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RSB [31]
1 year ago
14

a child in a tree house uses a rope attached to a basket to lift a 26 nn dog upward through a distance of 5.0 mm into the house.

Physics
1 answer:
agasfer [191]1 year ago
4 0

Total work done is 0.13 Joules

<h3>What is work done ?</h3>

The sum of the displacement and the component of the applied force of the object in the displacement direction is the work done by a force.

According to the given information

We need to find the work done

work done  = force × distance

We are given,

force  = 26 N

Distance = 0.0005 meter

hence ,

Work done  = 26 × 0.005  

                   = 0.13 Joules

Total work done is 0.13 Joules

To know more about Work done

brainly.com/question/13662169

#SPJ4

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Which pollution is a major problem resulting from the _______ industry in eastern Alberta, Canada
Likurg_2 [28]
Water pollution is a major problem resulting from the oil industry in eastern Alberta, Canada.
Hope it helps!
5 0
3 years ago
A positively-charged object with a mass of 0.129 kg oscillates at the end of a spring, generating ELF (extremely low frequency)
9966 [12]

Answer:308 N/m

Explanation:

Given

mass\left ( m\right )=0.129 kg

wavelength\left ( \lambda \right )=3.86\times 10^7

We know frequency =\frac{c}{\lambda }=\frac{3\times 10^8}{3.86\tmes 10^7}

f=7.772 Hz

As the frequency of radio waves is same as the frequency at which object oscillates

f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}

7.772=\frac{1}{2\pi }\sqrt{\frac{k}{0.129}}

7.772\times 2\times \pi =\sqrt{\frac{k}{0.129}}

k=307.70\approx 308 N/m

7 0
3 years ago
Which of the following lies in the ecliptic plane?
babymother [125]
<h2>Answer: Earth's orbital path around the Sun</h2><h2></h2>

The <u>Ecliptic</u> refers to the orbit of the Earth around the Sun. Therefore, <u>for an observer on Earth it will be the apparent path of the Sun in the sky during the year, with respect to the "immobile background" of the other stars.</u>

<u />

It should be noted that the ecliptic plane (which is the same orbital plane of the Earth in its translation movement) is tilted with respect to the equator of the planet about 23\° approximately. This is due to the inclination of the Earth's axis.

Hence, the correct option is Earth's orbital path around the Sun.

7 0
3 years ago
A satellite has a mass of 5832 kg and is in a circular orbit 4.13 × 105 m above the surface of a planet. The period of the orbit
elena55 [62]

Answer:

W = 28226.88 N

Explanation:

Given,

Mass of the satellite, m = 5832 Kg

Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m

The time period of the orbit, T = 1.9 h

                                            = 6840 s

The radius of the planet, R = 4.38 x 10⁶ m

The time period of the satellite is given by the formula

                             T = 2\pi \sqrt{\frac{(R+h)^{3} }{R^{2} g} }  second

Squaring the terms and solving it for 'g'

                             g = 4 π² \frac{(R+h)^{3} }{R^{2}T^{2}  }   m/s²

Substituting the values in the above equation

                   g = 4 π² \frac{(4.38X10^6+4.13X10^5)^{3} }{(4.38X10^6)^{2}X6840^{2}}  

                                    g = 4.84 m/s²      

Therefore, the weight

                                     w = m x g   newton

                                         = 5832 Kg x 4.84 m/s²

                                         = 28226.88 N

Hence, the weight of the satellite at the surface, W = 28226.88 N                

8 0
3 years ago
5. Describe the shape of the waveform in the secondary coil for a sine, square and triangle wave in the primary coil. How does t
Volgvan

Answer:

When primary coil is exited by sin wave,this will result in sin wave in secondary coil as well.According to law,flux induced in the secondary coil will have same waveform as in the primary coil.

5 0
3 years ago
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