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Citrus2011 [14]
3 years ago
12

If we have two objects with the same mass but different densities (Lets assume object 1 is denser,therefore volume is lower rela

tive to object 2).Assume both objects have a lower density than water.Now if we put both in water,both will float.Since they float, buoynacy force is equal to the weight of the object,so buoynacy force in 1 and 2 should be equal .However if we use F(boyancy)=pvg--> then becuase the submerged v is less for the denser object and p fluid is the same,then we can conclude that F bouynacy are not the same for both objects. True or false?

Physics
1 answer:
o-na [289]3 years ago
7 0

take the volume that is submerged. That's the volume of water displaced.

but the volume of submerged is different in those two

so buoyancy force is different in those two

weights are the same

probably. since the densities* are different.

In question it says wights are the same and diffferent volumes

so it seems that the one with more density should have a lower buyonacy force

but you said that when an object floats the buoyancy force equals weight,so since both objects have the same weight,then buoynacy force should be equal in those two


The more dense object will float with a greater percentage of its volume immersed, not less.

2) If they have the same MASS, the more dense one will have less VOLUME

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qaws [65]

Answer:

≅50°

Explanation:

We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:

Δx=V₀t+at²/2

And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:

Δx=(V₀cosθ)t+at²/2

Now luckily we are given everything we need to solve (or you found the info before posting here):

  • Δx=760 m
  • V₀=87 m/s
  • t=13.6 s
  • a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!

With that we can plug the values in to get:

760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}

760=(1183.2)(cos\theta)

cos\theta=\frac{760}{1183.2}

\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}

3 0
3 years ago
It has been suggested that rotating cylinders about 10 mi long and 5.9 mi in diameter be placed in space and used as colonies. T
tekilochka [14]

Answer:

ω = 0.05 rad/s

Explanation:

We consider the centripetal force acting as the weight force on the surface of the cylinder. Therefore,

Centripetal Force = Weight\\\frac{mv^{2}}{r} = mg\\\\here,\\v = linear\ speed = r\omega \\therefore,\\\frac{(r\omega)^{2}}{r} = g\\\\\omega^{2} = \frac{g}{r}\\\\\omega = \sqrt{\frac{g}{r}}\\

where,

ω = angular velocity of cylinder = ?

g = required acceleration = 9.8 m/s²

r = radius of cylinder = diameter/2 = 5.9 mi/2 = 2.95 mi = 4023.36 m

Therefore,

\omega = \sqrt{\frac{9.8\ m/s^{2}}{4023.36\ m}}\\\\

<u>ω = 0.05 rad/s</u>

7 0
3 years ago
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omeli [17]

The acceleration of the ball after leaving the hand is 9.8 m/s^2 downward

Explanation:

In order to find the acceleration of the ball during its motion, we have to study which forces are acting on it.

After the ball leaves the hand, if we neglect air resistance, there is only one force acting on the ball: the force of gravity, whose magnitude is

F=mg

where m is the mass of the ball and g is the acceleration of gravity (g=9.8 m/s^2), acting in the downward direction.

According to Newton's second law, the acceleration of the ball is given by

a=\frac{\sum F}{m}

where

\sum F is the net force acting on the ball

After the ball leaves the hand, the only force acting on it is the force of gravity, so we can substitute (mg) into the previous equation:

a=\frac{mg}{m}=g=9.8 m/s^2

This means that the acceleration of the ball remains 9.8 m/s^2 downward for the entire motion, after leaving the hand.

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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kifflom [539]
The Answer Is A 
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Answer:

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Explanation:

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