True or FalseIf an object is submerged in water, on earth, at a depth of 3km, the Pressure on the object is2.94 x 10^7 Pa
1 answer:
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
,• The acceleration of gravity is 9.8 m/s^2
,• The depth of the object is 3 km, that is, 3000 m.
,• 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.
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Answer:
T = 480.2N
Explanation:
In order to find the required force, you take into account that the sum of forces must be equal to zero if the object has a constant speed.
The forces on the boxes are:
(1)
T: tension of the rope
M: mass of the boxes 0= 49kg
g: gravitational acceleration = 9.8m/s^2
The pulley is frictionless, then, you can assume that the tension of the rope T, is equal to the force that the woman makes.
By using the equation (1) you obtain:
The woman needs to pull the rope at 480.2N
Complete question:
Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.
Answer:
The magnitude of this field is 826 N/C
Explanation:
Given;
The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m
PEsinθ = T
where;
E is the magnitude of the electric field
P is the dipole moment
First, we determine the magnitude dipole moment;
Magnitude of dipole moment = q*r
P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m
Finally, we determine the magnitude of this field;
E = 826 N/C (in three significant figures)
Therefore, the magnitude of this field is 826 N/C
First we gotta use an equation of motion:
Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!
Now we just need to solve for t:
Hit the calculators, and you'll get 4.5 seconds!
Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!
Now we just need to solve for t:
Hit the calculators, and you'll get 4.5 seconds!