The required lift force is approximately 866.92 N. To determine this, we first establish the shark's mass at 92 kg and its density at 1040 kg/m³. The volume of the shark is calculated by dividing mass by density, yielding 0.08846 m³. The buoyant force acting on the shark is then determined by multiplying the volume by the density of water and gravity, resulting in a lift force of 866.92 N.
<span>Response:
Chlorine has 17 electrons, thus, for 1+ and 2+ ions, we require elements with 18 and 19 electrons, which are argon and potassium: Ar+ and K 2+.
For 1- and 2- ions, we need elements with 16 and 15 electrons, namely sulfur and phosphorus, represented as S- and P 2-.
It’s important to note that + ions indicate electron loss, while - ions reflect electron gain.</span>
Answer:
Electric flux is calculated as 
Explanation:
We start with the given parameters:
The electric field impacting the circular surface is 
Our objective is to ascertain the electric flux passing through a circular region with a radius of 1.83 m situated in the xy-plane. The area vector is oriented in the z direction. The formula for electric flux is expressed as:
Applying properties of the dot product, we calculate the electric flux as:
Consequently, the electric flux for the circular area is . Thus, this represents the required answer.
Answer:
A rock weighing 50kg should be positioned at a distance of 0.5m from the pivot of the seesaw.
Explanation:
τchild=τrock
We will utilize the formula for torque:
(F)child(d)child)=(F)rock(d)rock)
The gravitational force acts equally on both objects.
(m)childg(d)child)=(m)rockg(d)rock)
We can eliminate gravity from both sides of the equation for simplification.
(m)child(d)child)=(m)rock(d)rock)
Now employing the given masses for the rock and child. The seesaw's total length is 2 meters, with the child sitting at one end, placing them 1 meter from the center of the seesaw.
(25kg)(1m)=(50kg)drock
Solve for the distance where the rock should be positioned in relation to the seesaw's center.
drock=25kg⋅m50kg
drock=0.5m
