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Charra [1.4K]
3 years ago
15

Two large parallel metal plates are 1.2 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces

. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then 6.6 V, what is the electric field in the region between the plates?
Physics
1 answer:
zheka24 [161]3 years ago
3 0

Answer:

E=1100V/m

                                                                     

Explanation:

Given           required           <u>solution</u>

V=6.6v         E=?                   V=Ed ;    V is the potential difference between                                                              

d=D/2=1.2cm/2=0.6cm=0.006m             the   halfway                  

                                                               E is the electric field between the two                                                                              

                                                                       plates.

                                                              d is the distance between the halfway.

So we can use the above formula to calculate the electric field.

    V=Ed   from this E=V/d  substitute the values from the given equation.

E=6.6v/0.006m

E= 1100 v/m

                                                                     

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Answer:  75 mph

Explanation:

The Relative Speed for a mobile is equal to the diference between the object and the observer:

Relative Speed (Rs) = Object's Velocity  - Observer's Velocity

Thinking on those terms, we would need to have a universal observer to do any understandable measurement on daily basics. This is why we all use earth as a Static Observer for every measurement we do everyday.

Using Earth as an observer, the Velocity for the Patrolman is:

Patrolman Velocity (Vp) = 60 mph

Because the radar gun does measure the Relative Speed for the object, which is 135 mph, we need to work with the equation to find the Velocity using Earth as a reference.

Object's Relative Velocity = Object's Velocity - Patrolman's Velocity

Object's Velocity = Object's Relative Velocity + Patrolman's Velocity

We need to keep in mind, the Patrolman is going on the opposite direction. Because of this the sign for his velocity should be negative.

Object's Velocity = 135 mph + ( -60 mph)

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8 0
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mass of the planet is 12 times that of earth and its radius is thrice that of earth , then find the escape velocity on that plan
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Answer:

The escape velocity on the planet is approximately 178.976 km/s

Explanation:

The escape velocity for Earth is therefore given as follows

The formula for escape velocity, v_e, for the planet is v_e = \sqrt{\dfrac{2 \cdot G \cdot m}{r} }

Where;

v_e = The escape velocity on the planet

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m = The mass of the planet = 12 × The mass of Earth, M_E

r = The radius of the planet = 3 × The radius of Earth, R_E

The escape velocity for Earth, v_e_E, is therefore given as follows;

v_e_E = \sqrt{\dfrac{2 \cdot G \cdot M_E}{R_E} }

\therefore v_e = \sqrt{\dfrac{2 \times G \times 12 \times M}{3 \times R} } =  \sqrt{\dfrac{2 \times G \times 4 \times M}{R} } = 16 \times \sqrt{\dfrac{2 \times G \times M}{R} } = 16 \times v_e_E

v_e = 16 × v_e_E

Given that the escape velocity for Earth, v_e_E ≈ 11,186 m/s, we have;

The escape velocity on the planet = v_e ≈ 16 × 11,186 ≈ 178976 m/s ≈ 178.976 km/s.

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Answer:

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Explanation:

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f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀    (1)

Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²

Equating (1) and (2) we ave

2(√mg/μ)/f = T²g/4π²

Making g subject of the formula

g = 2π√(2√(m/μ)/f)/T

The period T = 316 s/100 = 3.16 s

Substituting the other values into , we have

g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16

g = 2π√(2 × 35.877/200 Hz)/3.16

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g = 2π√(0.358)/3.16

g = 2π × 0.599/3.16

g = 1.19 m/s²

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