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3 years ago

Lines of force never intersect why

1 answer:

5 0

If the lines cross each other at a given location, then there must be two distinctly different values of electric field with their own individual direction at that given location. This could never be the case. Therefore the lines representing the field cannot cross each other at any given location in space.

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58.5 million excess electrons are inside a closed surface. What is the net electric flux through the surface?

Anni [7]

The biggest thing you're doing wrong is ignoring the units
when you're working with the quantities.

Now let's look at the rest of the problem:

The formula you used is correct:

   Net flux through the surface = (net charge inside) / ε₀

and      ε₀ = 8.85 x 10⁻¹² farad/meter.

What's the net charge inside the surface in this problem ?

It's (5.85 x 10⁷ electrons) x (the charge on each electron)

   = (5.85 x 10⁷ electrons) x (-1.6 x 10⁻¹⁹ coulomb/electron)

   =    -9.36 x 10⁻¹² coulomb .

Finally,    (net charge inside) / ε₀

   = (-9.36 x 10⁻¹² coulomb) / (8.85 x 10⁻¹² farad/meter)

   = -1.058 newton-m²/coulomb .

The sign and the significant figures in your answer are correct, so
we can see that you know what you're doing. The only thing left is
the order of magnitude. You most likely took one of the negative
exponents and made it positive. You got an answer that's 10²² too
small. Big deal. You could claim "that's close", and see whether you
can convince a teacher.

8 0

3 years ago

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru

r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

= 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

$s = u t +\dfrac{1}{2}at^2$

$-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2$

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

$t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}$

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0

3 years ago

Quels branchement dois-tu faire pour mesurer la tension entre les bornes de la pile?

natulia [17]

Answer:

This is not physics

Explanation:

8 0

3 years ago

PLEASE ANSWER ASAP

kirill [66]

Answer:

55.44.4 hjejjejjejej be

Explanation:

hehhehebebebbebeb

4 0

3 years ago

How much Power is needed to lift a 200 N block

vova2212 [387]

Answer:

10 W

Explanation:

P = F x v

F = 200N

v = x2 - x1 / t

= 6 - 3 / 60

= 3 / 60

= 1 / 20

= 0.05 m/s

P = 200 x 1/20

= 10 W

4 0

2 years ago