Determine e when I = 0.50 A and R = 12 W.

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:

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:

The second one I think

Explanation:

B

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:

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.

To know more about Friction:

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They could determine it by counting their total amount of miles mark went by the directions he went to walk those miles.

Answer:

true

Explanation:

Answer:

angle of reflection and angle of incident is always equal

Answer:

0.8 m/s

Explanation:

F(avg) * Δt = I, where

F(avg) = Average force

Δt = change in time

I = impulse

From the question, we know that the

average force is given as -17600 N

time change is given as 55 milliseconds

speed of the player, v = 8 m/s

Mass of the player is given as 110 kg

Impulse, I = F(avg) * Δt

I = -17600 * 0.055

I = -968

Now, using the Impulse-Momentum theory, we have that

Δp = I and thus,

Δp = m [v(f) - v(i)], where

Δp = change in the momentum

v(f) = final speed of the player

v(i) = initial speed of the player

Substituting the values, we have

I = m [v(f) - v(i)]

-968 = m.v(f) - m.v(i)

m.v(f) = m.v(i) - 968

110v(f) = 110 * 8 - 968

110v(f) = 880 - 968

110v(f) = -88

v(f) = -88 / 110

v(f) = -0.8 m/s

Answer:

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Explanation: