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3 years ago

Directed Reading for Section 2 - Acceleration

Physics

1 answer:

erastovalidia [21]3 years ago

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Abs and piper have a great time and

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Where are the variables from the lab?

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2 years ago

In one scene in the movie The Godfather II, a solid gold phone is passed around a large table for everyone to see. Suppose the v

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Volume of gold in the phone = 10 cm^3
= 0.00001 m^3 
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4 years ago

propane, the gas used in barbeque grills, is made of carbon and hydrogen. Will the atoms that make up propane form covalent bond

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Hydrocarbons are molecules that contain only carbon and hydrogen. Due to carbon's unique bonding patterns, hydrocarbons can have single, double, or triple bonds between the carbon atoms. The names of hydrocarbons with single bonds end in "-ane," those with double bonds end in "-ene," and those with triple bonds end in "-yne". The bonding of hydrocarbons allows them to form rings or chains.

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3 years ago

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As the elephant falls from 10 m does it lose or gain KE? Explain.

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4 years ago

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A person of mass 55 kg swings on a rope length 4 m from rest (when the rope makes an angle of 30 degrees with the vertical) and

vovangra [49]

Answer:

θ = 19.66°

Explanation:

To determine the angle that the rope makes with the vertical for the two people, you first take into account the potential energy of the first person before he swings on the rope:

$U=mgh$

h: distance to the ground

g: gravitational acceleration = 9.8m/s^2

m: mass of the first person = 55 kg

In the image attache below you can notice that the height h is:

$h=4-4cos(30\°)=0.53m$

Then, the potential energy is:

$U=(55kg)(9.8m/s^2)(0.53m)=285.67J$

When the first person picks up the second person (when the rope is exactly vertical), all the potential energy becomes kinetic energy. Next, when both people reaches the maximum height h' the energy must be equal to the initial potential energy of the first person:

$U'=(m_1+m_2)gh'=285.67\ J$

From the previous equation you can get h':

$h'=\frac{285.67J}{(55kg+70kg)(9.8m/s^2)}=0.2332m$

Finally, you obtain the angle between the rope at the height h,' and the vertical, by calculating the following:

$h'=4-4cos(\theta)\\\\\theta=cos^{-1}(\frac{4-h'}{4})=cos^{-1}(\frac{4-0.2332}{4})=19.66\°$

hence, the angle between the rope and the vertical, when the two people are in the rope is 19.66°

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3 years ago