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BaLLatris [955]
3 years ago
12

How can we drop eggs the fewest amount of times without breaking it?​

Physics
1 answer:
enot [183]3 years ago
4 0

Answer:

drop in water

Explanation:

follow me m mm m

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A piece of wire 29 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria
skelet666 [1.2K]

Answer:

Explanation:

Total length of the wire is 29 m.

Let the length of one piece is d and of another piece is 29 - d.

Let d is used to make a square.

And 29 - d is used to make an equilateral triangle.

(a)

Area of square = d²

Area of equilateral triangle = √3(29 - d)²/4

Total area,

A = d^{2}+\frac{\sqrt3}{4}\left ( 29-d \right )^{2}

Differentiate both sides with respect to d.

\frac{dA}{dt}=2d- \frac{\sqrt3}{4}\times 2(29-d)

For maxima and minima, dA/dt = 0

d = 8.76 m

Differentiate again we get the

\frac{d^{2}A}{dt^{2}}= + ve

(a) So, the area is maximum when the side of square is 29 m

(b) so, the area is minimum when the side of square is 8.76 m

8 0
3 years ago
A skydiver has a mass of 65 kg.<br> Calculate the weight of the skydiver.
Vinil7 [7]
F=m*a
F=65 kg *9.8 m/s^2
F=637 N (Newtons) — this is the weight
3 0
3 years ago
approximation to the average velocity in that time interval, what should be the sequence of calculations?Update the (vector) pos
Klio2033 [76]

Answer:

The steps are outlined in the explanation below.

Explanation:

The average velocity is derived midpoint from the initial to the final velocity. Here is the proof:

Find the total displacement:

let the displacement be given by the letter s

Then since the average velocity is defined as:  v_{av}  = \frac{x - x_{0} }{t - t_{0} }

where t = final time

           t₀ = initial time

           v = final speed

           v₀ = initial time

where x denotes the position, then

v_{ave} = \frac{v+v_{0} }{2}

where v = \frac{dx}{dt} and dx = change in distance with respect to time.

6 0
3 years ago
NEED HELP ASAP WILL MARK BRAINLIEST.
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G is the answer for apex vs / Chehhhh
3 0
3 years ago
If a particle undergoes shm with amplitude 0.21 m, what is the total distance it travels in one period?
Ymorist [56]
If a particle undergoes simple harmonic motion with an amplitude of 0.21 meters, this means that the maximum displacement of the particle from its resting position is 0.21. For one period, it traveled from its starting position which is twice the amplitude and then back to its original position which is another distance that is twice the amplitude as well. Therefore, the total distance it traveled is 2*amplitude + 2*amplitude = 2*0.21 + 2*0.21 = 0.42 + 0.42 = 0.84 meters.
7 0
3 years ago
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