If two solutions have unequal concentrations of a solute, the solution with the lower concentration is called

Calculate how far a ball will travel horizontally if the ball reaches a high of 1.5 m above the ground and is thrown at a horizo

nataly862011 [7]

67.5

hope this is right

6 0

2 years ago

Look at the two wave diagram.Which best describes the difference between wave A and wave B

Goshia [24]

Answer:

B. Wave B has greater intensity and transfer more energy.

Explanation:

Amplitude of wave B is more and it seems to be compact in less period of time as compared to Wave A.

Yeah_Boi ;)

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5 0

2 years ago

A ball is dropped from rest at point O . It passes a window with height 3.8 m in time interval tAB = 0.02 s.Identify the correct

Taya2010 [7]

Answer:

VB − VA = g tAB & (VA + VB)/2 = h / tAB

Explanation:

s = h = Displacement

tAB = t = Time taken

VA = u = Initial velocity

VB = v = Final velocity

a = g = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{3.8-\frac{1}{2}\times 9.8\times 0.02^2}{0.02}\\\Rightarrow u=189.902\ m/s$

$v=u+at\\\Rightarrow v=189.902+9.8\times 0.02\\\Rightarrow v=190.098\ m/s$

$\frac{v+u}{2}=\frac{190.098+189.902}{2}=190\ s$

$\frac{h}{t}=\frac{3.8}{0.02}=190\ s$

Hence, the equations VB − VA = g tAB & (VA + VB)/2 = h / tAB will be used

3 0

2 years ago

Define Velocity with an example

kolbaska11 [484]

Velocity stands for Displacement w.r.t time

$\\ \bull\tt\longmapsto Velocity=\dfrac{Displacement}{Time}$

Or

$\\ \bull\tt\longmapsto v=\dfrac{ds}{dt}$

It has units m/s.

8 0

2 years ago

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HELP PLZZZ!!!! Hurry

Mashcka [7]

Answer:

3. When the number of turns, N is doubled, the strength of the electromagnet is also doubled

4. Doubling the voltage, doubles the strength of the electromagnet

5. The number of paper clips a 7.5 V battery would pick is approximately 28 paper clips

The number of paper clips a 7.5 V battery would pick is 59 paperclips

6. For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is approximately 7 paperclips

For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is 16 paperclips

Explanation:

3. The Magnetomotive Force, MMF = The Number of Turns on the Coil, N × The Current I Flowing in the Coil, I

∴ MMF = N × I

When the number of turns, N is doubled, the magnetomotive force, MMF is also doubled, and the strength of the electromagnet is doubled

4. Given that the voltage, V applied to the coil = The current, I flowing × The resistance, R of the coil, we have

V = I × R

Therefore, for a given constant resistance in the coil, doubling the voltage, doubles the current and therefore doubles the strength of the electromagnet

5. The average slope for the 25-coil electromagnet = (23 - 12)/(6 - 3) = 3. $\bar 6$

The number of paper clips a 7.5 V battery would pick = 12 + (7.5 - 3) × 11/3 = 28.5 paperclips ≈ 28 paper clips

The average slope for the 50-coil electromagnet = (48 - 26)/(6 - 3) = 7. $\bar 3$

The number of paper clips a 7.5 V battery would pick = 26 + (7.5 - 3) × 22/3 = 59 paperclips

6. The slope calculated from a start point of approximately 0.4 V, is given as follows;

The slope for the 25-coil electromagnet = (12 - 6)/(3 - 0.4) = 30/13

Therefore, for the 25-coil electromagnet, the average number of paper clips a 1 V battery would pick = 6 + (1 - 0.4) × 30/13) = 96/13 ≈ 7 paperclips

The slope for the 50-coil electromagnet = (26 - 13)/(3 - 0.4) = 5

Therefore, for the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick = 13 + (1 - 0.4) × 5 = 16 paperclips

8 0

3 years ago

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