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Nady [450]
3 years ago
13

At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what

distance from the line will the field strength to be 2000 N/C?
a. D/√2.
b. √2D.
c. D/2.
d. 2D.
e. D/4.
Physics
1 answer:
Alla [95]3 years ago
5 0

Answer:

The correct option is (a).

Explanation:

We know that, the E is inversely proportional to the distance as follows :

E=\dfrac{k}{d^2}

We can write it as follows :

\dfrac{E_1}{E_2}=(\dfrac{d_2}{d_1})^2

Put all the values,

\dfrac{1000}{2000}=(\dfrac{d_2}{d})^2\\\\\sqrt{\dfrac{1000}{2000}}=(\dfrac{d_2}{D})\\\\0.7071=\dfrac{d_2}{d}\\\\d_1=0.7071D\\\\d_1=\dfrac{D}{\sqrt2}

So, the correct option is (a).

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NO,,The answer is A) potassium oxide

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3 years ago
A sled slides down a hill 42 meters high with a slope of 27 o. The total mass of the sled plus contents is 256 kg. It starts fro
kkurt [141]

Answer:

v = 28.7 m/s

Explanation:

  • According to the Work Energy Theorem, the work done by external forces on the system, is equal to the change in the kinetic energy of the system.
  • In absence of friction, the only force that does work , producing a displacement in the direction of the force, is the component of gravity parallel to the slide:

       Fg_{p } = m*g* sin \theta (1)

  • The displacement d,along the slide, can be found from the definition of the sine of an angle, as follows:

        sin \theta =\frac{h}{d}

  • Solving for d, we get:

       d = \frac{h}{sin \theta} (2)

  • Now, the work done by Fgp, is just the product of the force times the displacement, as follows:

        W = Fgp * d

  • From (1) and (2) we can find W as follows:

       W =Fg_{p }* d = m*g* sin \theta * \frac{h}{sin \theta} = m*g*h

  • This expression must be equal to ΔK, as follows:

        \Delta K = \frac{1}{2} * m *v^{2} = m*g*h

  • Simplifying common terms, we can solve for v (the velocity of the sled at the bottom of the slide), as follows:

        v =\sqrt{2*g*h} = \sqrt{2*9.8 m/s2*42 m} = 28.7 m/s

  • The velocity of the sled at the bottom of the slide is 28.7 m/s, taking as positive the direction down the slide.
7 0
3 years ago
Dwight uses a spring (k = 70 N/m) on a horizontal surface to make a launcher for model cars. The spring is attached to a holder
kramer

Part A)

As we know that spring force is given by

F = kx

here x = stretch in the spring from natural length

So here  when spring reaches to its natural length

Force due to spring = 0

so acceleration = 0

Part b)

When spring is compressed from its natural length it will have elastic potential energy in it

so it is given by

U = \frac{1}{2}kx^2

now we know that there is no friction in it so maximum kinetic energy of the launcher must be equal to the elastic potential energy of the spring

KE = \frac{1}{2}kx^2

here we have

k = 70 N/m

x = 0.4 m

KE = \frac{1}{2}(70)(0.4)^2

KE = 5.6 J

Part c)

Now to find the speed we know that

KE = \frac{1}{2} mv^2

5.6 = \frac{1}{2}0.3v^2

v = 6.11 m/s

so its speed is 6.11 m/s

8 0
3 years ago
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When salt and water are mixed, they form a solution like in the ocean. When sand and water are mixed, they form a mixture where
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Answer:

D

Explanation:

The answer is D. Solutions you cannot see the separate substances. however in a mixture you can see the different substances. So the correct choice is D

3 0
3 years ago
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omeli [17]

Answer:

14 cm

Explanation:

F = (frac{uv}{u – v})

F = +ve

v = -ve

30 = (frac {25 {times} (-v)}{25 – (-v)})

v = (frac {25 {times} (-v)}{25+v})

v = 14cm

(Note that either negative or positive values go to show the positioning and hence, they are not a strong necessity in your final answer.)

So happy that i could help you!

Now this question could turn out to be easy for you!!

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