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3 years ago

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Three equal point charges, each with charge 1.25 μC , are placed at the vertices of an equilateral triangle whose sides are of l

ength 0.450 m . What is the electric potential energy U of the system

1 answer:

natali 33 [55]3 years ago

7 0

Answer:

electric potential energy is 9.36 × $10^{-2}$ J

Explanation:

given data

charge q = 1.25 μC =1.25× $10^{-6}$ C

length L = 0.450 m

to find out

electric potential energy

solution

we apply electric potential energy U formula

U = U1 + U2 + U3 .................1

we know magnitude of all three charge is

q = q1 = q2 = q3

so here from 1

U = 1 / 4π∈ × (q²/L + q²/L +q²/L )

U = 1 / 4π∈ × 3(q²/L)

put here all value and we know ∈ = 8.85 × $10^{-12}$ F/m

U = 1 / 4π( 8.85 × $10^{-12}$ ) × ( 3 (1.25× $10^{-6}$ )² / 0.450 )

U = 8.99× $10^{9}$ × ( 3 (1.25× $10^{-6}$ )² / 0.450 )

U = 9.36 × $10^{-2}$

so electric potential energy is 9.36 × $10^{-2}$ J

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a baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground. (g=32 feet/

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Answer:

hmax=81ft

Explanation:

Maximum height of the object is the highest vertical position along its trajectory.

The vertical velocity is equal to 0 (Vy = 0)

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we isolate th (needed to reach the maximum height hmax)

$th = \frac{v_{0}*sin(\alpha)}{g}$

The formula describing vertical distance is:

$y = Vy * t-g* t^{2} / 2$

So, given y = hmax and t = th, we can join those two equations together:

$hmax = Vy* th-g*th^{2}/2$

$hmax =Vo^{2}*sin(\alpha )^{2}/(2*g)$

if we launch a projectile from some initial height h all you need to do is add this initial elevation

$hmax =h+Vo^{2}*sin(\alpha)^{2}/(2*g)$

$hmax =3+100^{2}*sin(45)^{2}/(2 * 32)=81 ft$

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3 years ago

A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N1 turns. A second toroidal solenoid w

lilavasa [31]

Answer:

Mutual inductance, $M=2.28\times 10^{-5}\ H$

Explanation:

(a) A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N₁ turns. A second thyroidal solenoid with N₂ turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius.

Mutual inductance is given by :

$M=\dfrac{\mu_oN_1N_2A}{2\pi r}$

(b) It is given that,

$N_1=550$

$N_2=290$

Radius, r = 10.6 cm = 0.106 m

Area of toroid, $A=0.76\ cm^2=7.6\times 10^{-5}\ m^2$

Mutual inductance, $M=\dfrac{4\pi \times 10^{-7}\times 550\times 290\times 7.6\times 10^{-5}}{2\pi \times 0.106}$

$M=0.0000228\ H$

or

$M=2.28\times 10^{-5}\ H$

So, the value of mutual inductance of the toroidal solenoid is $2.28\times 10^{-5}\ H$ . Hence, this is the required solution.

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Mrs. Perez added a room temperature copper cube and an aluminum cube she just removed from the freezer to a beaker of boiling wa

faust18 [17]

I think the answer is A

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3 years ago

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm . The explorer finds that

larisa86 [58]

Answer:

12.4 m/s²

Explanation:

L = length of the simple pendulum = 53 cm = 0.53 m

n = Number of full swing cycles = 99.0

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g = magnitude of gravitational acceleration on the planet

Time period of the pendulum is given as

$T = \frac{t}{n}$

$T = \frac{128}{99}$

T = 1.3 sec

Time period of the pendulum is also given as

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3 years ago

Due to the wave nature of light, light shined on a single slit will produce a diffraction pattern? Green light (520 nm) is shine

TiliK225 [7]

Answer:

Yes, it will produce a diffraction pattern.

a. 3.9 mm b. 1.95 mm

Explanation:

The light shined from a single slit will produce a diffraction pattern because, the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.

Given

wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m

width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m

Distance of slit from central maximum , D = 1.65 m

Distance of first minimum from central maximum, y = ?

a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...

The relationship between y and D is given by $tanθ = y/D$

Since θ is small, sinθ ≈ θ ≈ tanθ

so, dθ = mλ ⇒ θ = mλ/d = y/D

Therefore, y = mλD/d

Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ = λD/d and y₋₁ = -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d

Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ = 3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm

b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ = λD/d and y₂ = 2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have

λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm

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3 years ago