Which is an example of a passive solar energy system?

A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

Kitty [74]

Answer:

$E_{net} = 6.44 \times 10^5 N/C$

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

$E = \frac{2k \lambda}{r}$

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

$E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}$

now plug in all data

$\lambda_1 = 4.68 \muC/m$

$\lambda_2 = 2.48 \mu C/m$

$r = 0.200 m$

now we have

$E_{net} = \frac{2k}{r}(4.68 + 2.48)$

$E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})$

$E_{net} = 6.44 \times 10^5 N/C$

8 0

3 years ago

calculate the spring constant if a weight of 250N is added to a spring which increases in length by 20cm

ZanzabumX [31]

Since, F = k . ∆x

Therefore, k = F / ∆x = 250 / 0.2 = 1250 N/m

(ps: convert 20 cm into 0.2 m)

8 0

3 years ago

Which is true about the way air flows

JulsSmile [24]

Answer:

A High-to-Low

Explanation:

its like water running down a hill.

4 0

3 years ago

A stone is dropped from a cliff at time t = 0 and strikes the ground below.

Dafna1 [17]

Answer:

S=

2

1

gt

2

........(1)

And that of the other stone is

S=u(t−n)+

2

1

[g(t−n)

2

]........(2)

Since both the stones meet at the distance so equation (1)and equation(2) will be equal

2

1

gt

2

=u(t−n)+

2

1

[g(t−n)

2

]

gt

2

=2ut−2un+gt

2

+gn

2

−2gnt

t(2gn−2u)=gn

2

−2un

t=

(gn−u)

n(

2

gn

−u)

now putting value of t in equation(1)

s=

2

g

⎣

⎢

⎡

gn−u

2

[n(

2

gn

−u)]

⎦

⎥

⎤

2

S=

2

g

⎣

⎢

⎡

(gn−u)

2

n(

2

gn

−u)

⎦

⎥

⎤

Explanation:

plzzz mark me brainliest and follow me

l

i will also follow u

5 0

3 years ago

A 34.9-kg child starting from rest slides down a water slide with a vertical height of 16.0 m. (Neglect friction.)

aalyn [17]

Answer:

a) 12.528 m/s

b) 15.344 m/s

Explanation:

Given:

Mass of the child, m = 34.9 kg

Height of the water slide, h = 16.0 m

Now,

a) By the conservation of energy,

loss in potential energy = gain in kinetic energy

mgh = $\frac{\textup{1}}{\textup{2}}\textup{m}\times\textup{v}^2$

where,

g is the acceleration due to the gravity

v is the velocity of the child

thus,

at halfway down, h = $\frac{\textup{16}}{\textup{2}}$ = 8 m

therefore,

34.9 × 9.81 × 8 = $\frac{\textup{1}}{\textup{2}}\textup{34.9}\times\textup{v}^2$

or

v = 12.528 m/s

b)

at three-fourth way down

height = $\frac{\textup{3}}{\textup{4}}\times16$ = 12 m

thus,

loss in potential energy = gain in kinetic energy

34.9 × 9.81 × 12 = $\frac{\textup{1}}{\textup{2}}\textup{34.9}\times\textup{v}^2$

or

v = 15.344 m/s

5 0

3 years ago

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