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Tanzania [10]
4 years ago
11

What is the average velocity of an arrow that travels 12 m [E] in 0.15 s?

Physics
1 answer:
sergij07 [2.7K]4 years ago
6 0
Velocity is defined as Distance divided by Time.
In other words, V = D/T.

Now that we have our formula, we can solve.
Let's plug in the numbers we have.

We have 12m [East (direction not necessary when solving yet)] for our distance, and 0.15s as our time.

Divide the distance (12 /) by the time (0.15)
12 / 0.15 = 80.

Your velocity is 80 m/s [E]

I hope this helps!
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Why electrons flow in a wire when connected to a battery?
vesna_86 [32]

Answer:

The metal atoms in the wire can't move, but their outer electrons can. The force pushes those electrons and they move to further parts of the wire, trying to reach the other end. As the electrons move away, new electrons flow into the wire through the battery to take their place.

Explanation:

6 0
3 years ago
In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur
Margaret [11]

Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)

First, we isolate C from equation (2):

2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)

using this value of C from equation (3) in equation (1):

3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}

<u>D = - 0.57</u>

Put this value in equation (3), we get:

C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C =  \frac{17}{7}\\

<u>C = 2.43</u>

5 0
3 years ago
The energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard
gogolik [260]

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

|\Delta E_{ij}| = |E_i-E_j|

Our states are given by

E_1 = -11.7eV

E_2 = -4.2eV

E_3 = -3.3eV

In this way the energy balance for the states would be given by,

|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\

|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV

|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV

Therefore the states of energy would be

Lowest : 0.9eV

Middle :7.5eV

Highest: 8.4eV

8 0
4 years ago
Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of
WINSTONCH [101]

a₀).  You know ...
         -- the object is dropped from 5 meters
             above the pavement;
         --  it falls for 0.83 second.

a₁).  Without being told, you assume ...
         -- there is no air anyplace where the marshmallow travels,
             so it free-falls, with no air resistance;
         -- the event is happening on Earth,
            where the acceleration of gravity is  9.81 m/s² .

b).  You need to find how much LESS than 5 meters
       the marshmallow falls in 0.83 second.
    
c).  You can use whatever equations you like.
       I'm going to use the equation for the distance an object falls in
       ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d).  To see how this all goes together for the solution, keep reading:


The distance that an object falls in ' T ' seconds
when it's dropped from rest is

                                 (1/2 G) x (T²) .

On Earth, ' G ' is roughly  9.81 m/s², so in 0.83 seconds,
such an object would fall

                               (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

                         (5.00 - 3.38)  =  1.62 meters

above the pavement.  That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.

7 0
3 years ago
witch of the following is the least reliable source of background information for a scientific project.
Tems11 [23]

Primary sources are most often produced around the time of the events you are studying.


hopwe this helps

3 0
4 years ago
Read 2 more answers
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