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Troyanec [42]
3 years ago
14

In his novel From the Earth to the Moon (1866), Jules Verne describes a spaceship that is blasted out of 12,000 yards/s. the Col

umbiad is 900 ft long, but part of it is packed with poweder, so the spaceship accelerates over a distance of only 700 ft. Estimate the acceleration experienced by the occupants of the spaceship during launch. Give your answer in m/s2. (Verne realized that the "travelers would...encounter a violent recoil," but he probably didn't know that people generally lose consciousness if they experience accelerations greater than about 7g ~70 m/s2.)
Physics
1 answer:
saw5 [17]3 years ago
5 0

Answer:

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

Explanation:

Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration (a), measured in meters per square second, is estimated by this kinematic formula:

a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s } (1)

Where:

\Delta s - Travelled distance, measured in meters.

v_{o}, v - Initial and final speeds of the spaceship, measured in meters.

If we know that v_{o} = 0\,\frac{m}{s}, v = 10968\,\frac{m}{s} and \Delta s = 212.8\,m, then the acceleration experimented by the spaceship is:

a = \frac{\left(10968\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (212.8\,m)}

a = 282652.782\,\frac{m}{s^{2}}

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

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A football punter accelerates a .55 kg football
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Answer:

17.6 N

Explanation:

The force exerted by the punter on the football is equal to the rate of change of momentum of the football:

F=\frac{\Delta p}{\Delta t}

where

\Delta p is the change in momentum of the football

\Delta t=0.25 s is the time elapsed

The change in momentum can be written as

\Delta p = m(v-u)

where

m = 0.55 kg is the mass of the football

u = 0 is the initial  velocity (the ball starts from rest)

v = 8.0 m/s is the final velocity

Combining the two equations and substituting the values, we  find the force exerted on the ball:

F=\frac{m(v-u)}{\Delta t}=\frac{(0.55)(8.0-0)}{0.25}=17.6 N

5 0
3 years ago
An object with a non-zero speed must be _________.
pogonyaev
It would be either A or C if its still moving and not stopping
8 0
3 years ago
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Zero field spot for opposite unequal charges
xz_007 [3.2K]

Answer:

The zero field location has to be on the line running between the two point charges because that's the only place where the field vectors could point in exactly opposite directions. It can't be between the two opposite charges because there the field vectors from both charges point toward the negative charge.

7 0
3 years ago
A shopper pushes a 5.32 kg grocery cart
Juli2301 [7.4K]

Answer:

\text { acceleration of the cart is } 10.94 \mathrm{m} / \mathrm{s}^{2}

Explanation:

According to “Newton's second law”

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Force = mass × acceleration

\text { Acceleration }=\frac{\text { force }}{\text { mass }}

Given that,

Mass = 5.32 kg

\text { Force }=12.7 \mathrm{N} \text { forces at }-28.7^{\circ}

x=-28.7^{\circ}

F = 12.7N

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“g” being the "acceleration of gravity",

“x” being the "angle of the cart"

\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}\text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^ 2 \text { on Earth })

To find normal force substitute the values in the formula,

Normal force = 5.32 × 9.8 + 12.7 × sin(-28.7)

Normal force = 52.136 + 12.7 × 0.480

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Normal force = 58.232 N

<u>Acceleration of the cart</u>:

\text { Acceleration }=\frac{\text {Normal force}}{\text { mass }}

\text { Acceleration }=\frac{58.232}{5.32}

\text { Acceleration }=10.94 \mathrm{m} / \mathrm{s}^{2}

\text { Therefore, "acceleration of the cart" is } 10.94 \mathrm{m} / \mathrm{s}^{2}

7 0
3 years ago
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