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2 years ago

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read the excerpt below and answer the question. "no roving foot shall crush thee here, no busy hand provoke a tear." what type o

f figurative language does freneau employ in these lines from “the wild honeysuckle”? a.metaphor b.onomatopoeia c.paradox d.personification

1 answer:

Aneli [31]2 years ago

3 0

The type of figurative language that Freneau employ in these lines from "The Wild Honeysuckle" is personification. The correct answer would be option D. Why is it personification? Personification is a figure of speech that uses human attributes to something that is not living. In this line, the author attributes the words "roving" and "busy" to foot and hand, respectively.

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An ant travels 2.78 cm [W] and then turns and travels 6.25 cm [S 40 degrees E]. What is the ant's total displacement?

Sunny_sXe [5.5K]

Answer is 6.84 approx
reason:-
(2.78^2+6.25^2)^1/2=6.84 approx

4 0

2 years ago

A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of

olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through $90^{o}$ then position of two mirrors will be changed. The resultant path difference will be as follows.

$\frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}$

Formula for change in fringe shift is as follows.

n = $\frac{2lv^{2}}{\lambda c^{2}}$

$v^{2} = \frac{n \lambda c^{2}}{2l}$

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

l = 11 m

$\lambda = 5.9 \times 10^{-7} m$

c = $3.0 \times 10^{8}$ m/s

Hence, putting the given values into the above formula as follows.

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

= $\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}$

= $2.41363 \times 10^{9} m/s$

Thus, we can conclude that velocity deduced is $2.41363 \times 10^{9} m/s$ .

3 0

1 year ago

A segment of wire of total length 2.0 m is formed into a circular loop having 5.0 turns. If the wire carries a 1.2-A current, de

docker41 [41]

Answer:

Magnetic field at the center of the loop $B=5.89\times 10^{-5}\ T.$

Explanation:

It is given that total length of wire is 2 m and number of circular loop is 5 turns.

Therefore ,

$5\times ( 2\pi r)=2 \ m .\\\\r=\dfrac{1}{5 \pi}=0.064\ m.$

We know , magnetic field at the center of loop is given by :

$B=N\dfrac{\mu_o i}{2r}$

Putting all values in above equation we get :

$B=5\times \dfrac{4\pi\times 10^{-7}\times 1.2}{2\times 0.064}\\\\B=5.89\times 10^{-5}\ T.$

Hence , this is the required solution.

8 0

1 year ago

A large crate is suspended from the end of a vertical rope. Is the tension in the rope greater when the crate is at rest or when

choli [55]

Answer:

Part a)

the tension force is equal to the weight of the crate

Part b)

tension force is more than the weight of the crate while accelerating upwards

tension force is less than the weight of crate if it is accelerating downwards

Explanation:

Part a)

When large crate is suspended at rest or moving with uniform speed then it is given as

$F_t - mg = ma$

here since speed is constant or it is at rest

so we will have

$a = 0$

$F_t = mg$

so the tension force is equal to the weight of the crate

Part b)

Now let say the crate is accelerating upwards

now we can say

$F_t - mg = ma$

$F_t = mg + ma$

so tension force is more than the weight of the crate

Now if the crate is accelerating downwards

$F_t - mg = -ma$

$F_t = mg - ma$

so tension force is less than the weight of crate if it is accelerating downwards

4 0

2 years ago

The deuterium nucleus starts out with a kinetic energy of 1.24 × 10-13 joules, and the proton starts out with a kinetic energy o

MrMuchimi

Answer:

The total kinetic energy of both particles is $2.43\times10^{-13}$

Explanation:

Given that,

Kinetic energy of nucleus $K.E= 1.24\times10^{-13}\ J$

Kinetic energy of proton $K.E= 2.47\times10^{-13}\ J$

Radius of proton $r= 0.9\times10^{-15}\ m$

We need to calculate the final potential energy

Using formula of final potential energy

$U=\dfrac{kq^2}{r}$

Put the value into the formula

$U_{f}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times0.9\times10^{-15}}$

$U_{f}=1.28\times10^{-13}\ J$

We need to calculate the initial energy of both the particles

Using formula of energy

$E_{i}=(K.E_{n}+K.E_{p})+U_{i}$

$E_{i}=1.24\times10^{-13}+2.47\times10^{-13}+0$

$E_{i}=3.71\times10^{-13}\ J$

We need to calculate the total kinetic energy of both particles

Using conservation of energy

$E_{i}=E_{f}$

$E_{i}=K.E_{f}+U_{f}$

$3.71\times10^{-13}=K.E_{f}+1.28\times10^{-13}$

$K.E_{f}=3.71\times10^{-13}-1.28\times10^{-13}$

$K.E_{f}=2.43\times10^{-13}$

Hence, The total kinetic energy of both particles is $2.43\times10^{-13}$

3 0

1 year ago