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Zielflug [23.3K]
3 years ago
14

When creating any map, what four basic elements should you include?

Physics
2 answers:
melisa1 [442]3 years ago
5 0
Title,Scale,Date of Publication,North Arrow (Legend,Location Information,and Source of Information)
hoa [83]3 years ago
4 0
You must include a key, title, date of publication, and scale.
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Which of the following situations describes a change that will result in a new kind of matter with different characteristics?
taurus [48]

Answer:

Chemical property - characteristic of something that allows it to change to something new.

Explanation:

please make me braniest if I'm r8

6 0
2 years ago
The charges and coordinates of two charged particles held fixed in the xy plane are: q1 = +3.3 µc, x1 = 3.5 cm, y1 = 0.50 cm, an
Readme [11.4K]
1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
substituting the coordinates of the two charges, we get
d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m

2) Then, we can calculate the electrostatic force between the two charges q_1 and q_2, which is given by
F=k_e  \frac{q_1 q_2}{d^2}
where k_e=8.99\cdot10^{9} Nm^2C^{-2} is the Coulomb's constant.
Substituting numbers, we get 
F=8.99\cdot10^{9} Nm^2C^{-2}  \frac{(3.3\cdot10^{-6}~C) (-4\cdot10^{-6}~C)}{(0.056~m)^2} =-37.8~N
and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.
7 0
3 years ago
Can a body have increasing velocity with decreasing acceleration?
Sergio039 [100]
Sure.  The acceleration may be decreasing, but as long as it stays
in the same direction as the velocity, the velocity increases.

I think you meant to ask whether the body can have increasing velocity
with negative acceleration.  That answer isn't simple either.

If the body's velocity is in the positive direction, then positive acceleration
means speeding up, and negative acceleration means slowing down.

BUT ... If the body's velocity is in the negative direction, then positive
acceleration means slowing down, and negative acceleration means
speeding up.

I know that's confusing. 

-- Take a piece of scratch paper, write a 'plus' sign at one edge and
a 'minus' sign at the other edge.  Those are the definitions of which
direction is positive and which direction is negative. 

-- Then sketch some cars ... one traveling in the positive direction, and
one driving in the negative direction.  Those are the directions of the
velocities.

-- Now, one car at a time:
. . . . . first push on the back of the car, in the direction it's moving;.
. . . . . then push on the front of the car, against its motion.
Each push causes the car to accelerate in the direction of the push.

When you see it on paper, all the positive and negative velocities
and accelerations will come clear for you.
3 0
3 years ago
Suppose the acceleration of the particle moving gin a circle of radius "r" with the uniform speed "v" is proportional to some po
garik1379 [7]
So i say the power or the radius cortex in my equinox when i look at this V is the determination of V
5 0
2 years ago
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
3 years ago
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