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Sever21 [200]
3 years ago
11

Archimedes discovered relationships between _____. pressure and volume of gases temperature and volume of gases density and floa

ting pressure and surface area
Physics
2 answers:
Naily [24]3 years ago
5 0

Archimedes discovered relationships between density and floating. He discovered that the mass of an object will have density to the volume being displaced.

Slav-nsk [51]3 years ago
3 0

Answer: The correct answer is density and floating.

Explanation:

Buoyant forces is forces exerted by the fluid on the object immersed in it.

Archimedes discovered the relationship between the density and floating.

Archimedes principle states as: the buoyant force on the object is equal to weight of the fluid is displaces.

F_B=w_f

F_B = Buoyant force on the object

w_f = Weight of fluid displaced by the object

Density is the measure of mass present in per unit volume of the object.

  • If the density of the object is greater than the density of the fluid in which it is immersed then it will get sink. This because more the mass more weight of the fluid it will displace
  • If the density of the object is less than the density of the fluid in which it is immersed then it will float. This is because lesser the mass lesser the weight of fluid it will displace.

Hence, the correct answer is density and floating.


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3 years ago
A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it
Savatey [412]

The charge of the object must be 1.11 \times e^{-5} \text { coulomb }

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

         Electric field, E=\frac{\text { Force }(F)}{q}

Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

By substituting the given values, we get

      q=\frac{F}{E}=\frac{0.05 N}{4500 \mathrm{N} / \mathrm{c}}=1.11 \times e^{-5} \text { coulomb }

6 0
3 years ago
A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, wi
grin007 [14]

Explanation:

It is given that,

Magnitude of charge, q=15\ \mu C=15\times 10^{-6}\ C

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field, B=0.08\ j

Velocity, v=(5\ cos25)i+(5\ sin25)j

v=[(4.53)i+(2.11)j]\ m/s

We need to find the magnitude of force on the charge. Magnetic force is given by :

F=q(v\times B)

F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]

<em>Since</em>, i\times j=k\ and\ j\times j=0

F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]

F=0.00000543\ kN

F=5.43\times 10^{-6}\ kN

So, the force acting on the charge is 5.43\times 10^{-6}\ kN and is moving in positive z axis. Hence, this is the required solution.

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