Answer:

Answer:

Check attachment for better understanding

Explanation:

Given that,

Current in wire I =2.2A

Capacitor plate dimension is 2cm by 2cm

s=2cm=2/100 = 0.02m

Rate at which electric field Is changing dE/dt?

The current in the wires must also be the displacement current in the capacitor. We find the rate at which the electric field is changing from

ID = ε0•A•dE/dt

Where ε0 is a constant

ε0= 8.85×10^-12C²/Nm²

Area of the square plate is

A =s² =0.02² = 0.0004m²

Then,

Make dE/dt the subject of formula

dE/dt = ID/ε0A

dE/dt = 2.2 / (8.85×10^-12 ×4×10^-4)

dE/dt = 6.215×10^14 V/m-s

Or

dE/dt = 6.215×10^14 N/C.s

The rate at which the electric field is changing between the plates is 6.215×10^14 N/C.s

Your friend is trying to construct a clock for a craft show and asks you for some advice. She has decided to construct the clock with a pendulum. The pendulum will be a very thin, very light wooden bar with a thin, but heavy, brass ring fastened to one end. The length of the rod is 80 cm and the diameter of the ring is 10 cm. She is planning to drill a hole in the bar to place the axis of rotation 15 cm from one end. She wants you to tell her the period of this pendulum.

An electromagnetic wave is traveling the +y direction. The maximum magnitude of the electric field associated with the wave is Em, and the maximum magnitude of the magnetic field associated with the wave is Bm. At one instant, the electric field has a magnitude of 0.25 Em and points in the +x direction. At this same instant, the wave's magnetic field has a magnitude which is ... and it is in the ... direction.

A 750-kg automobile is moving at 26.2 m/s at a height of 5.00 m above the bottom of a hill when it runs out of gasoline. The car coasts down the hill and then continues coasting up the other side until it comes to rest. Ignoring frictional forces and air resistance, what is the value of h, the highest position the car reaches above the bottom of the hill?

Instantaneous speed is...a) A speed of 1000 km/hb) The speed attained at a particular instant in time.c) The speed that can be reached in a particular amount of time.PLEASE HURRY

A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that when placed in the field the sphere is in a static situation (all the forces on the sphere cancel). If the thread is horizontal, find the magnitude and direction of the electric field. The sphere has a mass of 0.018 kg and contains a charge of + 6.80 x 103 C. The tension in the thread is 6.57 x 10-2 N. Show your work and/or explain your reasoning. (20 pts)

An electromagnetic wave is traveling the +y direction. The maximum magnitude of the electric field associated with the wave is Em, and the maximum magnitude of the magnetic field associated with the wave is Bm. At one instant, the electric field has a magnitude of 0.25 Em and points in the +x direction. At this same instant, the wave's magnetic field has a magnitude which is ... and it is in the ... direction.

A 750-kg automobile is moving at 26.2 m/s at a height of 5.00 m above the bottom of a hill when it runs out of gasoline. The car coasts down the hill and then continues coasting up the other side until it comes to rest. Ignoring frictional forces and air resistance, what is the value of h, the highest position the car reaches above the bottom of the hill?

Instantaneous speed is...a) A speed of 1000 km/hb) The speed attained at a particular instant in time.c) The speed that can be reached in a particular amount of time.PLEASE HURRY

A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that when placed in the field the sphere is in a static situation (all the forces on the sphere cancel). If the thread is horizontal, find the magnitude and direction of the electric field. The sphere has a mass of 0.018 kg and contains a charge of + 6.80 x 103 C. The tension in the thread is 6.57 x 10-2 N. Show your work and/or explain your reasoning. (20 pts)

**Answer:**

time is 32 s and speed is 304.3 m/s

**Explanation:**

Height, h = 146 m

speed, u = 14 m/s

Angle, A = 43 degree

Let it hits the ground after time t.

Use second equation of motion

Time cannot be negative so the time is t = 32 s .

The vertical velocity at the time of strike is

v' = u sin A - g t

v' = 14 sin 43 - 9.8 x 32 = 9.5 - 313.6 = - 304.1 m/s

horizontal velocity

v'' = 14 cos 43 =10.3 m/s

The resultant velocity at the time of strike is

**Answer:**

The work done in stretching the spring is 0.875 J.

**Explanation:**

**Given that,**

Force = 140 N

Natural length = 60-40 = 20 cm

Stretch length of the spring = 65-60 = 5 cm

**We need to calculate the spring constant**

**Using formula of Hooke's law**

**We need to calculate the work done**

On integration

**Hence, The work done in stretching the spring is 0.875 J.**

**Answer:**

**Explanation:**

Let the charge on bead A be q nC and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} =

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

**Answer; 10.6 i think**

**Explanation:**

(a) At the top of the hill, the coaster has total energy (potential and kinetic)

*E* = (1000 kg) *g* (10 m) + 1/2 (1000 kg) (6 m/s)² = 116,000 J

As it reaches its lowest position, its potential energy is converted to kinetic energy, and some is lost to friction, making its speed *v* such that

1/2 (1000 kg) *v* ² = 116,000 J - 1700 J = 114,300 J

===> *v* ≈ 15.2 m/s

If no energy is lost to friction as the coaster makes its way up the second hill, all of its kinetic energy would be converted to potential energy at the maximum possible height *H*.

1/2 (1000 kg) (15.2 m/s)² = (1000 kg) *g**H*

===> *H* ≈ **11.7 m**

(b) At the top of the second hill with minimum height *h*, and with maximum speed 4.6 m/s, the coaster has energy

*E* = *P* + *K* = (1000 kg) *g**h* + 1/2 (1000 kg) (4.6 m/s)²

Assuming friction isn't a factor again, the energy here should match the energy at the lowest point in part (a), 114,300 J.

(1000 kg) *g h* + 1/2 (1000 kg) (4.6 m/s)² = 114,300 J

===> *h* ≈ **10.6 m**

**Answer:**

Gravity

**Explanation:**

Gravity is constantly pulling objects downward. Without it, everything would float out into space.

I hope this answer helps :)

**Answer:**

The answer for the given question above would be option C. GRAVITATIONAL FORCE. Based on the given scenario above of a leaf that falls to the ground when Tonya let it go, the force that pulled the leaf to the ground is the gravitational force. This kind of force is a force that attracts any object with mass.

Hope this helps!!!

**Complete question:**

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:

a) the midpoint between the two rings?

b) the center of the left ring?

**Answer:**

a) the electric field strength at the midpoint between the two rings is **0**

b) the electric field strength at the center of the left ring is **2712.44 N/C**

**Explanation:**

Given;

distance between the two rings, d = 25 cm = 0.25 m

diameter of each ring, d = 10 cm = 0.1 m

radius of each ring, r =

the charge on each ring, q = 20 nC

Electric field strength for a ring with radius r and distance x from the center of the ring is given as;

The electric field strength at the midpoint;

the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m

The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;

The electric field strength at the midpoint;

(b)

The distance from the right ring to center of the left ring, x = 0.25 m.