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vaieri [72.5K]
3 years ago
14

Consider a small child sliding down two differently shaped slides that have the same coefficient of kinetic friction. How must t

he second slide be shaped to provide the same change in gravitational potential energy of the child-Earth system but more change in kinetic energy for the child while they slide down
Physics
1 answer:
timama [110]3 years ago
6 0

Answer:

Having a bigger angle above the horizontal

Explanation:

Applying the energy conservation theorem:

K_1+U_1+W_{ext}=K_2+U_2

The kinetic energy is reduced because of the work done by the friction force.

The friction force is given by:

F_f=F_N*\µ

so the friction force depends on the Normal force, because the slide has an angle the normal force is given by:

F_N=m.g*cos(\theta)

So when the angle of the slide is bigger, the friction force decreases, for example:

for 45 degrees:

F_N=m.g*cos(45)\\F_N=0.70(m.g)\\

for 75 degrees:

F_N=m.g*cos(75)\\F_N=0.26(m.g)\\

as you can see if the angle is bigger above the horizontal, the friction force is reduced and so the work done by that force. We didn't have to change the height of the slide, so the potential gravitational energy remains the same.

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Select the correct answer.
Pachacha [2.7K]

Answer:

c

Explanation:

It's c the last one u see

8 0
2 years ago
Find the cube roots of 27(cos 327° + i sin 327° ). Write the answer in trigonometric form.
Sati [7]

Answer:

z^{\frac{1}{3} }= -0.978 + i\cdot 2.836, z^{\frac{1}{3} }= -1.967 - i\cdot 2.265, z^{\frac{1}{3} }= 2.945 - i\cdot 0.571

Explanation:

The cube root of the complex number can determined by the following De Moivre's Formula:

z^{\frac{1}{n} } = r^{\frac{1}{n} }\cdot \left[\cos\left(\frac{x + 2\pi\cdot k}{n} \right) + i\cdot \sin\left(\frac{x+2\pi\cdot k}{n} \right)\right]

Where angles are measured in radians and k represents an integer between 0 and n - 1.

The magnitude of the complex number is 27 and the equivalent angular value is 1.817\pi. The set of cubic roots are, respectively:

k = 0

z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{1.817\pi}{3} \right)+i\cdot \sin\left(\frac{1.817\pi}{3} \right)]

z^{\frac{1}{3} }= -0.978 + i\cdot 2.836

k = 1

z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{3.817\pi}{3} \right)+i\cdot \sin\left(\frac{3.817\pi}{3} \right)]

z^{\frac{1}{3} }= -1.967 - i\cdot 2.265

k = 2

z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{5.817\pi}{3} \right)+i\cdot \sin\left(\frac{5.817\pi}{3} \right)]

z^{\frac{1}{3} }= 2.945 - i\cdot 0.571

5 0
3 years ago
The pulley system below uses a gasoline engine to raise a drill head up through a smooth drill pipe. The engine provides a const
polet [3.4K]

NB: The diagram of the pulley system is not shown but the information provided is sufficient to answer the question

Answer:

Power = 2702.56 W

Explanation:

Let the power consumed be P

Energy expended = E = mgh

height, h = 5 m

E = 80 * 9.8 * 5

E = 3920 J

Power = \frac{Energy}{time}

To calculate the time, t

From F = ma

F = 900 N

900 = 80 a

a = 900/80

a = 11.25 m/s²

From the equation of motion, s = ut + 0.5at^{2}

The drill head starts from rest, u = 0 m/s

5 = 0 * t + (0.5*11.25*t^{2} )\\5 = 5.625t^{2}\\t^{2} = 5/5.625\\t^{2} = 0.889\\t = 0.943 s

Power, P = E/t

P = 3920/0.0.943

P = 4157.79 W

But Efficiency, E = 0.65

P = 0.65 * 4157.79

Power = 2702.56 W

6 0
2 years ago
Your cousin is moving into an apartment in San Francisco. This apartment is on a street that is angled at 25∘ above the horizont
Vesnalui [34]

Answer:

70

Explanation:

5 0
2 years ago
Microwave ovens emit microwave energy with a wavelength of 12.2 cm. What is the energy of exactly one photon of this microwave r
Iteru [2.4K]

Answer:

1.63\cdot 10^{-24} J

Explanation:

The energy of a photon is given by:

E=\frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda is the wavelength of the photon

In this problem, we have a microwave photon with wavelength

\lambda=12.2 cm=0.122 m

Substituting into the equation, we find its energy:

E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{0.122 m}=1.63\cdot 10^{-24} J

5 0
3 years ago
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