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Basile [38]
3 years ago
12

An object is placed in a fluid and then released. Assume that the object either floats to the surface (settling so that the obje

ct is partly above and partly below the fluid surface) or sinks to the bottom. (Note that, for Parts A through D, you should assume that the object has settled in equilibrium.)
A.Consider the following statement: The magnitude of the buoyancy force is equal to the weight of fluid displaced by the object.
Under what circumstances is this statement true?

A. always
B. only for an object that floats
C. only for an object that sinks
D. never
Physics
2 answers:
Irina18 [472]3 years ago
7 0

Answer:

A. Always true

Explanation:

This is because, the buoyancy force is always present whenever and object is placed in a fluid. The magnitude of this  buoyancy force is always equal to the weight of the fluid    displaced by the object according to Archimedes' principle. This principle is true irrespective of whether the object floats or not. When any object is inserted in a fluid, the buoyancy force is always present irrespective of whether it floats or not.  

sp2606 [1]3 years ago
3 0

Answer:

The correct option is;

A. always

Explanation:

Archimedes' principle states that when an object is immersed in a fluid, and it is either partially or wholly submerged, the upward force acting on the object, that is the buoyancy force, is equal to the weight of the fluid displaced by the object.

That is the object immersed in the fluid is observed to have an apparent reduction in weight that is equal to the weight of the volume of the fluid displaced by the  immersed  object.

Therefore, the above principle always occurs.

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Two go carts, A and B race eachother around a 1.0 km track. Go cart A travels at a constant speed of 20.0 m/s. Go cart B acceler
jarptica [38.1K]

We have that Since Cart A spends t=77.5secs and Cart B spends t=50sec

Therefore

The Cart A wins the Race and by 25.5( 77.7-50)secs

From the question we are told

  • Two go carts, A and B race each other around a 1.0 km track.
  • Go cart A travels at a constant speed of 20.0 m/s.
  • Go cart B accelerates uniformly from rest at a rate of 0.333 m/s

For Cart A

Generally the equation for the Velocity  is mathematically given as

v=\frac{d}{t}\\\\t=\frac{d}{v}\\\\t=\frac{1000}{20}\\\\t=50sec

For Cart B

Generally the Newtons equation for the Motion  is mathematically given as

S=ut+1/2at^2\\\\Therefore\\\\S=ut+1/2at^2\\\\1000=0+1/2*(0.33)t^2\\\\t=\frac{1000}{1/2*(0.33}}\\\\t=77.5secs

Since

Cart A spends t=77.5secs and Cart B spends t=50sec

Therefore

The Cart A wins the Race and by 25.5( 77.7-50)secs

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5 0
3 years ago
Write the nuclear reaction equation for the beta decay of Iodine-131
den301095 [7]

Answer:

_{53}^{131}I \rightarrow _{54}^{131}Xe + e + \bar{\nu}

Explanation:

In a beta (minus) decay, a neutron in a nucleus turns into a proton, emitting a fast-moving electron (called beta particle) alongside with an antineutrino.

The general equation for a beta decay is:

^A_Z X \rightarrow _{Z+1}^AY+^0_{-1}e+ ^0_0\bar{\nu} (1)

where

X is the original nucleus

Y is the daughter nucleus

e is the electron

\bar{\nu} is the antineutrino

We observe that:

  • The mass number (A), which is the sum of protons and neutrons in the nucleus, remains the same in the decay
  • The atomic number (Z), which is the number of protons in the nucleus, increases by 1 unit

In this problem, the original nucles that we are considering is iodine-131, which is

_{53}^{131}I

where

Z = 53 (atomic number of iodine)

A = 131 (mass number)

Using the rule for the general equation (1), the dauther nucleus must have same mass number (131) and atomic number increased by 1 (54, which corresponds to Xenon, Xe), therefore the equation will be:

_{53}^{131}I \rightarrow _{54}^{131}Xe + e + \bar{\nu}

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3 years ago
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Answer:

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