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larisa86 [58]
3 years ago
9

Observing a spacecraft land on a distant asteroid, scientists notice that the craft is falling at a rate of 5 m/s. When it is 10

0 m closer to the surface of the asteroid, the craft reports a velocity of 8 m/s. According to their data, what is the approximate gravitational acceleration on this asteroid?
A. 0 m/s2
B. 0.03 m/s2
C. 0.20 m/s2
D. 0.65 m/s2
E. 33 m/s2
Physics
1 answer:
denpristay [2]3 years ago
4 0

Answer:

(c) 0.20 m/s²

Explanation:

Using The equation of motion,

V² = u² + 2gs...................... equation 1

Where V = final velocity, u = initial velocity, g = gravitational acceleration,

s = distance.

Making g the subject of the equation in equation 1 above,

g =( V² - u²)/2s............................ equation 2

Where V= 8 m/s, u = 5 m/s, s= 100 m.

g = (8²-5²)/(2×100)

g = (64-25)/200

g = 39/200

g =0.195 m/s²

g ≈ 0.20 m/s².

The gravitational acceleration = 0.20 m/s²

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2 years ago
A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput
Nikitich [7]
There is a missing data in the text of the problem (found on internet):
"with force constant<span> k=</span>450N/<span>m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
</span>E=U+K=  \frac{1}{2}kx^2 + \frac{1}{2} mv^2
<span>where
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v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
</span>E=K_{max} =  \frac{1}{2}mv_{max}^2
<span>Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
</span>E=U_{max}= \frac{1}{2}k A^2
<span>
Since the mechanical energy must be conserved, we can write
</span>\frac{1}{2}mv_{max}^2 =  \frac{1}{2}kA^2
<span>from which we find the maximum speed
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<span>
b) </span><span> the </span>speed<span> of the </span>glider<span> when it is at x= -0.015</span><span>m

We can still use the conservation of energy to solve this part. 
The total mechanical energy is:
</span>E=K_{max}=  \frac{1}{2}mv_{max}^2= 0.36 J
<span>
At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
</span>U= \frac{1}{2}kx^2 =  \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J
<span>And since 
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</span>K=E-U=0.36 J - 0.05 J = 0.31 J
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v=  \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s
<span>
c) </span><span>the magnitude of the maximum acceleration of the glider;
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For a simple harmonic motion, the magnitude of the maximum acceleration is given by
a_{max} = \omega^2 A
where \omega= \sqrt{ \frac{k}{m} } is the angular frequency, and A is the amplitude.
The angular frequency is:
\omega =  \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s
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d) <span>the </span>acceleration<span> of the </span>glider<span> at x= -0.015</span><span>m

For a simple harmonic motion, the acceleration is given by
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<span>
e) </span><span>the total mechanical energy of the glider at any point in its motion. </span><span>

we have already calculated it at point b), and it is given by
</span>E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J
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