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Bad White [126]
3 years ago
5

When does an atom become an ion?

Physics
2 answers:
Mice21 [21]3 years ago
6 0

Answer:

When a stable atom gains or loses an electron

Explanation:

Sonbull [250]3 years ago
5 0
The answer above is correct
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A 1.5-kilogram cart initially moves at 2.0 m/s. It is brought to rest by a constant net force in 0.3 seconds. What is the magnit
ser-zykov [4K]
We know, F = m.a
Here, m = 1.5 Kg
a = v/t = 2/0.3 = 6.67 m/s

Substitute their values, 
F = 1.5 * 6.67
F = 10 N

In short, Your Answer would be 10 Newtons

Hope this helps!
5 0
4 years ago
Numeria
Vinil7 [7]

Answer:

hope it helps

plz mark as brainliest

5 0
3 years ago
Read 2 more answers
A uniform electric field with a magnitude of 125 000 N/C passes through a rectangle with sides of 2.50 m and 5.00 m. The angle b
jeyben [28]

Answer:

6.60\cdot 10^5 Nm^2/C

Explanation:

The electric flux through the rectangle is given by

\Phi = E A cos \theta

where

E is the electric field strength

A is the area of the rectange

\theta is the angle between the direction of the electric field and of the vector normal to the plane of the rectangle

In this problem we have

E = 125 000 N/C

The area of the rectangle is

A=2.50 m \cdot 5.00 m=12.5 m^2

and the angle is

\theta=65.0^{\circ}

so, the electric flux is

\Phi = (125,000 N/C)(12.5 m^2)(cos 65^{\circ})=6.60\cdot 10^5 Nm^2/C

8 0
4 years ago
What is bricks matter rigidity ​
expeople1 [14]

Answer:

In a tabular column represent the following characteristics of matter-rigidity, compressibility, shape, kinetic energy and density of following substances- brick, honey, air, block of wood, water

Explanation:

3 0
3 years ago
13. If you shorten the length of string by half that holds an object in rotation at the same tangential
Dmitrij [34]

13. doubles

The tension in the string corresponds to the centripetal force that holds the object in rotation, so:

T=F=m\frac{v^2}{r}

where m is the mass of the object, v is the tangential speed, and r is the distance of the object from the centre of rotation (therefore it corresponds to the length of the string). The problem tells us that the tangential speed remains the same (v), while the length of the string is halved, so r'=r/2. Therefore, the new tension in the string will be

T'=m\frac{v^2}{r'}=m\frac{v^2}{r/2}=2m\frac{v^2}{r}=2T

so, the Tension doubles.

14. Variations of centripetal forces

Both revolution and rotation refer to the rotational motion of an object, therefore they both involve the presence of a centripetal force, which keeps the object in circular motion. The only difference between the two is:

- Revolution is the circular motion of an object around a point external to the object (for instance, the motion of the Earth around the Sun)

- Rotation is the circular motion of an object around its centre, so around a point internal to the object (for instance, the rotation of the Earth around its axis)

15. Rotational speed

For a uniform object in circular motion, all the points of the object have same rotational speed. In fact, the rotational speed is defined as

\omega=\frac{\Delta \theta}{\Delta t}

where \Delta \theta is the angular displacement covered in a time interval of \Delta t. Since all the points of the wheel are coeherent (they move together), they all cover the same angular displacement in the same time, so they all have same rotational speed.

16. away from the center of the path.

The tension in the string is responsible for keeping the tin can in circular motion. Therefore, the tension in the string represents the centripetal force, and so it is directed towards the centre of the path. According to Newton's third law, the tin can exerts a force on the string which is equal in magnitude (so, same magnitude of the tension), but opposite in direction: therefore, away from the centre of the path.

17. weight of the bob.

There are two forces acting on the bob in the vertical direction: the weight of the bob (downward) and the vertical component of the string tension (upward). Since there is no acceleration along the vertical direction, the net force must be zero, so these two forces must be equal: it means that the vertical component of the string tension is equal to the weight of the bob. Along the horizontal direction, instead, the horizontal component of the string tension corresponds to the centripetal force that keeps the bob in circular motion.

18. horizontal component of string tension.

Along the horizontal direction, there is only one force acting on the bob: the horizontal component of the string tension. Since the bob is moving of circular motion along the horizontal direction, this means that this force (the horizontal component of the string tension) must correspond to the centripetal force that keeps the pendulum in circular motion.

19. inward, toward the center of swing.

The force that the can exerts on the bug is the force that keeps the bug in circular motion (since it prevents the bug from moving away). Therefore, it must corresponds to the centripetal force.

20. speed of the car. AND radius of curvature.

The normal force exerted on a car executing a turn on a banked track is given by the expression:

N=\frac{mg}{cos \theta - \mu sin \theta}

where m is the mass of the car, g is the gravitational acceleration, \theta is the angle of the bank, and \mu is the coefficient of friction.

From the formula, we see that the normal force depends on \theta (the angle of the bank) and \mu (the coefficient of friction), while it does not depend on the speed of the car or on the radius of curvature. Therefore, these two are the correct answers.

3 0
3 years ago
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