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Scrat [10]
3 years ago
10

A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At

a point P that is 1.25 cm outside the sheet, the magnitude of the electric field due to the sheet is E. If the sheet is now stretched so that its sides have length 2d, what is the magnitude of the electric field at P?
A) 4E B) 2E C) E D) E/2 E) E/4
Physics
1 answer:
Umnica [9.8K]3 years ago
8 0

Answer:

Explanation:

Given

Charge Q is uniformly spread over large non-conducting Elastic sheet      

Electric field due to non-conducting Elastic sheet

E=\frac{\sigma }{2\epsilon }

where \sigma =surface charge density=\frac{q}{d^2}

E=\frac{\frac{q}{d^2}}{2\epsilon }

for side 2d Electric Field is given by

E'=\frac{\frac{q}{2d^2}}{2\epsilon }

E'=\frac{1}{4}\times \frac{\frac{q}{d^2}}{2\epsilon }

E'=\frac{E}{4}

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MariettaO [177]

Answer:

F = 6666.7 N

Explanation:

Given that,

Mass of a chip, m = 0.1 mg

Initial speed, u = 0

Final speed,v=4\times 10^{3}\ m/s

Time of collision,t=6\times 10^{-8}\ s

We know that,

Force, F = ma

Put all the values,

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.1\times 10^{-6}\times (4\times 10^3-0)}{6\times 10^{-8}}\\\\F=6666.7\ N

So, the required force is 6666.7 N.

3 0
3 years ago
Please help me to do this problem
Novosadov [1.4K]

Answer:

we got time and velocity over time.

so the distance is again the area underneath the graph

for a triangle with known base and height it's

4*10 / 2

distance traveled is 20

deceleration occurs when velocity decreases. that happens from t=2 till t=4

in 2 time-units we loose 10 units of velocity, so we decelerate by 5 units per 1 time

a (from t=2 to t=4) = -5v/t

7 0
3 years ago
A wave has a speed of 30 m/s, a frequency of 6 Hz, and a wavelength of 5 m. If the wavelength remains constant, and the frequenc
mixer [17]

Answer: 60m/s

Explanation:

The wavespeed is the distance covered by the wave in one second. It is measured in metre per second, and represented by the symbol V

Wavespeed (V) = Frequency F x wavelength λ

i.e V = F λ

In the first case:

Wavespeed = 30 m/s

Frequency of sound = 6Hz

Wavelength = 5m

In the second case:

Wavespeed = ?

Frequency of sound = (2x 6Hz = 12Hz)

Wavelength = 5m (remains constant)

Apply V = F λ

Wavespeed = 12 Hz x 5m

Wavespeed = 60m/s

Therefore, when frequency is doubled, the speed is also doubled. Thus, the new speed of the wave is 60m/s

8 0
3 years ago
Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s
alekssr [168]

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

a)

  • Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
  • Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
  • vₓ₀ = v * cos θ (1)
  • where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
  • Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

        x_{f} = v_{ox} * t = v_{o} * cos \theta * t  (2)

  • Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

        cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)

  • ⇒θ = cos⁻¹ (0.526) = 58.3º (4)

b)

  • At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
  • Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
  • vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

c)

  • At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

d)

  • Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

       \Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)

  • Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
  • v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
  • Replacing (7) in (6), we get:

       \Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)

e)

  • When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
  • The horizontal component, since it keeps constant, is just v₀x:
  • v₀ₓ = 13.7 m/s
  • The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

       v_{fy} = v_{oy} - g*t  (9)

  • Replacing by the time t (a given), g, and  v₀y from (7), we can solve (9) as follows:

       v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s  (10)

  • Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

       v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)

3 0
2 years ago
Which one of the following SI base units is not matched to the correct unit of measurement?
laiz [17]
<span><span>A.    </span>Length: km<span>


The Si units of all the left measurements are true thus, in length the typical and standard SI unit is meter. SI units, also called as International Standard of Units, is composed of the following systematized units of physical measure which include, mole (amount of substance), candela (luminous intensity), kelvin (temperature), ampere (electric current), second (time), kilogram (mass), and meter (length). Hence, these measures spheres in multiplication or division by power of 10, increase or decrease in measure.</span></span>  



3 0
3 years ago
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