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dmitriy555 [2]
3 years ago
8

The electric field on the surface of an irregularly shaped conductor varies from 66.0 kN/C to 26.0 kN/C. (a) Calculate the local

surface charge density at the point on the surface where the radius of curvature of the surface is greatest. nC/m2 (b) Calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is smallest. nC/m2
Physics
1 answer:
Bess [88]3 years ago
4 0

Answer:

Explanation:

We know that at pointed corner of a conductor , charge accumulate ie electric field is  maximum . Ar pointed corner , radius of curvature will be least . That is why charge leakage is greatest at pointed corner of a charged conductor.

a ) Electric field will be 26.0 kN/C at the point where radius of curvature is greatest .

b ) So electric field will be 66 kN /C at the point where radius of curvature is smallest .

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When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reache
Masteriza [31]

Answer:

b) True.    the force of air drag on him is equal to his weight.

Explanation:

Let us propose the solution of the problem in order to analyze the given statements.

The problem must be solved with Newton's second law.

When he jumps off the plane

     fr - w = ma

Where the friction force has some form of type.

     fr = G v + H v²

Let's replace

     (G v + H v²) - mg = m dv / dt

We can see that the friction force increases as the speed increases

At the equilibrium point

      fr - w = 0

      fr = mg

      (G v + H v2) = mg

For low speeds the quadratic depended is not important, so we can reduce the equation to

     G v = mg

     v = mg / G

This is the terminal speed.

Now let's analyze the claims

a) False is g between the friction force constant

b) True.

c) False. It is equal to the weight

d) False. In the terminal speed the acceleration is zero

e) False. The friction force is equal to the weight

3 0
3 years ago
Which of the following is considered a physical change?
zmey [24]

Answer:

C)evaporating of water

Explanation:that is because waters physical property is lost after evaporating

4 0
3 years ago
Read 2 more answers
From an electromagnetic wave the electric and magnetic field is what direction toward each other?
liubo4ka [24]
What are the choices?
3 0
3 years ago
At noon, ship A is 140 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 25 km/h. How fast (in
Luba_88 [7]

Answer:

The rate of change of distance between the two ships is 18.63 km/h

Explanation:

Given;

distance between the two ships, d = 140 km

speed of ship A = 30 km/h

speed of ship B = 25 km/h

between noon (12 pm) to 4 pm = 4 hours

The displacement of ship A at 4pm = 140 km - (30 km/h x 4h) =

140 km - 120 km = 20 km

(the subtraction is because A is moving away from the initial position and the distance between the two ships is decreasing)

The displacement of ship B at 4pm = 25 km/h x 4h = 100 km

Using Pythagoras theorem, the resultant displacement of the two ships at 4pm is calculated as;

r² = a²   +  b²

r² = 20²  +  100²

r = √10,400

r = 101.98 km

The rate of change of this distance is calculated as;

r² = a²   +  b²

r = 101.98 km, a = 20 km, b = 100 km

2r(\frac{dr}{dt} ) = 2a(\frac{da}{dt} )  + 2b(\frac{db}{dt} )\\\\r(\frac{dr}{dt} ) = a(\frac{da}{dt} )  + b(\frac{db}{dt} )\\\\101.98(\frac{dr}{dt} ) = 20(-30 )  + 100(25 )\\\\101.98(\frac{dr}{dt} ) = -600 + 2,500\\\\101.98(\frac{dr}{dt} ) = 1900\\\\\frac{dr}{dt}  = \frac{1900}{101.98} = 18.63 \ km/h

5 0
3 years ago
60m27Co → 6027Co Predict the type of radioactive emission produced from the decay of metastable cobalt-60 to cobalt-60. Describe
lara [203]
Some one already asked this question and you can copy paste and google it but I believe it is c you may want to double check
4 0
3 years ago
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