HELP NOW PLS !!
If you exert 950 N*s of impulse on a 12 kg frictionless cart over the course of 5 seconds, how far will it travel during those seconds?

Answers

Answer 1

Answer:

The value is [tex]s =197.88 \ m[/tex]  

Explanation:

From the question we are told that

    The  impulse is  [tex]I = 950 \ N\cdot s[/tex]

    The mass is  [tex]m = 12 \ kg[/tex]

     The time t =  5 s

Generally impulse is mathematically represented as

         [tex]I = F * t[/tex]

=>       [tex]F = \frac{I}{t}[/tex]

=>       [tex]F = \frac{950 }{ 5 }[/tex]

=>       [tex]F = 190 \ N[/tex]

Generally force is mathematically represented as

      [tex]F = m * a[/tex]

=>    [tex]a = \frac{F }{ m }[/tex]

=>     [tex]a = \frac{190 }{ 12 }[/tex]

=>     [tex]a = 15.83 \ m/s^2[/tex]

Generally from kinematic equation , the distance covered is  

      [tex]s = ut + \frac{1}{2} * at^2[/tex]

Here  u is the initial velocity of the cart and the value is [tex]u = m/s[/tex]

=>   [tex]s = 0 * 5 + \frac{1}{2} * 15.83 * 5^2[/tex]  

=>   [tex]s =197.88 \ m[/tex]  


Related Questions

An ambulance is traveling east at 62.4 m/s. Behind it a car travels along the same direction at 34.5 m/s. The ambulance driver hears his siren with a wavelength of 0.47 m. What wavelength would a stationary observer behind the ambulance measure for the sound? The velocity of sound in air is 343 m/s.

Answers

Answer:

The answer is "0.5555 m"

Explanation:

Where the reference leaves the list and the viewer is at rest:

[tex]\lambda'=\frac{v-v_s}{v} \times \lambda\\\\[/tex]

   [tex]=\frac{343 \frac{m}{s} - (-62.4 \frac{m}{s})}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{343 \frac{m}{s} + 62.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{405.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{405.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m[/tex]

   [tex]=0.5555 \ m[/tex]

Pls Help physics 8th class question ​

Answers

Answer:

angle of reflection and angle of incident is always equal

what belongs in the center section

Answers

Answer:

The second one I think

Explanation:

B

HELP PLZ!!!!!!!!!!!!!!

Answers

Answer:

Since binary is only 1 and 0, you can use a flashlight to display something similar to Morse code (see explanation below)

Explanation:

In binary, 1 means "on" and 0 means "off". A way you can use visible light is through turning on and off a flashlight. If the flashlight is turned on, it would represent a 1. If the flashlight is turned off, it would represent a 0. To make the message easier and more accurately understood for the receiver make sure to flash the lights in a consistent pattern (ex. each flash lasts no longer than half a second, one second between each digit, etc.)

For example, let's say you're trying to send the message "11001"

  on     on    off     off     on

0       1       2       3       4       5      Numbers represent seconds

As you can see above the message starts at 0 seconds. Between 0 and 1 seconds the flashlight is turned on once. Between 1 and 2 seconds the flashlight is turned on again, Between 2 and 3 seconds as well as 3 and 4 seconds the flashlight is not turned on at all. And finally between 4 and 5 seconds the flashlight is turned on.

Drag each item to indicate whether it is a characteristic of dark matter or baryonic matter. (2 points)
Dark Matter
Baryonic Matter
:: has been observed directly
:: interacts with baryonic matter
:: interacts with electromagnetic radiation
has not been observed directly
:: does not interact with electromagnetic radiation
++ does not interact with baryonic matter

Answers

Answer:

Dark matter:

- Doesn't interact with baryonic matter.

- It has not been observed directly

- dark matter does not absorb, reflect or even emit light, thereby making it to be extremely hard to spot. Therefore, it does not interact with electromagnetic radiation.

Explanation:

Dark matter:

- Doesn't interact with baryonic matter.

- It has not been observed directly

- dark matter does not absorb, reflect or even emit light, thereby making it to be extremely hard to spot. Therefore, it does not interact with electromagnetic radiation.

Baryonic matter:

- Has been observed directly because it includes nearly all the matter that we see in the world daily.

- It interacts with baryonic matter

- interacts with electromagnetic radiation

Dark Matter:

It has not been observed directly.It does not interact with baryonic matter.It does not interact with electromagnetic radiation.

Baryonic Matter:

It has been observed directly.It interacts with baryonic matter.It interacts with electromagnetic radiation.

Dark matter can be defined as often invisible substances that are difficult to spot because they don't absorb, emit or reflect light.

Hence, dark matter do not affect human view because they do not interact or interfere with electromagnetic radiation (force).

Although, humans can see right through the (weakly interacting) dark matter but it has not been observed directly.

Baryonic matter can be defined as a dark matter that is made up of baryons such as neutrons, and protons. Also, they are ordinary matter (both fermions and hadrons), as distinct from exotic forms.

In conclusion. baryonic matter has been observed directly and it can interact with electromagnetic radiation.

Read more: https://brainly.com/question/15677526

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