Ray run 30 N. 30 feet with in 30 feet so what is the distance

Given the necessary information about the two components of projectile motion, your task is to explain the difference or even de

ivann1987 [24]

The two components of projectile motion include the following:

Horizontal motion

Vertical motion.

<h3>What is a Projectile?</h3>

This is defined as a missile propelled by the application of an external force and allowed to move freely under the influence of gravity and air resistance.

The equation for Horizontal motion Vx = V * cos(α)

Vertical velocity component: Vy = V * sin(α)

Read more about Projectile here brainly.com/question/24216590

8 0

2 years ago

If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that t

Anastaziya [24]

Answer:

f1 = -3.50 m

Explanation:

For a nearsighted person an object at infinity must be made to appear to be at his far point which is 3.50 m away. The image of an object at infinity must be formed on the same side of the lens as the object.

∴ v = -3.5 m

Using mirror formula,

i/f1 = 1/v + 1/u

Where f1 = focal length of the contact lens, v = image distance = -3.5 m, u = object distance = at infinity(∞) = 1/0

∴ 1/f1 = (1/-3.5) + 1/infinity

Note that, 1/infinity = 1/(1/0) = 0/1 =0.

∴ 1/f1 = 1/(-3.5) + 0

1/f1 = 1/(-3.5)

Solving the equation by finding the inverse of both side of the equation.

∴ f1 = -3.50 m

Therefore a converging lens of focal length f1 = -3.50 m

would be needed by the person to see an object at infinity clearly

8 0

3 years ago

Two parallel plate capacitors 1 and 2 are identical except that capacitor 1 has charge +q on one plate and charge −q on the othe

Grace [21]

Answer:

a) the capacitance is the same for both capacitors.

b) The potential difference between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q.

c) The electric field magnitude between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q

d) The energy stored between the plates for the capacitor with charge +2q, is 4 times the value for the one with charge +q

Explanation:

a) The capacitance of a capacitor, by definition, is as follows:

$C = \frac{q}{V}$

Appying Gauss' Law to one of plates, it can be showed, that the capacitance (for a parallel plates capacitor) can be expressed as follows:

C = ε*A / d

As it can be seen, it does not depend on the charge. so we conclude that the capacitance must be the same for both capacitors, due to they are identical except for the value of the charge on the plates.

b) By definition, as we said above, the capacitance is equal to the proportion between the charge of one of the plates, and the potential difference between them.

If this proportion must remain the same, and one of the capacitors has the double of the charge than the other, the potential difference must be the double also.

c) Applying Gauss' law, to the surface of one of the plates, and assuming a constant surface charge density σ, it can be showed that the electric field can be calculated as follows:

E*A = Q/ε₀ as σ=Q/A

⇒ E = σ/ε₀

As σ is directly proportional to the charge (being the area A the same), we conclude that the electric field for the capacitor with charge +2q must be the double than the one for the capacitior with charge +q.

c) The electric potential energy, stored between plates of a capacitor, can be written as follows:

$Ue = \frac{1}{2} *\frac{q^{2}}{C}$

If the capacitance remains the same, we can conclude that the electric potential energy for the capacitor with charge +2q, as the charge is raised to the 2nd power, must be 4 times the one for the capacitor with charge +q.

4 0

3 years ago

If you have observed something about the world and want to

ExtremeBDS [4]

Answer: a

Explanation: :)

5 0

2 years ago

Find the electric potential VP at point P. [Hint: To input a natural logarithm into the answer box, simply type the letters "ln"

Alik [6]

Answer:

After finding the electric potential VP at point P = Q/Чπϵ₀L ㏑(1+ $\frac{L}{d}$ )