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2 years ago

How do make friends and other people like being around u also im lonely

Physics

2 answers:

nadya68 [22]2 years ago

6 0

A good way to do that is to talk abt anything, TALK MORE NOT LESS:)

Roman55 [17]2 years ago

5 0

Answer:

I'm a lonely person too. It would be good if you found someone that are respectful.If someone approaches you, accept them.

Explanation:

But its not that easy to make friends especially when your shy and a introvert. Slowly make a progress to socialize.

You can try to impress them with your talent or kindness and humor.

As long as they are a good person. Always help people and they would automatically help you back and I believe a friendship will be bonded.

Hope this helped :D

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After the move, Objects 1 and 2 attract each other with a force of (18/16) = 1.125 units .

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A 912-kg car is being driven down a straight, level road at a constant speed of 31.5 m/s. When the driver sees a police cruiser

sergejj [24]

Answer:

786.6 N

Explanation:

mass of car, m = 912 kg

initial velocity of car, u = 31.5 m/s

final velocity of car, v = 24.6 m/ s

time, t = 8 s

Let a be the acceleration of the car

Use first equation of motion

v = u + a t

24.6 = 31.5 + a x 8

a = - 0.8625 m/s^2

Force, F = mass x acceleration

F = 912 x 0.8625

F = 786.6 N

Thus, the force on the car is 786.6 N.

8 0

2 years ago

two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is t

Tomtit [17]

The charges have opposite sign and magnitude $6 \mu C$

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have:

F = 3.60 N is the force between the two charges

r = 30 cm = 0.30 m is their separation

The two charges have same magnitude, so

$q_1 = q_2 = q$

So we can rewrite the equation as

$F=\frac{kq^2}{r^2}$

And solving for q:

$q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C$

Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

$q_1 = 6 \mu C\\q_2 = -6 \mu C$

Learn more about electric force:

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