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3 years ago

What can i find when im only given the initial and final speed of an object

Physics

1 answer:

Novosadov [1.4K]3 years ago

8 0

Answer:

you can you should find the velocity of the thing by which it is running

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Clutter is _______. annoying frustrating death

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Answer: _____= beautiful, yet annoying frustrating death

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What condition is necessary for the sustained flow of water in a pipe?

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3 years ago

A student has a small metal sphere that hangs from an insulating string. The student, in seeking to determine the charge on the

Leokris [45]

Answer:

the answer the correct one is c

Explanation:

Electric charges of different signs attract and those of the same sign repel. In addition, there are two types of insulating bodies, where the loads are fixed (immobile) and metallic (with mobile loads.

Let's analyze the situation presented

* A rod with positive approaches and the sphere is attracted, so the charge on the sphere is negative

* A rod with a negative charge approaches and the sphere is attracted, therefore the charge of the sphere must be positive.

For this to happen, the sphere must be unloaded and the charge that creates the phenomenon are induced charges because the mobile charges of the same sign as the sphere are repelled.

when checking the answer the correct one is c

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3 years ago

Can anyone please answer this cq..

ZanzabumX [31]

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1 year ago

The star Rho1 Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a

s344n2d4d5 [400]

Answer:

82780.42123 m/s

14.45 days

Explanation:

m = Mass of the planet

M = Mass of the star = $0.85\times 1.989\times 10^{30}\ kg=1.69065\times 10^{30}\ kg$

r = Radius of orbit of planet = $0.11\times 149.6\times 10^{9}\ m=16.456\times 10^{9}\ m$

v = Orbital speed

The kinetic and potential energy balance is given by

$\frac{GMm}{r^2}=\frac{mv^2}{r}\\\Rightarrow v=\sqrt{\dfrac{Gm}{r}}\\\Rightarrow v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.69065\times 10^{30}}{16.456\times 10^{9}}}\\\Rightarrow v=82780.42123\ m/s$

The orbital speed of the star is 82780.42123 m/s

The orbital period is given by

$t=\frac{2\pi r}{v}\\\Rightarrow t=\dfrac{2\pi \times 16.456\times 10^{9}}{82780.42123}\\\Rightarrow t=1249040.48419\ seconds=\dfrac{1249040.48419}{24\times 60\times 60}=14.45\ days$

The orbital period is 14.45 days

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2 years ago