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QveST [7]
3 years ago
11

What is the rhyme scheme in Emily Dickinson’s poem, “Hope is the thing with feathers”?

Physics
1 answer:
kaheart [24]3 years ago
5 0
Answer is B. ABAB. Hope it helped you, and have a great day.
-Charlie
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Gina made a poster for plastic recycling week and included this information on her poster:
miskamm [114]

Answer:

Gina should put “rubber tires” under “Synthetic.”

Gina should put “starch” under “Natural.”

Explanation: I just did the assignment ;P

4 0
3 years ago
Read 2 more answers
Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.8 x 105 kg on springs that hav
Vikki [24]

Answer:

The force constant is  k =1.316 *10^{7} \  N/m

The energy stored in the spring is  E =  1.68 *10^{7} \ J

Explanation:

From the question we are told that

   The mass of the object is  M  = 4.8*10^{5} \ kg

    The period is T  = 1.2 \ s

The period of the spring oscillation is  mathematically represented as

         T  =2 \pi \sqrt{ \frac{M}{k}}

where  k is the force constant

   So making k the subject

       k = \frac{4 \pi ^2 M }{T^2}

substituting values

       k = \frac{4 (3.142) ^2 (4.8 *10^{5}) }{(1.2)^2}

      k =1.316 *10^{7} \  N/m

The energy stored in the spring is mathematically represented  as

       E =  \frac{1}{2} k x^2

Where x is the spring displacement which is given as

        x =  1.6 \ m

substituting values

      E =  \frac{1}{2} (1.316 *10^{7}) (1.6)^2

       E =  1.68 *10^{7} \ J

   

7 0
3 years ago
Un objeto, que se encuentra a nivel del suelo, es lanzado verticalmente hacia arriba con una velocidad de 160 km/hr. ¿Qué altura
LiRa [457]
I don’t speak Spanish?
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3 years ago
A train travelling at 18m/s accelerates uniformly at 2m/s² for 6 seconds. Calculate the final speed of the train.
kobusy [5.1K]

its going  to be 30 m/s

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3 years ago
What are the factors affect the Electric forces between two charges and What is the relationship between each factor and the Ele
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Explanation:

If the two charges are point charges - i.e., they don't have a size - the force between these charges depends on the

• Magnitude if each charge, q1 and q2

• Sign of each charge (+ or -)

• Distance between the charges, r

This is essentially Coulomb’s Law:

FE = (kq1q2)/r2

For collections of charges, you need to find the electric field E, and then use this fields to find a force on a small test charge q in the field. The test charge is always small to help you map the electric field, but not disturb it.

3 0
2 years ago
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