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Paraphin [41]
2 years ago
4

A constant horizontal pull acts on a sled on a horizontal frictionless ice pond. The sled starts from rest. When the pull acts o

ver a distance x, the sled acquires a speed v and a kinetic energy K. If the same pull instead acts over twice this distance, A) The sled’s speed will be 2v and its kinetic energy will be 2K.
B) The sled’s speed will be 2v and its kinetic energy will be K 2 .
C) The sled’s speed will be v 2 and its kinetic energy will be 2K.
D) The sled's speed will be v 2 and its kinetic energy will be K 2 .
E) The sled's speed will be 4v and its kinetic energy will be 2K.
Physics
1 answer:
Kamila [148]2 years ago
4 0

Answer:

K'= 2 K

v'= v√2

Explanation:

As we know that

From work power energy

Work done by all the force = Change in kinetic energy

F . x = K

When distance become 2 x

Then

F.2 x = K'

K' = 2 F . x

K'= 2 K

We know that

K=\dfrac{mv^2}{2}

K'=\dfrac{mv'^2}{2}

K'= 2 K

2\dfrac{mv^2}{2}=\dfrac{mv'^2}{2}

v'²= 2 v²

v'= v√2

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Answer:

1620

Explanation:

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5 0
2 years ago
You are working on a laboratory device that includes a small sphere with a large electric charge Q. Because of this charged sphe
madam [21]

Answer:

the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.

Explanation:

We can answer this exercise using Gauss's law

      Ф = ∫ e . dA = q_{int} / ε₀

field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell.  the flow must be zero since the charge of the sphere is equal  induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field

From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.

5 0
2 years ago
A particular material has an index of refraction of 1.25. What percent of the speed of light in a vacuum is the speed of light i
beks73 [17]

Answer:

80% (Eighty percent)

Explanation:

The material has a refractive index (n) of 1.25

Speed of light in a vacuum (c) is 2.99792458 x 10⁸  m/s

We can find the speed of light in the material (v) using the relationship

n = c/v, similarly

v = c/n

therefore v = 2.99792458 x 10⁸  m/s ÷ (1.25) = 239 833 966 m/s

v = 239 833 966 m/s

Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as

(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%

Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)

3 0
2 years ago
Read 2 more answers
Sandra's target heart rate zone is 135bpm—172bpm. Marissa's target heart rate zone is 143bpm—176bpm. They stop playing basketbal
Feliz [49]

Answer: Neither Sandra nor Marissa will be in her THR zone.


Explanation:


1) Actual pulse of both Sandra and Marissa : 144 bpm


2) Decrease of 20 bpm ⇒ 144 bpm - 20 bpm = 124 bpm


3) Sandra's TRH is in the range 135 - 172 bpm.


Since 124 < 135, she will be below the range.


4) Marissa's TRH range is 143 - 176 bpm.


Since, 124 < 143, she is below the range


In conlusion, neither Sandra nor Marissa will be in her THR zone.

6 0
2 years ago
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If a scale on Earth reads 650 N, what is your mass?
OlgaM077 [116]
If the scale reads 650N, then the mass of whoever it is standing on the scale is

         (weight) / (gravity)  =  (650N) / (9.8 m/s²)  =  66.3 kilograms  .

It's not MY mass, even if I'm the one standing on the scale. 
If I stand on a scale and it reads 650 N, the scale is broken.
4 0
2 years ago
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