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3 years ago

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You are given the melting points of three unknown substances and are asked to predict which one is an ionic compound. You would

select the compound with the ____________________ melting point.

1 answer:

nikitadnepr [17]3 years ago

4 0

The answer is <span>highest</span>

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You are bungee jumping from a bridge. Initially, while you are falling the slack bungee cord isn’t exerting any forces or torque

harina [27]

Answer:

he fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

Explanation:

Let's pose the solution of this problem, to be able to analyze the firm affirmations.

When the person is falling, the weight acts on them all the time, initially the rope has no force, but at the moment it begins to lash it exerts a force towards the top that is proportional to the lengthening of the rope.

The equation for this part is

Fe - W = m a

k x - mg = m a

As the axis of rotation is located at the top where they jump, there is a torque.

What is it

Fe y - W y = I α

angular and linear acceleration are related

a = α r

Fe y - W y = I a / r

In the fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

4 0

2 years ago

Monochromatic light of wavelength 385 nm is incident on a narrow slit. On a screen 3.00 m away, the distance between the second

LiRa [457]

To solve this problem it is necessary to apply the concepts related to the concept of overlap and constructive interference.

For this purpose we have that the constructive interference in waves can be expressed under the function

$a sin\theta = m\lambda$

Where

a = Width of the slit

d = Distance of slit to screen

m = Number of order which represent the number of repetition of the spectrum

$\theta =$ Angle between incident rays and scatter planes

At the same time the distance on the screen from the central point, would be

$sin\theta = \frac{y}{d}$

Where y = Represents the distance on the screen from the central point

PART A ) From the previous equation if we arrange to find the angle we have that

$\theta = sin^{-1}(\frac{y}{d})$

$\theta = sin^{-1}(\frac{1.4*10^{-2}}{3})$

$\theta = 0.2673\°$

PART B) Equation both equations we have

$a sin\theta = m\lambda$

$a \frac{y}{d} = m\lambda$

Re-arrange to find a,

$a = \frac{(2)(385*10^{-9})(3)}{(1.4*10^{-2})}$

$a = 1.65*10^{-4}m$

8 0

3 years ago

What is the speed of a wave that has a frequency of 173 Hz and a wavelength of 2.59 meters? Express your answer to the nearest w

devlian [24]

Answer:

448 m/s is the correct answer.

Explanation:

7 0

3 years ago

Read 2 more answers

The engineer of a train traveling at 30 m/sec sees a cow on the tracks. He applies the brakes and causes the train to accelerate

sineoko [7]

Answer:The train travels 105 meters after applying the brakes

Explanation:If he decelerates 1.5 every minute, then he went from 28,5 m/s, to 27.0 m/s, to 25.5 m/s, to 24.0 m/s, after 4 seconds. Add all this together and youll get 105 meters moved in 4 seconds after he hit the brakes, I dont have a notebook on me though sorry :/

8 0

2 years ago

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

$g=G\frac{m}{r^{2}}$

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

$g_{E}=G\frac{m}{r_{E}^{2}}$

$g_{P}=G\frac{m}{r_{P}^{2}}$

The problem tells us the radius of the planet is twice that of the radius on earth, so:

$r_{P}=2r_{E}$

If we substituted that into the gravity of the planet equation we would end up with the following formula:

$g_{P}=G\frac{m}{(2r_{E})^{2}}$

Which yields:

$g_{P}=G\frac{m}{4r_{E}^{2}}$

So we can now compare the two gravities:

$\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}$

When simplifying the ratio we end up with:

$\frac{g_{P}}{g_{E}}=\frac{1}{4}$

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0

3 years ago