B. Between points A and C

C. Between points B and E

D. Between points B and C

Answer:

**Answer:**

A

**Explanation:**

voltage between A and C is equal battery's voltage.

Look at the figure below and calculate the length of side y.A. 8.5B. 1712yC. 6O D. 1245Х

A 1 225.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 700.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east.(a) What is the velocity of the truck right after the collision? (Give your answer to five significant figures.) m/s east (b) What is the change in mechanical energy of the cartruck system in the collision? J (c) Account for this change in mechanical energy.

A 580-mm long tungsten wire, with a 0.046-mm-diameter circular cross section, is wrapped around in the shape of a coil and used as a filament in an incandescent light bulb. When the light bulb is connected to a battery, a current of 0.526 A is measured through the filament. (Note: tungsten has a resistivity of 4.9 × 10-8 Ω • m.). How many electrons pass through this filament in 5 seconds?

A race car travels on a circular track at an average rate of 125 mi/h. The radius of the track is 0.320 miles. What is the centripetal acceleration of the car? 391 mi/h2 40.0 mi/h2 5,000 mi/h2 48,800 mi/h2

A charge Q = 1.96 10-8 C is surrounded by an equipotential surface with a surface area of 1.18 m2. what is the electric potential at this surface?

A 1 225.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 700.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east.(a) What is the velocity of the truck right after the collision? (Give your answer to five significant figures.) m/s east (b) What is the change in mechanical energy of the cartruck system in the collision? J (c) Account for this change in mechanical energy.

A 580-mm long tungsten wire, with a 0.046-mm-diameter circular cross section, is wrapped around in the shape of a coil and used as a filament in an incandescent light bulb. When the light bulb is connected to a battery, a current of 0.526 A is measured through the filament. (Note: tungsten has a resistivity of 4.9 × 10-8 Ω • m.). How many electrons pass through this filament in 5 seconds?

A race car travels on a circular track at an average rate of 125 mi/h. The radius of the track is 0.320 miles. What is the centripetal acceleration of the car? 391 mi/h2 40.0 mi/h2 5,000 mi/h2 48,800 mi/h2

A charge Q = 1.96 10-8 C is surrounded by an equipotential surface with a surface area of 1.18 m2. what is the electric potential at this surface?

This question involves the concepts of **derivative**, **apparent temperature**, **actual temperature**,and **wind speed**.

The drop in **apparent temperature **will be **"1.25°C"**.

The **apparent temperature (W) **is given in terms of **actual temperature (T)** and **wind speed (v)** is given by the following function:

Taking the **derivative **with respect to **actual temperature**, we get:

where,

dW = drop in **apparent temperatures** = ?

dT = drop in **actual temperature** = - 1°C

v = **wind speed **= 18 km/h

Therefore,

**dW = - 1.25°C**

Learn more about **derivatives** here:

**Answer:**

** Δw=1.25°C**

**Explanation:**

Given that

Given that T= 12°C and v=19 km/h

Now to find the drop in the apparent temperature w

So

Now by putting the values v=19 km/hr and ΔT=1

** Δw=1.25°C**

**So we can say that when temperature is decrease by 1°C then apparent temperature will decrease by 1.25°C at given velocity.**

**Answer:**

The current flows in the second wire is

**Explanation:**

**Given that,**

Upward current = 24 A

Force per unit length

Distance = 7.0 cm

**We need to calculate the current in second wire**

**Using formula of magnetic force**

Where,

=force per unit length

I₁= current in first wire

I₂=current in second wire

r = distance between the wires

Put the value into the formula

**Hence, The current flows in the second wire is **

**Answer:**

r₁ = 20.5 cm

**Explanation:**

In this exercise we can use the conservation of energy

the gravitational power energy is always attractive, the electrical power energy is repulsive if the charges are of the same sign

starting point.

Em₀ = U_g + U_e + K =

the two in the kinetic energy is because they are two particles

final point. When it is detained

Em_f = U_g + U_e =

the energy is conserved

Em₀ = em_f

the charges and masses of the two particles are equal

sustitute the values

-6.67-11 (4.5 10-3) ² / 0.25 - 9, 109 (30 10-9) ² / 0.25 + 4.5 10-3 4² = - 6.67 10- 11 (4.5 10-3) ² / r1 -9 109 (30 10-9) ² / r1

-5.4 10⁻¹⁵ + 3.24 10⁻⁵ - 7.2 10⁻⁵ = -1.35 10⁻¹⁵ / r₁ + 8.1 10⁻⁶ / r₁

We can see that the terms that correspond to the gravitational potential energy are much smaller than the terms of the electric power, which is why we depress them.

3.24 10⁻⁵ - 7.2 10⁻⁵ = 8.1 10⁻⁶ / r₁

-3.96 10⁻⁵ = 8.1 10⁻⁶ / r₁

r₁ = 8.1 10⁻⁶ /3.96 10⁻⁵

r₁ = 2.045 10⁻¹ m

r₁ = 20.5 cm

Answer:

The value is

Explanation:

From the question we are told that

The value of the far point is

The distance of the lens to the eye is

Generally

Generally the power spectacle lens needed is mathematically represented as

Here is the object distance which for a near sighted person is

And is the image distance which is evaluated as

=>

=>

So

=>

=>

**Answer:**

E = 1.76 J

**Explanation:**

Given that,

Mass of an object, m = 0.4 kg

It moves by a vertical distance of 0.45 m in the Earth's gravitational field.

We need to find the change in its gravitational potential energy. It can be given by the formula as follow :

So, the change in its gravitational potential energy is 1.76 J.

**Answer:**

Use proportions to find the scale of the first photo, then use that scale and other given information to fill in the equation

S=f/(H-h)

Where:

S = scale of the photo

f = focal length of the camera (in feet)

H = flying height

h = average elevation