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Reil [10]
2 years ago
5

Lemonade powder is stirred into water until it dissolves. Which are true statements?

Physics
1 answer:
ASHA 777 [7]2 years ago
6 0
Also B. Water is the solvent is correct
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Hãy nêu sự tương tự giữa công của lực điện trong trường hợp này với công của trọng lực.​
statuscvo [17]

Answer:

tốt ý bạn là gì?

Explanation:

6 0
2 years ago
A car is traveling with a velocity of 5.5 m/s and has a mass of 1200 kg. What is the kinetic energy?!
vesna_86 [32]

Answer:

<h2>18150 J</h2>

Explanation:

The kinetic energy of the car can be found by using the formula

k =  \frac{1}{2} m {v}^{2}  \\

m is the Mass

v is the velocity

From the question we have

k =  \frac{1}{2}  \times 1200 \times  {5.5}^{2}  \\  = 600 \times 30.25

We have the final answer as

<h3>18150 J</h3>

Hope this helps you

3 0
2 years ago
How will the motion of the arrow change after it leaves the bow?
Pavel [41]

The string moves to the right, as it restores its original position with the median plane of the bow. As a result, the string "pulls" on the arrow with a force F2. 2. The tip of the arrow T moves slightly to the left.

pls thank me and brainliest me

4 0
3 years ago
Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.
shusha [124]

Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

- Then compute the angle that Force makes with the y axis:

                                 cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

                                 F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

                                F_net = F_1,2 + kq / y^2

- Hence,

                                F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

- Now for the limit y >>a:

                              F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2}  )

- Insert limit i.e a/y = 0

                              F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2})  \\\\F_n = 3*k*q/y^2

Hence the Electric Field is off a point charge of magnitude 3q.

8 0
3 years ago
What type of wave borders the violet end
wariber [46]
Violet light is at the end of the visible light section of the electromagnetic spectrum. Ultraviolet rays are directly next to violet rays on the EM Spectrum.
3 0
3 years ago
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