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anyanavicka [17]
2 years ago
10

A source charge of 5. 0 µC generates an electric field of 3. 93 × 105 at the location of a test charge. How far is the test char

ge from the source charge?.
Physics
1 answer:
lora16 [44]2 years ago
6 0

Answer:

r = 0.338 meters

Explanation:

Q = Source charge

q = Test charge

5μC= 5*10^-6 C

Start with the electric field equation:

E = \frac{F}{q}

and Coulombs law

F=\frac{k|Qq|}{r^2}

Combining these functions we get:

E=\frac{k|Q|}{r^2}

As you can see the test charge cancels so we do not need the value of it.

Plug in your numbers:

3.93*10^5=\frac{(8.99*10^9)(5*10^-^6)}{r^2}

solve for r

r=\sqrt{\frac{(8.99*10^9)(5*10^-^6)}{(3.93*10^5)} }

r= 0.338 meters

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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

7 0
4 years ago
A ski gondola is connected to the top of a hill by a steel cable of length 620 m and diameter 1.5 cm. As the gondola comes to th
xz_007 [3.2K]

Answer:

(a) 89 m/s

(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}

(b) The tension, T, is given by

v =\sqrt{\dfrac{T}{\mu}}

where v is the speed, T is the tension and \mu is the mass per unit length.

Hence,

T = \mu\cdot v^{2}

To determine \mu, we need to know the mass of the cable. We use the density formula:

\rho = \dfrac{m}{V}

where m is the mass and V is the volume.

m=\rho\cdot V

If the length is denoted by l, then

\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}

T = \dfrac{\rho\cdot V}{l} v^{2}

The density of steel = 8050 kg/m3

The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

V = \pi \dfrac{d^{2}}{4} l

T = \dfrac{\rho\cdot\pi d^2 l}{4l}v^2 = \dfrac{\rho\cdot\pi d^2}{4}v^2

T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2

T = 11159.4186\ldots \text{ N} = 11000 \text{ N}

4 0
4 years ago
An object dropped on Planet P falls 144 m in 6 seconds. What is the gravitational acceleration of Planet P ? Gravitational accel
Tju [1.3M]

Answer:

The gravitational acceleration of the planet is, g = 8 m/s²

Explanation:

Given data,

The distance the object falls, s = 144 m

The time taken by the object is, t = 6 s

Using the III equations of motion

                  S = ut + ½ gt²

∴                 g = 2S/t²

Substituting the given values,

                   g = 2 x 144 /6²

                      = 8 m/s²

Hence, the gravitational acceleration of the planet is, g = 8 m/s²

7 0
4 years ago
What does the area under a speed-time graph represent​
Tom [10]

Answer: It represents the whole distance traveled. Hope this helps!

Explanation:

4 0
3 years ago
A baseball pitcher throws a baseball horizontally at a linear speed of 42.5 m/s (about 95 mi/h). Before being caught, the baseba
Marina CMI [18]

Answer:

4.5m/s

Explanation:

Linear speed (v) = 42.5m/s

Distance(x) = 16.5m

θ= 49.0 rad

radius (r) = 3.67 cm

= 0.0367m

The time taken to travel = t

Recall that speed = distance / time

Time = distance / speed

t = x/v

t = 16.5/42.5

t = 0.4 secs

tangential velocity is proportional to the radius and angular velocity ω

Vt = rω

Angular velocity (ω) = θ/t

ω = 49/0.4

ω = 122.5 rad/s

Vt = rω

Vt = 0.0367 * 122.5

Vt =4.5 m/s

5 0
3 years ago
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