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3 years ago

a coin is dropped from the top of a tall building. determine the coin's (a) velocity and (b) displacement after 1.5 sec.

Physics

1 answer:

Lilit [14]3 years ago

8 0

1) v = gt = 10*1.5 = 15 m/s
2) r = gt^2 /2 = 10*(1.5)^2 / 2 = 11.25 meters

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What is the significance of the nose end marking on a rocket or missile?

Gemiola [76]

If you're referring to the different colors that usually occur at the tip of missles, rockets and some other aircraft, it either a) signifies the end of a particular plate of metal, fabricated specifically to be for the nose. Sometimes these can even be a different alloy or metal all together. or b) this shows where the curved surface begins, so in the case of damage or imperfections due to wear, they can be repaired and measured more easily. The shape of the nose is extremely important for smooth flight, and a dent or bump formed on it can make the aircraft unstable. If you can measure from where the curve starts by the difference in color, it makes repairing or re-fabricating the part much easier. Many of these curves aren't as simple as they appear.

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3 years ago

A system undergoes a two-step process. In the first step, the internal energy of the system increases by 222 J when 150 J of wor

joja [24]

Answer:0 J

Explanation:

Given

For first step

change in internal Energy of the system is $\Delta U_1=222 J$

Work done on the system $W_1=-150 J$

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change in internal Energy of the system is $\Delta U_2=123 J$

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$Q=\Delta U+W$

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3 years ago

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olga nikolaevna [1]

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Hope this helps! (:

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3 years ago

Read 2 more answers

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity v is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$