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Maksim231197 [3]
2 years ago
7

Energy conservation with conservative forces: Two identical balls are thrown directly upward, ball A at speed v and ball B at sp

eed 2v, and they feel no air resistance. Which statement about these balls is correct?A) At their highest point, the acceleration of each ball is instantaneously equal tozero because they stop for an instant.B) The balls will reach the same height because they have the same mass and thesame acceleration.C) At its highest point, ball B will have twice as much gravitational potential energyas ball A because it started out moving twice as fast.D) Ball B will go twice as high as ball A because it had twice the initial speed.E) Ball B will go four times as high as ball A because it had four times the initialkinetic energy.
Physics
1 answer:
MatroZZZ [7]2 years ago
6 0

Answer:

E) True.   Ball B will go four times as high as ball A because it had four times the initial kinetic energ

Explanation:

To answer the final statements, let's pose the solution of the exercise

Energy is conserved

Initial

          Em₀ = K

          Em₀ = ½ m v²

Final

         Emf = U = mg h

         Em₀ = emf

        ½ m v² = mgh

        h = v² / 2g

For ball A

         h_A = v² / 2g

For ball B

        h_B = (2v)² / 2g

        h_B = 4 (v² / 2g) = 4 h_A

Let's review the claims

A) False. The neck acceleration is zero, it has the value of the acceleration of gravity

B) False. Ball B goes higher

C) False  has 4 times the gravitational potential energy than ball A

D) False.  It goes 4 times higher

E) True.

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Explanation:

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Mass of the brick, m = 1.15 kg

Radius of the circle, r = 1.44 m

The cable will break if the tension exceeds 43.0 N

Let v is the maximum sped can have at the bottom of the circle before the cable will break. At the bottom of the circle, the net force is equal to the centripetal force along with the weight of the brick. So,

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v=\sqrt{(\dfrac{T}{m}-g)r}

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8 0
3 years ago
What is the frequency of a photon with an energy of 4. 56 x 10^-19 j
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The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

<h3>What is a frequency?</h3>

The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.

Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma  = 6.88 \times 10^{14}\ s^{-1}

Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

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Answer the following. (a) What is the surface temperature of Betelgeuse, a red giant star in the constellation of Orion, which r
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(b) T = 19986.2 k

Explanation:

The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{\lambda_{max}}

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(a)

here,

\lambda_{max} = 970 nm = 9.7 x 10⁻⁷ m

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T = \frac{0.2898\ x\ 10^{-2}\ m.k}{9.7\ x\ 10^{-7}\ m}

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here,

\lambda_{max} = 145 nm = 1.45 x 10⁻⁷ m

Therefore,

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A horizontal spring with spring constant 85 N/m extends outward from a wall just above floor level. A 5.5 kg box sliding across
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