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Zanzabum
4 years ago
10

How did Bourne believe that an object could be made to rise or sink at will?

Physics
2 answers:
Readme [11.4K]4 years ago
7 0

Explanation:

When an object is placed inside water, an upward force is exerted by water called as buoyant force.

An object will float or sink, this completely depends on the density of object. Also, if the magnitude of gravitational force is more as compared to the buoyant force the object will sink otherwise float.

If the density of substance is more than water, it will sink and if the density of substance is less than water it will float.

Angelina_Jolie [31]4 years ago
4 0
Bourne believed that an object would float or sink at will as long as he could <span>manipulate the effect's of buoyancy which control and object to sink or float. Hope this helps!

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You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 28.4 NN . If you t
shtirl [24]

Answer:

2491.23 kg/m³

Explanation:

From Archimedes principle,

R.d = weight of object in air/ upthrust in water = density of the object/density of water

⇒ W/U = D/D' ....................... Equation 1

Where W = weight of the ceramic statue, U = upthrust of the ceramic statue in water, D = density of the ceramic statue, D' = density of water.

Making D the subject of the equation,

D = D'(W/U).................... Equation 2

Given: W = 28.4 N, U = lost in weight = weight in air- weight in water

U = 28.4 - 17.0 = 11.4 N,

Constant: D' = 1000 kg/m³.

Substitute into equation 2,

D = 100(28.4/11.4)

D = 2491.23 kg/m³

Hence the density of the ceramic statue = 2491.23 kg/m³

7 0
3 years ago
Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



7 0
4 years ago
What is the distance between two spheres (20 kg and 30 kg) attracted by a force of 2 x 10^-5 Newtons.
creativ13 [48]
2 times 10 to the power of 5 is your answer.
6 0
4 years ago
Multiply 5.036×102m by 0.078×10−1, taking into account significant figures.
Anika [276]

Answer : The significant digit is 6

Explanation :

Multiply 5.036\times10^{2}m by 0.078\times10^{-1}m

Now, on multiplying

5.036\times10^{2}m \times0.078\times10^{-1}m = 0.392808 \times10^{1}\ m^{2}

5.036\times10^{2}m \times0.078\times10^{-1}m = 0.0392808\ m^{2}

Now, the significant digit is 6.

Hence, this is the required solution.

3 0
3 years ago
What's the value of 1,152 Btu in joules? A. 1,964,445 J B. 2,485,664 J C. 987,875 J D. 1,215,360 J
OLga [1]

Answer: Option (d) is correct.

Explanation:

Given,  1,152 British thermal units

1 British thermal unit = 1055.06 joules

So, in 1,152 British thermal units there will be :

1152\times 1055.06 Joules=1215429.12 Joules=1.21542912\times 10^{8} Joules

Hence, from the given options the closest answer is of option (d). So, option (d) is correct.

4 0
4 years ago
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