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Katarina
17 days ago
9

A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's int

erior, which statement is correct?
A.)There is no electric field on the interior of the conducting sphere.
B.)The interior field points in a direction parallel to the exterior field.
C.)The interior field points in a direction opposite to the exterior field.
D.)The interior field points in a direction perpendicular to the exterior field.
Physics
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A model of a spring/mass system is 4x'' + e−0.1tx = 0. By inspection of the differential equation only, discuss the behavior of
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Displacement stabilizes over time. It is known that exponentials raised to infinity approach zero, hence the system model will yield as time approaches infinity, resulting in 4x'' + e−0.1tx = 0. As time approaches infinity, we deduce that 4x'' equals zero. Consequently, upon integrating, we derive 4x' = c, and further integration leads to the conclusion 4x = cx + d.
4 0
2 months ago
You're driving your new sports car at 85 mph over the top of a hill that has a radius of curvature of 525 m.
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Given: Speed of the sports car, v = 85 mph = 37.99 m/s. Radius of curvature, r = 525 m. Let normal weight be denoted as n and apparent weight as a. The apparent weight can be described by:... or... Consequently, this provides the necessary solution.
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2 months ago
A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of
Sav [3153]

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

6 0
3 months ago
A student solving a physics problem for the range of a projectile has obtained the expression r= v20sin(2θ)g where v0=37.2meter/
ValentinkaMS [3465]

The formula for range is:

R = \frac{v_o^2 sin2\theta}{g}

Given values are:

v_0=37.2m/s

where θ equals 14.1 degrees

g=9.80m/s^2

Using the equation above,

R = \frac{37.2^2 sin2*14.1}{9.80}

The calculated range is 66.7 meters.

Therefore, the range is approximately 66.1 meters.

5 0
4 months ago
Read 2 more answers
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Sav [3153]

Complete Question

An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.

At time 5 seconds, I = 1.2 A.

We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

The question indicates that

    The wire’s diameter is  d = 0.205cm = 0.00205 \ m

     The radius of the wire is  r = \frac{0.00205}{2} = 0.001025 \ m

     Aluminum's resistivity is 2.75*10^{-8} \ ohm-meters.

       The electric field variation is described as

         E (t) = 0.0004t^2 - 0.0001 +0.0004

     

The charge is effectively given by the equation

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area expressed as

       A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

 Thus,

       \frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

By substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.

The question states that t =  5 seconds

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

5 0
4 months ago
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