Why does the closed top of a convertible bulge when the car is riding along a highway? Why does the closed top of a convertible

Answer:

a) Q = C*emf

b)  Decrease in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are linked in series with the battery

According to Kirchoff's voltage law, the total voltage in the circuit must equal zero

Let represent the Voltage across the Resistor

and

represent the Voltage across the capacitor

Implementing KVL;

.........................(1)

Since this is a series connection, the same current traverses through the circuit

Integrating and into equation (1)

Initially, as the capacitor reaches full charge, the current will drop to zero because of equilibrium

b) When the plastic sheet is inserted between the plates, current begins to flow because both the electric field intensity and electric potential decrease. As a result, charge diminishes, leading to current flow

c) The current through the resistor equates to the total current within the circuit (given the series connection)

Substituting the values of t and I₀ into the aforementioned formula for I

d) Note: The initial charge on the capacitor equals C * emf

Following the insertion of the plastic, the new charge will be:

The derived frequency equals 2.63 Hz. Explanation: For an object weighing 8.0 kg with a spring stretching 3.6 cm, calculations involving the spring constant and oscillation frequency lead to this specific oscillation rate.

Answer:

        h = 12.8 cm

Explanation:

The initial parameters are as follows:

distance = 6.4 cm

• when the object descends, its weight matches the spring's force

        weight = spring force

         mg = ky... equation 1

• potential energy stored in a stretched spring = work done by the spring

        mgh = 0.5 x k x h^{2}....equation 2

• Substituting from equation 1 into equation 2

                kyh =  0.5 x k x h^{2}

                y =  0.5 x h

                2y = h

• where y is 6.4, yielding the maximum elongation as

          h = 2 x 6.4 = 12.8 cm

Utilizing the equation F = ma, where F represents the force applied by the machine, A denotes acceleration (equivalent to v/t, with v as velocity and t as time), and M symbolizes mass, we can calculate as follows: F = mv/t. Thus, F = (0.15kg) (30 – 0 m/s) / 0.5 s, resulting in F = 9 N.

Answer:

F=126339.5N

Explanation:

To compute the force required to escape, a free-body diagram for the hatch must be drawn. We will equate the downward and upward forces, thus applying the following equation:

Fw=W+Fi+F

where

Fw=   force or weight exerted by the water column above the submarine.

To calculate Fw, we can use:

Fw=h. γ. A

h=height

γ= specific weight of seawater = 10074N / m ^ 3

A=Area

Fw=28x10074x0.7=197467N

w represents the hatch weight = 200N

Fi denotes the internal pressure force in the submarine, which is 1 atm = 101325Pa. We can calculate this force using:

Fi=PA=101325x0.7=70927.5N

Finally, the force needed to open the hatch is determined by the original equation:

Fw=W+Fi+F

F=Fw-W+Fi

F=197467N-200N-70927.5N

F=126339.5N