The best name for the ionic bond that forms between them is Beryllium Bromide.
We have been provided with data,
Beryllium charge, q = 2
Bromine charge, q = -1
As we know the valance electron of Be is +2 and the valance electron of bromine is -1. Since one is metallic and the other is non-metallic.
Now, when they combine they exchange valance electron, and bromine change into bromide so they form Beryllium Bromide.
So, the best name for the ionic bond that forms between them is Beryllium Bromide.
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Answer:
I think C
Explanation:
Since the bus is moving away from John.
{C - V}.
Magnitude of the force of tension: 139 N
Explanation:
The surface of the ramp here is assumed to be the positive x-direction.
To solve this problem and find the magnitude of the force of tension, we have to analyze only the situation along the x-direction, since the force of tension lie in this direction.
There are three forces acting along the x-direction:
- The force of tension,
, acting up along the plane - The force of friction,
, acting down along the plane - The component of the weight in the x-direction,
, acting down along the plane
We know that the magnitude of the weight is

So its x-component is

The net force along the x-direction can be written as

And therefore, since the net force is 98 N, we can find the magnitude of the force of tension:

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Answer:
All
Explanation:
I'm not sure what you meant but Newton's third law which basically states that every action has an equal and opposite reaction applies to <em>all</em> objects. So I think the answer is all.
Answer:
3.28 cm
Explanation:
To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:
r = 
Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.
The mass and charge of a proton are:
m = 1.67 * 10^-27 kg
q = 1.6 * 10^-19 C
So, we get that the radius r will be:
r =
= 0.0328 m, or 3.28 cm.