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3 years ago

Discuss in detail the manners of interaction with opposite gender

Engineering

1 answer:

siniylev [52]3 years ago

6 0

Answer:

8 Tips on Better Communication with the Opposite Sex

Put emotions away. Ladies, this one is more aimed at us, for the most part. ...

Forget your pride. In discussions, especially these days, people always want to be the one that prevails. .

Put yourself in their shoes. .

Listen. ...

Respond. ...

Actually communicate. ...

Be detailed. ...

Don't communicate too much.

Explanation:

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. What's an object in your everyday life that has a lot of Kinetic Energy? How do you know it has a lot?

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3 years ago

A woman approaches a room attendant with a letter. She requests that the attendant leave the letter in Room 1201. Which is the m

MrRa [10]

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3 years ago

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Genrish500 [490]

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2 years ago

A ball is dropped from rest from the top of a cliff that is 30 m high. From ground

trasher [3.6K]

The distance below the top of the cliff that the two balls cross paths is 7.53 meters.

<u>Given the following data:</u>

Initial velocity = 0 m/s (since the ball is dropped from rest).

Height = 30 meters.

<u>Scientific data:</u>

Acceleration due to gravity (a) = 9.8 $m/s^2$ .

To determine how far (distance) below the top of the cliff that the two balls cross paths, we would apply the third equation of motion.

<h3>How to calculate the velocity.</h3>

Mathematically, the third equation of motion is given by this formula:

$V^2 = U^2 +2aS$

<u>Where:</u>

V is the final velocity.

U is the initial velocity.

a is the acceleration.

S is the distance covered.

Substituting the parameters into the formula, we have;

$V^2 = 0^2 +2(9.8) \times 30\\\\V^2 = 588\\\\V=\sqrt{588}$

V = 24.25 m/s.

<u>Note:</u> The final velocity of the first ball becomes the initial velocity of the second ball.

The time at which the two balls meet is calculated as:

$Time = \frac{S}{U} \\\\Time = \frac{30}{24.25}$

Time = 1.24 seconds.

The position of the ball when it is dropped from the cliff is calculated as:

$y_1 = h-\frac{1}{2} at^2\\\\y_1 = 30-\frac{1}{2} \times 9.8 \times 1.24^2\\\\y_1 = 30-7.53\\\\y_1=22.47\;meters$

Lastly, the distance below the top of the cliff is calculated as:

$Distance = 30-22.47$

Distance = 7.53 meters.

Read more on distance here: brainly.com/question/10545161

4 0

3 years ago

Estimate the energy (head) loss a short length of a pipe conveying 300 litres of water per second and suddenly enlarging from a

Ludmilka [50]

Known :

Q = 300 L/s = 0.3 m³/s

D1 = 350 mm = 0.35 m

D2 = 700 mm = 0.7 m

g = 9.81 m/s²

Solution :

A1 = πD1² / 4 = π(0.35²) / 4 = 0.096 m²

A2 = πD2² / 4 = π(0.7²) / 4 = 0.385 m²

hL = (kL / 2g) • (U1² - U2²)

hL = (kL / 2g) • Q² (1/A1² - 1/A2²)

hL = (1 / 2(9.81)) • (0.3²) • (1/(0.096²) - 1/(0.385²))

hL = 0.467 m

5 0

3 years ago