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Pani-rosa [81]
2 years ago
11

A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s

Physics
1 answer:
Natasha2012 [34]2 years ago
6 0

Explanation:

There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

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lesantik [10]

Explanation:

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              \frac{3.19}{8.46}

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Hence,  frequency (f) = 0.38 Hz

Now, formula to calculate the speed is as follows.

               v = f \times \lambda

or,        \lambda = \frac{v}{f}                  

                = \frac{0.6 cm/s}{0.38}

                = 1.57 cm

Thus, we can conclude that the wavelength is 1.57 cm.

3 0
3 years ago
What is acceleration of a body moving with uniform velocity? (convert 50km/hr into m/s)​
gregori [183]

Answer:

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Explanation:

5 0
2 years ago
A 1.50-m-long rope is stretched between two supports with a tension that makes the speed of transverse waves 48.0 m/s. What are
astra-53 [7]

Answer: a) 16Hz, 3m b) 48Hz, 1mc) 80Hz, 0.6m

Explanation:

a) Fundamental frequency in string is represented as Fo = V/2L where;

Fo is the fundamental frequency

V is the speed of the transverse wave = 48m/s

L is the length of the wire. = 1.50m

Substituting this values in the formula given we have;

Fo = 48/2(1.5)

Fo = 48/3

Fo = 16Hz

The fundamental tone is therefore 16Hz

Using v =f¶

Where f is the frequency and ¶ is the wavelength, the wavelength of the fundamental note will be;

¶ = v/fo

¶ = 48/16 = 3m

b) Overtones or harmonics is the multiple integral of the fundamental frequency. The multiples are I'm arithmetical progression.

First overtone f1 = 2fo

Second overtone f2 = 3fo etc.

Since fo = 16Hz

Second overtone f2 = 3×16 = 48Hz

¶ = v/f2 = 48/48

¶ = 1m

c) Fourth harmonic or overtone will be f4 = 5fo

F4 = 5×16 = 80Hz

The fourth harmonic is therefore 80Hz

¶ = v/f4 = 48/80

¶ = 0.6m

4 0
3 years ago
Rank the six combinations of electric charges on the basis of the electric force acting on q1. Define forces pointing to the rig
g100num [7]

I have attached the image showing the 6 combination of charges

Answer:

Smallest are: q1 = +1 nc, q2 = +1 nc, q3 = +1 nc and q1 = -1 nc, q2 = -1 nc, q3 = -1nc

Third Largest: q1 = +1 nc, q2 = +1nc, q3 = -1 nc

Second Largest: q1 = +1 nc, q2 = -1 nc, q3 = +1 nc

Largest are: q1 = +1 nc, q2 = -1 nc, q3 = -1 nc and q1 = -1 nc, q2 = +1 nc, q3 = +1 nc

Explanation:

The electrostatic force between 2 charges is; F = k•q1•q2/r²

Where r is the distance between the 2 charges q1 and q2 and k is coulombs constant.

Thus;

F ∝ q1•q2/r²

The net force on the charge q1 is the sum of the forces on charge q2 and q3.

If the two charges are opposite, the force is an attractive force while if the two charges are similar, the force is a repulsive force.

Thus, if the combination has same type of charges, i.e;(+1,+1,+1) or (-1,-1,-1), the force will be very small.

So, now the order of forces from smallest to largest is;

Smallest are: q1 = +1 nc, q2 = +1 nc, q3 = +1 nc and q1 = -1 nc, q2 = -1 nc, q3 = -1nc

Third Largest: q1 = +1 nc, q2 = +1nc, q3 = -1 nc

Second Largest: q1 = +1 nc, q2 = -1 nc, q3 = +1 nc

Largest are: q1 = +1 nc, q2 = -1 nc, q3 = -1 nc and q1 = -1 nc, q2 = +1 nc, q3 = +1 nc

3 0
2 years ago
A grape is thrown straight up in the air. If it was thrown at 2 m/s, how high will it go in the air?
ser-zykov [4K]

Answer:

Answer: <u>Height</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>2</u><u>0</u><u>4</u><u> </u><u>m</u>

Explanation:

At the highest point, it is called the maximum height.

• From third newton's equation of motion:

{ \rm{ {v}^{2}  =  {u}^{2}  + 2gs}}

• At maximum height, v is zero

• u is initial speed

• g is -9.8 m/s²

• s is the height

{ \rm{0 {}^{2}  =  {2}^{2}   -  (2 \times 9.8 \times s)}} \\  \\ { \rm{4 = 19.6s}} \\  \\ { \rm{s = 0.204 \: m}}

4 0
2 years ago
Read 2 more answers
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