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Afina-wow [57]
3 years ago
8

PLEASE HELP!!!!!!!! More science questions coming soon!!

Physics
1 answer:
Naily [24]3 years ago
7 0

Answer:

1.) A - 1,2,1,2

2.) Both forces can act without objects touching.

3.) C

Both can be modeled as waves having amplitude, frequency and wave lengths.

4.) A, B and C

Explanation:

1.) SO2 + ___ H2 ----------> _____ S + _______H2O

Looking at above equation, we have one molecule of surfur at the right hand side (RHS), and one molecule of surfur at the left hand side (LHS).

Two atoms of hydrogen and both RHS and LHS

But oxygen is not balanced. We have two atoms of oxygen at the RHS while having one at the LHS.

So let's make oxygen 2 atoms at LHS by adding 2 to water molecules and hydrogen molecules at the RHS.

SO2 + ___ 2H2 ----------> _____ S + _______2H2O

The correct answer is A - 1,2,1,2

2.) Both magnetic force and gravitational force obey inverse square law with distance. They are not directly proportional but inversely proportional to the square distance. B is the correct answer because Both forces can act without objects touching.

3.) Light waves are transverse waves which can travel through a vacuum without a medium while sound waves are longitudinal waves which cannot travel through vacuum without a medium. But both can be modeled as waves having amplitude, frequency and wave lengths.

4.) If an object is slowing down, due to conservation of energy,

The potential energy could be increasing, like a ball thrown into the air.

The kinetic energy could be lost to friction or air resistance.

The ball could be returning to its natural resting state.

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Help i hope this actually works
KIM [24]

Answer:

it does tbh.

Explanation:

6 0
3 years ago
An electron moving with a velocity = 5.0 × 10 7 m/s enters a region of space where perpendicular electric and a magnetic fields
inna [77]

Answer:

The magnetic field is 2 \times 10^{-4} T

Explanation:

Given:

Velocity of electron v = 5 \times 10^{7} \frac{m}{s}

Electric field E = 10^{4} \frac{V}{m}

The force on electron in magnetic field is given by,

 F = qvB \sin \theta                      ......(1)

The force on electron in electric field is given by,

 F = qE                               ......(2)

Compare both equation,

   qE = qvB \sin \theta

Here \sin \theta = 1

  E = vB

  B= \frac{E}{v}

  B = \frac{10^{4} }{5 \times 10^{7} }

  B = 2 \times 10^{-4} T

Therefore, the magnetic field is 2 \times 10^{-4} T

5 0
3 years ago
11.
jolli1 [7]
<h3>Reducing Surface Area.</h3>

If the surface area becomes smaller, the pressure becomes larger.

8 0
3 years ago
Items such as graphics, charts, or spreadsheets that can be inserted into Word documents are called:
11Alexandr11 [23.1K]
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5 0
3 years ago
The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5
ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E =  K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) =  7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) =  4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.89e5)^{2}  + (7.89e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) net magnitude =  1.115e6\frac{N}{C} @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

 E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.88e5)^{2}  + (7.88e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) and (b) the net magnitude =  1.242e6\frac{N}{C} @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0
3 years ago
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