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3 years ago

Substance that does not contain any free ions and will not be an electrically conductive medium is what?

Physics

2 answers:

muminat3 years ago

5 0

The answer to your question is a nonelectrolyte.

Sedaia [141]3 years ago

3 0

Answer:

Nonelectrolyte

Explanation:

This is a substance which when dissolved in water does not dissociate Ito ions

They are held together by covalent bonds rather than ionic bond

Example is sugar when dissolved in water does not dissociate

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How fast would the International Space Station (ISS) have to travel to maintain a circular orbit a distance of 1400 km above the

professor190 [17]

Answer:

The International Space Station move at 7.22 km/s.

Explanation:

Orbital speed of satellite is given by $v=\sqrt{\frac{GM}{r}}$ , where G is gravitational constant, M is mass of Earth and r is the distance to satellite from centre of Earth.

r = R + h = 6350 + 1400 = 7750 km = 7.75 x 10⁶ m

G = 6.673 x 10⁻¹¹ Nm²/kg²

M = 5.98 x 10²⁴ kg

Substituting

$v=\sqrt{\frac{6.673\times 10^{-11}\times 5.98\times 10^{24}}{7.75\times 10^6}}=7223.86m/s=7.22km/s$

The International Space Station move at 7.22 km/s.

4 0

2 years ago

1) Arrange the following spectral regions in order of increasing energy: infrared, microwave, ultraviolet, visible.

Anton [14]

Answer:

The order of increasing energy is as follows

"microwave < infrared < visible < ultraviolet"

Option (A) is correct.

Explanation:

Given:

Arrange the following spectral regions in order of increasing energy: infrared, microwave, ultraviolet, visible.

From the formula of energy in terms of frequency.

$E = hf$

Where $h =$ planck constant, $f =$ frequency of light.

From above formula we can conclude that higher frequency means higher energy.

In our case ultraviolet has higher frequency and microwave has lower frequency.

So ultraviolet has higher energy and microwave has lower energy.

microwave < infrared < visible < ultraviolet

Therefore, the order of increasing energy is as follows

"microwave < infrared < visible < ultraviolet"

7 0

2 years ago

In the example of jumping off a chair, what is the impulse that will stop your fall?

liq [111]

Answer:

You will reach both your arms out to break your fall and save your head.

Explanation:

It common sense you don't want your head injured. Do you?

5 0

3 years ago

The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is this case is 0 as the road is level

Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0

3 years ago

What can you say about the speed of the car?

iren2701 [21]

I believe the answer is C

5 0

3 years ago