Mark walked 2 miles east than 1 mile north how would you determine his total displacement

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:

Baryonic Matter:

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.

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Answer:

angle of reflection and angle of incident is always equal

Answer:

By using [tex]v^{2}_{f} = v^{2}_{i} + 2a\Delta y[/tex], with [tex]v_{i} = -12m/s[/tex] and [tex]\Delta y = -40m[/tex]:

[tex]v^{2}_{f} = v^{2}_{i} + 2a/Delta y[/tex]

[tex]v^{2} = (-12m/s)^{2} + 2(-9.80m/s^{2})(-40m)[/tex]

[tex]v = -30m/s[/tex]

Explanation:

Hope this helped!

A rock is thrown downward from the top of a 40 m tall tower, with The speed of the rock just before hitting the ground will be equal to -30 m/s.

Friction is the resistance to a thing moving or rolling over another solid object. Although frictional forces can be advantageous, such as the traction required to walk while slipping, they also provide a significant amount of resistance to motion. In order to overcome frictional resistance in the moving parts, about 20% of an automobile's engine power is used.

The forces of attraction, also referred to as adhesion, between the contact zones of the surface, which are always minutely uneven, seem to be the main contributor to friction between metals.

From the given information in the question,

v²(f) = v²(i) + 2aΔy

v² = (-12 m/s)² + 2(-9.8)(-40)

v = -30 m/s.

Therefore, the velocity of the rock is -30 m/s.

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Answer: The height of the cliff is 104.59 m

Explanation:

The horizontal speed of the car when it leaves the cliff is 13 m/s, and it hits the ground 60m from the shoreline.

Here we can use the relationship:

Time*Speed = Distance.

To find the time that the car is in the air, we know that:

speed = 13m/s

distance = 60m

time = T

13m/s*T = 60m

T = (60m)/13m/s = 4.62 s

This means that the car is falling for 4.62 seconds.

Now let's analyze the vertical problem.

As the car leaves the cliff, it only has horizontal velocity, this means that the vertical initial velocity will be zero

The only force acting in the vertical axis is the gravitational force, this means that the acceleration will be equal to the gravitational acceleration, which is:

g = 9.8m/s^2

then:

a = -9.8m/s^2

Where the negative sign is because the acceleration is pulling the car downwards.

To get the vertical velocity, we could integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the constant of integration and the initial vertical velocity, that we already know that is equal to zero, then the vertical velocity as a function of time can be written as:

v(t) = (-9.8m/s^2)*t

To get the vertical position equation, we need to integrate again over the time:

P(t) = (1/2)*(-9.8m/s^2)*t^2 + H

Where H is the constant of integration and the initial vertical position, then H will be the height of the cliff.

We know that the car needs 4.62 seconds to hit the ground, this means that:

P(4.6s) = 0m

Then:

P(t) = (1/2)*(-9.8m/s^2)*(4.62s)^2 + H = 0

            (-4.9m/s^2)*(4.62s)^2 + H = 0

          H =  (4.9m/s^2)*(4.62s)^2 = 104.59 m

This means that the cliff is 104.59 meters high

Answer:

The frequency stays the same and the wavelength decreases.

Explanation:

When sound wave enters a new medium where sound travels faster, its frequency will remain same because it depends only on the source.

The relation between wavelength and speed is inverse, it means when the speed of sound increases, its wavelength will decrease.

So, the frequency stays the same and the wavelength decreases. Hence, the correct option is (d).

Answer:

2.51 kg * m/s

Explanation:

In order to find momentum, use the equation below:

momentum = mass * velocity.

Since neither mass nor velocity was given, you must solve for both variables.

In order to solve for mass, use the force equation for its weight / gravitational force.

Fg (gravitational force) = 3.2 N = ma = 9.8m

mass = 3.2 N / 9.8 m/s^2 = 0.326531 kg

In order to solve for velocity, use the equation:

velocity = displacement / time

velocity = 10m / 1.3 s = 7.69231 m/s

Momentum = mass * velocity = 0.326531 kg *  7.69231 m/s = 2.51177 kg * m/s = 2.51 kg * m/s

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.

Answer:

The second one I think

Explanation:

B

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]

Answer:

true

Explanation: