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Bas_tet [7]
3 years ago
11

You make the following measurements of an object 42kg and 22m3 what would the objects density be

Physics
2 answers:
miv72 [106K]3 years ago
4 0
The density of the object is approximately 1.91 kg per m³.

42 kg is a measure of mass, and 22 m³ is a measure of volume. Knowing this, you can use the relationship $$density = mass / volume$$ to solve for the object's density.

42 kg \div 22 m³ \approx 1.91 kg per m³.
sammy [17]3 years ago
4 0
<h2>Answer: </h2>

Density = 1.909 kg/m3

<h3>Explanation: </h3>

Given data:

Mass of the object = 42 kg

Volume of the object = 22 m3

Density of the object = ?

solution

Density = mass/volume

Density =42/22 = 1.909

Density = 1.909 kg/m3

Hence the density of the object is 1.909 and its unit is kg/m3 which is system international unit of density.


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Consider a rectangular ice floe 5.00 m high, 4.00 m long, and 3.00 m wide. a) What percentage of the ice floe is below the water
artcher [175]

Answer:

(a) 92 %

(b) 6.76 %

Explanation:

length, l = 4 m, height, h = 5 m, width, w = 3 m, density of water = 1000 kg/m^3

density of ice = 920 kg/m^3, density of mercury = 13600 kg/m^3

(a) Let v be the volume of ice below water surface.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of water x g = Total volume of ice block x density

                                                                      of ice x g

v x 1000 x g = V x 920 x g

v / V = 0.92

% of volume immersed in water = v/V x 100 = 0.92 x 100 = 92 %

(b) Let v be the volume of ice below the mercury.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of mercury x g = Total volume of ice block x  

                                                                      density of ice x g

v x 13600 x g = V x 920 x g

v / V = 0.0676

% of volume immersed in water = v/V x 100 = 0.0676 x 100 = 6.76 %

4 0
3 years ago
Let w(x)=3x-7.If w(x)=14, find x
dlinn [17]

Answer:

7

Explanation:

We are given:

    w(x) = 3x   -   7

     w(x)  = 14

The problem here entails us to solve for x;

To solve for x; equate the two expressions:

         

          3x  - 7  = 14

           3x  = 14 + 7

           3x  = 21  

             x = 7

So the value of x  = 7

6 0
3 years ago
What observations can the geologist make by working outdoors instead of in a lab?
vredina [299]

Answer:

Geology is the study of the Earth that involves the process at Earth, materials of which it is made, and its history.

<u>Geologists combine both laboratory and field data to illustrate the results of their research. Some observations that can the geologist make by working outdoors instead of in a lab are as follows:</u>

  • Understanding and exploring the earth's surface closely using geophysical tools.
  • Collecting samples by own and make some interpretations at the same time.
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8 0
3 years ago
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
aivan3 [116]

Answer:

Explanation:

Given that,

Initial angular velocity is 0

ωo=0rad/s

It has angular velocity of 11rev/sec

ωi=11rev/sec

1rev=2πrad

Then, wi=11rev/sec ×2πrad

wi=22πrad/sec

And after 30 revolution

θ=30revolution

θ=30×2πrad

θ=60πrad

Final angular velocity is

ωf=18rev/sec

ωf=18×2πrad/sec

ωf=36πrad/sec

a. Angular acceleration(α)

Then, angular acceleration is given as

wf²=wi²+2αθ

(36π)²=(22π)²+2α×60π

(36π)²-(22π)²=120πα

Then, 120πα = 8014.119

α=8014.119/120π

α=21.26 rad/s²

Let. convert to revolution /sec²

α=21.26/2π

α=3.38rev/sec

b. Time Taken to complete 30revolution

θ=60πrad

∆θ= ½(wf+wi)•t

60π=½(36π+22π)t

60π×2=58πt

Then, t=120π/58π

t=2.07seconds

c. Time to reach 11rev/sec

wf=wo+αt

22π=0+21.26t

22π=21.26t

Then, t=22π/21.26

t=3.251seconds

d. Number of revolution to get to 11rev/s

∆θ= ½(wf+wo)•t

∆θ= ½(0+11)•3.251

∆θ= ½(11)•3.251

∆θ= 17.88rev.

5 0
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Maurinko [17]

Answer:

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Explanation:

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