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andre [41]
3 years ago
5

(100 pts)Why can't you use distance divide by time to calculate the instantaneousspeed?​

Physics
2 answers:
Fiesta28 [93]3 years ago
4 0

Answer:

Instantaneous speed means speed at any instant

that means Speed is changing with time

You know speed is distance/time

So that means distance is also changing with time

So we take infinitesimal small distance per infinitesimal small time As we assume speed is constant in infinitesimal small time dt

So, we take speed = ds/dt

ds = infinitesimal small distance

dt = infinitesimal small time

As its ratio is equal to speed at any instant

Note : We are taking infinitesimal small distance

But :) we are taking infinitesimal small time also

As you know if denominator is small fraction is large So fraction always give large value

So it's not O ( this makes confuse to most of students)

So, thanks

Good question

Keep thinking like this :)

Korvikt [17]3 years ago
3 0

<u><em>mph/mps</em></u>

there you go kind sir

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What is true of an object pulled inward in an electric field?
slava [35]

Answer:

option b

Explanation:

There is an object pulled inward in an electric field.

We have to find out of the four options given which is true.

a) The object has a neutral charge is false since when electric field pulls the object inward, there is a charge inside.

b) The object has a charge opposite that of the field, this option is correct since there will be an equal and opposite charge created by the object

c) The object has a negative charge will be correct only if the original charge was positive hence wrong

d) The object has a charge the same as that of the field is incorrect since this would be opposite the charge

So only option b is right

5 0
3 years ago
The emission spectrum of iodine is shown below.<br> Which is the absorption spectrum?
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3 years ago
Which of the following factors will negatively impact the rate of diffusion of a molecule?
Maurinko [17]

Answer:

A low difference in the concentration of the molecule across the media

Explanation:

Diffusion is a type of passive transport where the molecules move in the influence of concentration gradient of diffusing molecules i.e. from the higher concentration region to the lower concentration region. There are some factors which affect the rate of diffusion, these are written below -

  1. Mass of diffusing molecule - lighter molecules diffuse faster and heavier one diffuse relatively slower.
  2. Concentration gradient - rate of diffusion is higher if the difference in concentration of the diffusing particles is larger in the two regions.
  3. Distance traveled - molecules diffuse faster if they need to travel little distance during diffusion.
  4. Temperature - rate of diffusion will be greater at higher temperatures because the movement of diffusing molecules gets increased.
  5. Solvent density - rate of diffusion tend to be lower if the solvent has higher density.

Looking at these factors we can conclude that the second statement in the question tells about a negative impact regarding the diffusion because due to low difference in concentration across the two media, the rate of diffusion will be lower.

3 0
3 years ago
A narrow region between two air masses of different densities is a
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A front is a narrow region  between two air masses of different densities.
5 0
2 years ago
A ball of moist clay falls 17.3 m to the ground. It is in contact with the ground for 24.0 ms before stopping. (a) What is the a
gizmo_the_mogwai [7]

Answer:

Acceleration,  767.08\ m/s^2

Explanation:

Given that,

Height from a ball falls the ground, h = 17.3 m

It is in contact with the ground for 24.0 ms before stopping.

We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

v^2=u^2+2ah

u = initial velocity=0 m/s

a = acceleration=g

v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 17.3} \\\\v=18.41\ m/s

It is in negative direction, u = -18.41 m/s

Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18.41}{24\times 10^{-3}}\\\\a=767.08\ m/s^2

So, the average acceleration of the ball during the time it is in contact is 767.08\ m/s^2.

4 0
3 years ago
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