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Ray Of Light [21]
3 years ago
12

An object floats in water with 5 8 of its volume submerged. The ratio of the density of the object to that of water is

Physics
1 answer:
ki77a [65]3 years ago
3 0

Answer:

\dfrac{5}{8}

Explanation:

m = Mass of object = \rho v

m' = Mass of water = \rho' v'

\rho = Density of object

\rho' = Density of water

Weight of the water displaced is the force in the case of floating objects

According to the question

v'=\dfrac{5}{8}v

In the case of floating objects

W=W'\\\Rightarrow mg=m'g\\\Rightarrow \rho vg=\rho'v'g\\\Rightarrow \rho v=\rho' \dfrac{5}{8}vg\\\Rightarrow \rho=\rho' \dfrac{5}{8}\\\Rightarrow \dfrac{\rho}{\rho'}=\dfrac{5}{8}

The ratio of the density of the object to that of water is \dfrac{5}{8}

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goldfiish [28.3K]

Answer:

b

Explanation:

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3 years ago
A laser is placed at the point $(3,5)$. The laser beam travels in a straight line. Larry wants the beam to hit and bounce off th
Black_prince [1.1K]

Answer:

4units

Explanation:

To calculate the total distance the beam will travel along this path, you will use the formula for calculating the distance between two coordinates expressed as;

D = √(x2-x1)²+(y2-y1)²

Given the coordinate points

(3,5) and (7,5)

Substitute

D = √(7-3)²+(5-5)²

D = √(7-3)²+0²

D = √4²

D = √16

D = 4

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3 years ago
Two identical rifles are shot at the same time, and the sound intensity level is 64.7 dB. What would be the sound intensity leve
mixas84 [53]

Answer:

The sound intensity is still the same. The sound intensity level does not depend on the number of sound sources but only the wave properties of the wave medium and the properties of the sound source. The wave properties of the medium includes the density and also the temperature.

Explanation:

5 0
3 years ago
Read 2 more answers
Both igneous and sedimentary rock can become metamorphic rock if enough ________________ are applied.
atroni [7]
Enough heat and pressure
I hope this helps.
8 0
2 years ago
Standing at a crosswalk, you hear a frequency of 530 Hz from the siren of an approaching ambulance. After the ambulance passes,
malfutka [58]

Answer:

_s = 37.77 m / s

Explanation:

This is an exercise of the Doppler effect that the change in the frequency of the sound due to the relative speed of the source and the observer, in this case the observer is still and the source is the one that moves closer to the observer, for which relation that describes the process is

                    f ’= f₀  \frac{v}{v - v_s}

where d ’= 530 Make

when the ambulance passes away from the observer the relationship is

                    f ’’ = f₀ \frac{v}{v + v_s}

where d ’’ = 424 beam

let's write the two expressions

               f ’ (v-v_s) = fo v

               f ’’  (v + v_s) = fo v

let's solve the system, subtract the two equations

                v (f ’- f’ ’) - v_s (f’ + f ’’) = 0

                v_s = v \frac{ f' - f''}{ f' + f''}

the speed of sound is v = 340 m / s

let's calculate

                 v_s = 340 (\frac{ 530 -424}{530+424} )

                 v_s = 340 (\frac{106}{954})

                  v_s = 37.77 m / s

5 0
2 years ago
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