From her bedroom window a girl drops a water-filled balloon to the ground, 4.77 m below. If the balloon is released from rest, how long is it in the air?


Answer 1
Answer: We need to use the equation x = vt + (1/2)at^2. We know x = 4.77, v = 0, and a = 9.81m/s^2. Plug in the values. 4.77 = (0)t + (1/2)(9.81)t^2 Solve for t. 4.77 = (4.905)t^2 0.972 = t^2 t = (sq.rt)_/0.972 t = 0.985 So it's in the air 0.985 seconds.

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I do not know the answer sorry :(

If a material has an index of refraction of 1.61, Determine the speed of light through this medium



1.86 x 10^8 m/s


n = 1.61

The formula for the refractive index is given by

n = speed of light in vacuum / speed of light in material

n = c / v

v = c / n

v = (3 x 10^8) / 1.61

v = 1.86 x 10^8 m/s

Final answer:

The speed of light in a material with an index of refraction of 1.61 is calculated as approximately 1.86 * 10^8 m/s, using the equation v = c/n where c is the speed of light in vacuum and n is the index of refraction.


The speed of light in a given material can be calculated using the index of refraction of the material, as defined by the equation n = c/v, where n is the index of refraction, c is the speed of light in a vacuum, and v is the speed of light in the material.

Given that the index of refraction for the material in question is 1.61, and the speed of light in vacuum, c = 3.00 * 10^8 m/s, the speed of light v in this medium would therefore be calculated by rearranging the equation to v = c/n.


By substituting the given values into the equation, v = 3.00 * 10^8 m/s / 1.61, we find that the speed of light in the material is approximately 1.86 * 10^8 m/s.

Learn more about Speed of Light here:


For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the length is p1. If the length of the pipe is then doubled, what is the relation of the new pressure drop p2 to the original pressure drop p1 at the original mass flow rate?


Answer: ∆p2 = 2* ∆p1


Given that all other factors remain constant. The pressure drop across the pipeline is directly proportional to the length.

i.e ∆p ~ L


∆p2/L2 = ∆p1/L1

Since L2 = 2 * L1

∆p2/2*L1 = ∆p1/L1

Eliminating L1 we have,

∆p2/2 = ∆p1

Multiplying both sides by 2

∆p2 = 2 * ∆p1

What type of electromagnetic radiation is being shown in the picture?A. Gamma rays
B. Ultraviolet radiation
C. X-rays
D. Infrared radiation



I think D. Infrared radiation.


infrared radition



A person is pushing a lawnmower of mass m D 38 kg and with h D 0:75 m, d D 0:25 m, `A D 0:28 m, and `B D 0:36 m. Assuming that the force exerted on the lawnmower by the person is completely horizontal and that the mass center of the lawnmower is at G, and neglecting the rotational inertia of the wheels, determine the minimum value of this force that causes the rear wheels (labeled A) to lift off the ground. In addition, determine the corresponding acceleration of the mower.



The acceleration of the mower will be "4.7 m/s²".


Balance of vertical force will be:

⇒  Ra + Rb = mg

For wheel to take off at A,

⇒  Ra = 0



Balancing moments about G will be:

⇒  F* h = Rb* LB

As we know,

Force, F = (Rb* LB )/(h)

On putting the values, we get

⇒           = (38* 9.81* 0.36)/(0.75)

⇒           = 178.9 \ N


Acceleration, a = (F)/(m)

⇒                       = (178.9)/(38)

⇒                       = 4.7 \ m/s^2

I'm a little bit unsure about this question.



Option C. 4 Hz


To know the correct answer to the question given above, it is important we know the definition of frequency.

Frequency can simply be defined as the number of complete oscillations or circles made in one second.

Considering the diagram given above, the wave passes through the medium over a period of one second.

Thus, we can obtain the frequency by simply counting the numbers of complete circles made during the period.

From the diagram given above,

The number of circles = 4


The frequency is 4 Hz