot 10^{-2} m outlines its radius

By substituting all values into equation (1) and solving for B, we obtain:

\epsilon= \frac{NB\pi r^2 cos \theta}{\Delta t}\\B=\frac{\epsilon \Delta t}{\pi r^2 cos \theta}=\frac{(0.166)(2.77\cdot 10^{-3})}{(505)\pi (7.75\cdot 10^{-2})^2(cos 0^{\circ})}=4.8\cdot 10^{-5} T

For further reading on magnetic fields:

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2 months ago
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A student claims that earthquakes are more destructive than tsunamis. In one or two sentences, explain at least two reasons that
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An earthquake can impact individuals on land, causing greater destruction than a tsunami, which primarily affects those near the ocean.
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Hawks and gannets soar above the ground and, when they spot prey, they fold their wings and essentially drop like a stone. They
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Answer:

  v = 54.2 m/s

Explanation:

We can utilize conservation of energy to solve this issue.

Initial condition Higher

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Now let's perform the calculation

         v = √ (2 * 9.8 * 150)

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Use the terms "force", "weight", "mass", and "inertia" to explain why it is easier to tackle a 220 lb football player than a 288
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A person who weighs 220 lb has less mass than someone who weighs 288 lb, so accelerating the 220 lb player requires less force. The heavier player therefore carries greater momentum. Because 288 lb corresponds to more weight (and mass), that player has higher inertia and is harder to stop. For these reasons it is easier to tackle a 220 lb player than a 288 lb player. 
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