An astronaut stands by the rim of a crater on the Moon, where the acceleration of gravity is 1.62 m/s2 and there is no air. To d

Look at the image for the solution

Answer:

will be less than and will be greater than .

Explanation:

The law of conservation of mass states that the rate of fluid mass () entering a system equals the rate at which the fluid mass () exits the system.

The mass flow rate can be expressed as follows:

where denotes the fluid density, signifies the cross-sectional area through which fluid flows, and represents the fluid's velocity.

Based on the problem conditions, as the fluid's density remains constant, we can write:

where and are the cross-sectional areas for the fluid flow, while and are the corresponding velocities across those areas.

Given the conditions in the problem, , leading from the formula to .

Furthermore, fluid pressure arises from the fluid's movement through any specific area. When the fluid accelerates, part of its energy increases its speed in the direction of flow, resulting in lower pressure.

Thus, in this instance, the pressure in the larger cross-sectional area will exceed the pressure in the smaller cross-sectional area, implying

.

E_total = 5.8 x 10⁴ N/C Explanation: To determine the electric field at specified points, we must calculate the vectors individually for each charge and sum them. The electric field caused by each charged conductive sheet can be derived via Gauss's law with the understanding of scalar products between the electric field and relevant surfaces.

According to Newton's second law, provided F is directed horizontally,

• the net horizontal force acting on the larger block is

F - µmg = 3mA

where µmg represents the friction experienced by the larger block from contact with the smaller block, µ is the static friction coefficient between both blocks, and A indicates the acceleration of the block;

• the net vertical force on the larger block is

4mg - 3mg - mg = 0

where 4mg denotes the magnitude of the normal force exerted by the surface on the combined mass of both blocks, 3mg corresponds to the weight of the larger block, and mg indicates the weight of the smaller block;

• the net horizontal force acting on the smaller block is

µmg = ma

where µmg again signifies the friction between both blocks; however, it is important to note that this force aligns in the same direction as F. It is the sole force influencing the smaller block in the horizontal direction, thus (b) static friction causes the smaller block's acceleration;

• the net vertical force on the smaller block is

mg - mg = 0

where mg represents the force of both the normal force from the larger block pushing up against the smaller one, and the weight of the smaller block.

(You should be able to create your own free-body diagrams based on the forces discussed.)

(c) Solve the equations stated above to find A and a:

A = (F - µmg) / (3m)

a = µg

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