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4 years ago

10

cheAt an amusement park, a swimmer uses a water slide to enter the main pool. If the swimmer starts at rest, slides without fric

tion, and falls through a vertical height of 2.77 m, what is her speed at the bottom of the slide

1 answer:

KiRa [710]4 years ago

3 0

Answer:

$v \approx 7.371\,\frac{m}{s}$

Explanation:

The swimmer can be modelled by the Principle of Energy Conservation. The speed is derived herein:

$m_{swimmer}\cdot g \cdot h = \frac{1}{2}\cdot m_{swimmer}\cdot v^{2}$

$v = \sqrt{2\cdot g \cdot h}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.77\,m)}$

$v \approx 7.371\,\frac{m}{s}$

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A boy kicked off a cliff and lands 151m away 45s later. What was the initial velocity? How tall is the cliff?

Mnenie [13.5K]

S = 1/2 g t^2 = 1/2 * 9.8 * 45^2 = 9922 m
That’s the height of the cliff

The original sideways velocity = 151/45 = 3.35 m/s

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3 years ago

What does it mean to say that the forces acting on an object are balanced?

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Hello,

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3 years ago

fenix001 [56]

Answer:

(b) Water

Explanation:

The density of the liquid is found by dividing its mass by its volume.

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The mass of the liquid is the difference in mass between the full and empty graduated cylinder:

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<h3>volume</h3>

The cylinder is said to contain 100 mL of the liquid, so that is the volume of interest.

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2 years ago

To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105

-Dominant- [34]

Explanation:

LD₁ = 10⁵ mm⁻²

LD₂ = 10⁴mm⁻²

V = 1000 mm³

Distance = (LD)(V)

Distance₁ = (10⁵mm⁻²)(1000mm³) = 10×10⁷mm = 10×10⁴m

Distance₂ = (10⁹mm⁻²)(1000mm³) = 1×10¹² mm = 1×10⁹ m

Conversion to miles:

Distance₁ = 10×10⁴ m / 1609m = 62 miles

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3 years ago

Read 2 more answers

A proton moves along the x-axis in the laboratory with velocity uy = 0.6c. An observer moves with a velocity of v=0.8c along the

Dima020 [189]

Explanation:

Given that,

Velocity of the proton in lab frame $u_{x} = 0.6c$

Velocity of the observer v= 0.8c

We need to calculate the velocity of the proton with respect to the observer

Using formula of velocity

$u'=\dfrac{v-u_{x}}{1-\dfrac{u_{x}v}{c^2}}$

$u'=\dfrac{0.8c-0.6c}{1-\dfrac{0.6c\times0.8c}{c^2}}$

$u'=c(\dfrac{0.8-0.6}{1-(0.8\times0.6)})$

$u'=0.385c$

(a). We need to calculate the total energy of the proton in the lab frame

Using formula of kinetic energy

$K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u_{x}^2}{c^2}}}-1)$

Where, Proton mass energy = m₀c²

Put the value into the formula

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.6c)^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.6)^2}}-1)$

$K.E=234.57\ MeV$

(b). We need to calculate the kinetic energy of the proton in the observer

Using formula of kinetic energy

$K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u'^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.385)^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.385)^2}}-1)$

$K.E=78.366\ MeV$

(c). We need to calculate the momentum of the proton with respect to observer

Using formula of momentum

$P_{obs}=\dfrac{m_{0}u'^2}{\sqrt{1-\dfrac{u'^2}{c^2}}}$

We know that,

Proton mass energy = m₀c²

$m_{0}=\dfrac{938.28}{c^2}$

$P_{obs}=\dfrac{\dfrac{938.28}{c^2}\times(0.385c)^2}{\sqrt{1-\dfrac{(0.385c)^2}{c^2}}}$

$P_{obs}=\dfrac{938.28\times(0.385)^2}{\sqrt{1-0.385^2}}$

$P_{obs}=150.69\ MeV/c$

Hence, This is required solution.

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3 years ago