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Setler79 [48]
4 years ago
14

A space-walking astronaut has become detached from her spaceship. She's floating in space with her handy tool belt attached to h

er waist, thinking about how she might get back to the ship, which she can see 50 meters from her current location.
How can she get back to the ship? Use language from the Laws of Motion in your answer.
Physics
1 answer:
telo118 [61]4 years ago
6 0
I can tell you clearly you copied off what the question completely said from your homework. You can't really do that though you can ask people part of the question but not the hole entire thing, it makes you not feel smart by us giving you the answer an plegerizing it that should not be one. This is no offence i am just telling i dont want you to feel bad though. its just personal prefence on what im telling you.
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A force acts on a particle as the particle moves along an x axis, with in newtons, x in meters, and c a constant. At x = 0 m, th
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Answer:

Incomplete question... We are suppose to be given the force equation and i think it is Fvec = (cx - 3.00x^2)i

Explanation:

Check attachment for solution

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4 years ago
What is the half-life of an isotope if after 30 days you have 31.25 g remaining from a 250 g beginning sample size?
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Answer:The time required for half of the original population of radioactive atoms to decay is called the half-life. The relationship between the half-life, T1/2, and the decay constant is given by T1/2 = 0.693/λ.

Explanation:

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2 years ago
A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring,
Mademuasel [1]

Answer:

Explanation:

Given

mass of cannon and cart is m_1=6475.91 kg

Spring constant k=20000 N/m

mass of Projectile m_2=234 kg

Launch Velocity of cannon v_2=170 m/s

Launch angle \theta =31.3^{\circ}

As the External Force is zero therefore we can conserve momentum

Initially both Projectile and cannon is at rest

Conserving momentum in horizontal direction

m_1u_1+m_2u_2=m_1v_1+m_2v_2\cos 31.3

0+0=6475.91\times v_1+234\times 170\cdot \cos (31.3)

v_1=-\frac{234\times 170\cdot \cos (31.3)}{6475.91}

v_1=-5.24 m/s

i.e. cannon is moving in opposite direction of Projectile

6 0
4 years ago
Two people, one with mass m1 and the other with mass m2, stand on a stationary sled with mass M on a frozen lake. Assume that th
lozanna [386]

Answer:

Part a)

Velocity of sled

v = \frac{m_1 s}{m_1 + m_2 + M}

velocity of first man who jump off

v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}

Part b)

Velocity of sled

v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s

Also the speed of second person is given as

v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}

Part c)

change in kinetic energy of sled + two people is given as

KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2

Explanation:

As we know that here we we consider both people + sled as a system then there is no external force on it

So here we can use momentum conservation

since both people + sled is at rest initially so initial total momentum is zero

now when first people will jump with relative velocity "s" then let say the sled + other people will move off with speed v

so by momentum conservation we have

0 = m_1(v - s) + (m_2 + M)v

v = \frac{m_1 s}{m_1 + m_2 + M}

so velocity of the sled + other person is

v = \frac{m_1 s}{m_1 + m_2 + M}

velocity of first man who jump off

v_1 = \frac{m_1 s}{m_1 + m_2 + M} - s

v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}

Part b)

now when other man also jump off with same relative velocity

so let say the sled is now moving with speed vf

so by momentum conservation we have

(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) = m_2(v_f - s) + Mv_f

(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) + m_2s = (m_2 + M)v_f

Now we have

v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s

Also the speed of second person is given as

v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s - s

v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}

Part c)

change in kinetic energy of sled + two people is given as

KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2

here we know all values of speed as we found it in part a) and part b)

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3 years ago
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