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pshichka [43]
3 years ago
9

A meteor moving 468 km per minute traveling in a south-to-north direction passed near Earth in 2013. Because the meteor was only

45 m wide and was 27,700 km above Earth’s surface, it was not visible without the aid of a telescope.Which statement BEST describes the meteor’s motion?
A Its velocity was 7.8 km/s northward
B Its acceleration was 468 kilometers per second squared
C Its speed was 468 km/s northward
D Its acceleration was 7.8 kilometers per second squared
Physics
2 answers:
muminat3 years ago
7 0
The correct statement is <span>A Its velocity was 7.8 km/s northward. 
In fact, the problem says that the meteor was moving at 468 km per minute. We can convert this value into km/s, keeping in mind that there are 60 seconds in one minute: 1 min=60 s:
</span>v=468 km/min =(468 km/min)( \frac{1}{60} min/s )=7.8 km/s<span>
</span>
Ivenika [448]3 years ago
5 0

Answer:

a

Explanation:

ummmmmm.

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Answer:

The remaining percentage of drug concentration is about 88.7% 2 years after manufacture.

Explanation:

Recall the formula for the decay of a substance at an initial N_0 concentration at manufacture:

N(t)=N_0\,e^{-k\,\,t}

where k is the decay rate (in our case 0.06/year), and t is the elapsed time in years. Therefore, after 2 years since manufacture we have:

N=N_0\,e^{-0.06\,\,(2)}\\N=N_0\,e^{-0.12}\\N/N_0=e^{-0.12}\\N/N_0=0.8869

This in percent form is 88.7 %. That is, the remaining percentage of drug concentration is about 88.7% 2 years after manufacture.

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3 years ago
Atoms are attracted to each other because____
yaroslaw [1]

Answer:

A

Explanation:

Opposite attract. Since electrons are negative they will be attracted to the protons because they a positive

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3 years ago
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him
nataly862011 [7]

Answer:

a) u_c=0\ m.s^{-1}       &        m_c.v_c=m_b.v_b\times \cos\theta

b) v_c=0.0566\ m.s^{-1}

c) p_e=2.9218\ kg.m.s^{-1}

Explanation:

Given:

mass of the book, m_b=1.35\ kg

combined mass of the student and the skateboard, m_c=97\ kg

initial velocity of the book, v_b=4.61\ m.s^{-1}

angle of projection of the book from the horizontal, \theta=28^{\circ}

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

u_c=0\ m.s^{-1}

where:

u_c= initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

m_c.v_c=m_b.v_b\times \cos\theta

where:

v_c= final velcotiy of the student after throwing the book

b)

m_c.v_c=m_b.v_b\times \cos\theta

97\times v_c=1.35\times 4.61\cos28

v_c=0.0566\ m.s^{-1}

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

p_e=m_b.v_b\sin\theta

p_e=1.35\times 4.61\times \sin28^{\circ}

p_e=2.9218\ kg.m.s^{-1}

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Answer:

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