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3 years ago

How is an arranged marriage different from or similar to other ways people find and experience love today?

Physics

1 answer:

Zepler [3.9K]3 years ago

8 0

Answer:

The arrange marriage and love marriage is same or different from each other "one is about loving and then getting committed and another one is getting committed and then loving,so in the either way both are same."

Explanation:

There is no difference at all! Because eventually, in both the cases, people manage to stay together and share the responsibilities. What actually matters is that, if Love happened.

Let’s put it this way. Two people fell in love and they get married, but why is that the same couple would want to separate later? Let’s take another example. Two people who don’t have any idea of each other’s existence get married. why is that they choose to live together forever? Love becomes the reason for any split or patch.

When two people are in love everything becomes easy. A relationship doesn’t look hard, in fact I have friends who have a wonderful married life and people still don’t bother to find if that’s a love or an arranged match. Few of them choose partners of their own and few of them fell head over heels in love with their spouse, but all of them look as if they are on an extended holiday.

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You are getting bored as you are stuck at home, and want to find out the effective 'spring constant' of random objects. You find

Luda [366]

Given Information:

Mass of sock = 0.23 kg

Stretched length of sock = x = 2.54 cm = 0.0254 m

Required Information:

Spring constant = k = ?

Answer:

Spring constant = k = 88.82 N/m

Explanation:

We know from the Hook's law that

F = kx

Where k is spring constant, F is the applied force and x is length of sock being stretched.

k = F/x

Where F is given by

F = mg

F = 0.23*9.81

F = 2.256 N

So the spring constant is

k = 2.256/0.0254

k = 88.82 N/m

Therefore, the spring constant of the sock is 88.82 N/m

4 0

3 years ago

Read 2 more answers

For general projectile motion, the horizontal component of a projectile's velocity aSelect one:O a. continuously increases.O b.

liubo4ka [24]

ANSWER:

d. remains a non-zero constant.

STEP-BY-STEP EXPLANATION:

If we consider that there is no air resistance and that the horizontal component would be at x, the velocity remains a non-zero constant

3 0

1 year ago

An object is acted upon by a force of 22 newtons to the right and a force of 13 newtons to the left. What is the magnitude and d

Harman [31]

The magnitude and direction of the net force is 9 newtons to the right ⇒ answer A

Explanation:

The net force is the vector sum of all the forces that act upon an object

If two forces are horizontal where,

$F_{1}$ > $F_{2}$
They are in opposite direction
Net force F = $F_{1}$ - $F_{2}$ in the direction of $F_{1}$

An object is acted upon by a force of 22 newtons to the right

and a force of 13 newtons to the left

We need to find the magnitude and direction of the net force

→ $F_{1}$ = 22 newtons to the right

→ $F_{2}$ = 13 newtons to the left

$F_{1}$ > $F_{2}$ and they are in opposite direction

Substitute these values in the rule of the net force above

→ Net force F = 22 - 13 = 9

→ The direction of the net force is the same direction of the larger

force, the direction of the larger force is to the right

→ Net force is 9 newtons to the right

The magnitude and direction of the net force is 9 newtons to the right

Learn more:

You can learn more about the net force in brainly.com/question/4033012

#LearnwithBrainly

8 0

3 years ago

A ball is kicked horizontally at 4.6 m/s off of a cliff 13.4 m high. How far from the cliff will it land.

Mekhanik [1.2K]

<h3>Answer:</h3>

7.53 m

<h3>Explanation:</h3>

<u>We are given:</u>

Initial Horizontal Velocity of the Ball = 4.6 m/s

Initial Vertical Velocity of the Ball = 0 m/s

Height from which ball is kicked = 13.4 m

<u>Time taken by the ball to reach the ground:</u>

The ball has an initial vertical velocity of 0 m/s

it also has a downward acceleration of 10 m/s² due to gravity

<u>Solving for the time taken:</u>

s = ut + 1/2(at²) [second equation of motion]

replacing the values

13.4 = (0)(t) + 1/2 (10)(t²)

13.4 = 5t²

t² = 13.4/5 [dividing both sides by 5]

t² = 2.68

t = 1.637 seconds [taking the square root of both sides]

<u>Horizontal distance covered by the ball:</u>

Since there are no horizontal opposing forces on the ball,

the ball will more horizontally at a velocity of 4.6 m/s until it hits the ground

We calculated that the ball will hit the ground in 1.637 seconds

<u>Distance covered:</u>

s = ut + 1/2 (at²) [seconds equation of motion]

s = ut [since a = 0m/s² in the horizontal plane]

replacing the values

s = 4.6 * 1.637

s = 7.53 m

Hence, the ball landed 7.53 m from the cliff

5 0

3 years ago

35. What would a ship's position be if that ship started at

Tcecarenko [31]

Answer:

The final position of the ship after the given time period is 42 km West of B.

Explanation:

Given;

average velocity of the ship, v = 35 km/h

time taken for the ship to reach point D, t = 1.2 hours

The position of the ship after the given time period is calculated as follows;

x = v x t

x = (35 km/h) x 1.2 h = 42 km

x = 42 km West of B.

Therefore, the final position of the ship after the given time period is 42 km West of B.

5 0

3 years ago