let the mass of Venus is M then mass of Saturn is 100 M
similarly if the radius of Venus is R then the radius of Saturn is 10 R
now the force of gravity on a man of mass "m" at the surface of Venus is given by

now similarly the gravitational force on the man if he is at the surface of Saturn


so here if we divide the two forces

so here we can say
F1 = F2
so on both planets the gravitational force will be same
Pascal's law of fluid transfer states that when there is an increase in fluid pressure, the rest of the extrinsic variables also increases. For example, in a flow of liquid in an orifice, there is a contraction of diameter in the orifice part. The fluid that will go in there increases in pressure and thereby an increase in velocity as well.
Answer:
A+B; 5√5 units, 341.57°
A-B; 5√5 units, 198.43°
B-A; 5√5 units, 18.43°
Explanation:
Given A = 5 units
By vector notation and the axis of A, it is represented as -5j
B = 3 × 5 = 15 units
Using the vector notations and the axis, B is +15i. The following vectors ate taking as the coordinates of A and B
(a) A + B = -5j + 15i
A+B = 15i -5j
|A+B| = √(15)²+(5)²
= 5√5 units
∆ = arctan(5/15) = 18.43°
The angle ∆ is generally used in the diagrams
∆= 18.43°
The direction of A+B is 341.57° based in the condition given (see attachment for diagrams
(b) A - B = -5j -15i
A-B = -15i -5j
|A-B|= √(15)²+(-5)²
|A-B| = √125
|A-B| = 5√5 units
The direction is 180+18.43°= 198.43°
See attachment for diagrams
(c) B-A = 15i -( -5j) = 15i + 5j
|B-A| = 5√5 units
The direction is 18.43°
See attachment for diagram
Answer:

Explanation:
Hi!
The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.
The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

Then the electric field at the point of interest is estimated as:
