The pressure of a submerged object in a fluid is given by:
where ρ is the density of the fluis, g is the acceleration of gravity, h is the depth of the object and Patm is the pressure of the atmosphere. In this case we know that:
• The density of water is 1000 kg/m^3
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• The acceleration of gravity is 9.8 m/s^2
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• The depth of the object is 3 km, that is, 3000 m.
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• The atmospheric pressure is 101325 pascals.
Plugging these values in the equation given above we have:
Therefore, the pressure at this depth is 2.95x10^7 Pa.
Answer:
Explanation:
Consider two point charges located on the x axis: one charge, q1 = -19.5 nC , is located at x1 = -1.670 m ; the second charge, q2 = 34.0 nC , is at the origin (x=0).
Given that:
q1 = -19.5 nC , x1 = -1.670 m, q2 = 34.0 nC , x2=0, q3 = 54.5 nC, x3 = -1.145 m
From Coloumbs law, the force between two charges is given by the formula:
Hence the force of attraction of q1 on q3 is given by:
Hence the force of attraction of q2 on q3 is given by:
The force of attraction of q1 on q3 is in the negative direction and force of attraction of q2 on q3 is in the negative direction. Hence the net force is:
Answer:
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Explanation:
Answer:
Explanation:
Work is the product of force and distance, so the formula is:
The work is 6.00 Joules, but
- 1 Joule (J) is equal to 1 Newton meter (N*m).
- Therefore, the work of 6.00 J is equal to 6.00 N*m
The force is 2.50 Newtons.
The known values are:
Substitute the values into the formula.
We are solving for distance, so we must isolate the variable, d. It is being multiplied by 2.50 Newtons and the inverse of multiplication is division. Divide both sides by 2.50 N.
The units of Newtons cancel out.
The force was applied to the rock over a distance of <u>2.40 meters</u> and choice <u>C</u> is correct.
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