What factor has the greatest impact on flexibility?

please answer as soon as possible. How does the gravitational pull of different objects in space affect other objects in space?

elena55 [62]

Answer:

Every object in space exerts a gravitational pull on every other

explanation : gravity influences the paths taken by everything traveling through space. It is the glue that holds together entire galaxies. It keeps planets in orbit. It can also cause life-destroying asteroids to crash into planets..

6 0

2 years ago

Charge A and charge B are 2.2 m apart. Charge A is 1.0 C, and charge B is

Ipatiy [6.2K]

Its really hurts

Explanation:

Charge A and charge B are 2.2 m apart. Charge A is 1.0 C, and charge B is

2.0 C. Charge C, which is 2.0 C, is located between them and is in

electrostatic equilibrium. How far from charge A is charge C?

6 0

2 years ago

How can a soccer ball and a small tennis ball bouncing relates to newtons first law interita?

Anna35 [415]

Answer:

I think it has something to do with both inertia and gravity?

7 0

2 years ago

What’s the formula for work

padilas [110]

Answer:

Fd

Explanation:

Work is force times distance. If you push on an object really hard but it does not budge, you have still performed no work on it, because anything times zero is still zero.

6 0

2 years ago

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A straight segment of a current-carrying wire has a current element IL where I = 2.70 A and L = 3.20 cm i + 4.30 cm j. The segme

myrzilka [38]

The component of the force in negative z-direction is -0.144 N.

The given parameters;

current in the wire, I = 2.7 A
length of the wire, L = (3.2 i + 4.3j) cm
magnetic filed, B = 1.24 i

The force on the segment of the wire is calculated as follows;

$F = ILBsin(\theta)$

where;

θ is the angle wire and magnetic field

The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.

The angle between the wire and the magnetic field is calculated as follows;

$\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{4.3}{3.2} )\\\\\theta = 53.3 \ ^0$

The magnitude of the wire length is calculated as follows;

$|l | = \sqrt{3.2^2 + 4.3^2} = 5.36 \ cm = 0.0536 \ m$

The component of the force in negative z-direction is calculated as;

$F_z = -ILB sin(\theta)\\\\F_z = -2.7 \times 0.0536 \times 1.24 \times sin(53.3)\\\\F_z = -0.144 \ N$

Thus, the component of the force in negative z-direction is -0.144 N.

Learn more here:brainly.com/question/22719779

6 0

2 years ago