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Goshia [24]
3 years 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:
Elan Coil [88]3 years ago
5 0

Answer:

a=0.284\ m/s^2

Explanation:

Given that,

Initially, the spaceship was at rest, u = 0

Final velocity of the spaceship, v = 11 m/s

Distance accelerated by the spaceship, d = 213 m

We need to find the acceleration experienced by the occupants of the spaceship during the launch. It is a concept based on the equation of kinematics. Using the third equation of motion to find 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

So, the acceleration experienced by the occupants of the spaceship is 0.284\ m/s^2.

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A sinusoidal sound wave moves through a medium and is described by the displacement wave function
labwork [276]

By applying the wave equation we know that the displacement on the y-axis is 1.999 micrometers.

We need to know about wave equations to solve this problem. The displacement of the wave on the y-axis can be explained by the wave equation

y = A cos (kx - ωt)

where y is y-axis displacement, A is amplitude, k is wave number, x is x-axis displacement, ω is angular speed and t is time.

the wavenumber and angular speed of the wave equation can be determined respectively by

k = 2π / λ

ω = 2πf

where k is the wavenumber, λ is wavelength and f is frequency.

From the question above, we know that:

y = 2.00cos (15.7x - 858t)

(x in meters, t in second, y in micrometers)

x = 0.05 m

t = 3 ms

Convert time to second

t = 3ms = 0.003 s

By applying the wave equation, we get

y = 2.00cos (15.7x - 858t)

y = 2.00cos (15.7(0.05) - 858(0.003))

y  = 2 cos(-1.789)

y = 1.999 micrometers

For more on wave equation on: brainly.com/question/25699025

#SPJ4

7 0
1 year ago
Physics is explicitly involved in studying which of these activities?
Dmitrij [34]
Hey Friends
The answer to this question would be C
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8 0
3 years ago
Read 2 more answers
A 4-kilogram ball moving at 8 m/sec to the right collides with a 1-kilogram ball at rest. After the collision,
elena-14-01-66 [18.8K]
Conservation of momentum: total momentum before = total momentum after

Momentum = mass x velocity

So before the collision:
4kg x 8m/s = 32
1kg x 0m/s = 0
32+0=32

Therefore after the collision
4kg x 4.8m/s = 19.2
1kg x βm/s = β
19.2 + β = 32

Therefore β = 12.8 m/s
4 0
3 years ago
A young kid of mass m = 36 kg is swinging on a swing. The length from the top of the swing set to the seat is L = 3.5 m. The boy
lara31 [8.8K]

Answer:

Explanation:

Given

mass of boy=36 kg

length of swing=3.5 m

Let T be the tension in the swing

At top point mg-T=\frac{mv^2}{r}

where v=velocity needed to complete circular path

Th-resold velocity is given by mg-0=\frac{mv^2}{r}

v=\sqrt{gr}=\sqrt{9.8\times 3.5}=5.85 m/s

So apparent weight of boy will be zero at top when it travels with a velocity of v=\sqrt{gr}

To get the velocity at bottom conserve energy at Top and bottom

At top E_T=mg\times 2L+\frac{mv^2}{2}

Energy at Bottom E_b=\frac{mv_0^2}{2}

Comparing two as energy is conserved

v_0^2=4gl+gl

v_0^2=5gL

v_0=\sqrt{5gL}=13.09 m/s

Apparent weight at bottom is given by

W=\frac{mv_0^2}{L}-mg=\frac{36\times 13.09^2}{3.5}+36\times 9.8=2115.23 N

6 0
2 years ago
What is the difference between resilience and modulus of resilience
pshichka [43]
<span>Resilience is the amount of energy that can be put into a volume of material and still be stored elastically. ie When the energy goes away, the material regains its undeformed shape. The Mod of R is the amount that can be stored by a unit volume of that material. The Mod of R is heavily related to Youngs Modulus.</span>
6 0
3 years ago
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