Answer:
se me hace que son 8 protones
Answer:
1.7×10^5 ms-1
Explanation:
From
qE= qvB
q= charge on the electron
E = electric field
v= velocity
B= magnetic field
E= vB
v= E/B= 110×10^3/0.6
v= 1.7×10^5 ms-1
We are given a box that slides up a ramp. To determine the force of friction we will use the following relationship:

Where.

To determine the Normal force we will add the forces in the direction perpendicular to the ramp, we will call this direction the y-direction as shown in the following diagram:
In the diagram we have:

Adding the forces in the y-direction we get:

Since there is no movement in the y-direction the sum of forces must be equal to zero:

Now we solve for the normal force:

To determine the y-component of the weight we will use the trigonometric function cosine:

Now we multiply both sides by "mg":

Now we substitute this value in the expression for the normal force:

Now we substitute this in the expression for the friction force:

Now we substitute the given values:

Solving the operations:

Therefore, the force of friction is 15.01 Newtons.
With that information you can only suppose a uniformly accelerated motion. This is, acceleration is constant.
Then, acceleration = change in velocity / change in time = (58 -54)km/h / 2 h = 4km/h / 2 h = 2 km/h^2
Then the equation for velocity, V is
V = Vo + a*t = Vo + 2 (km/h^2) * t = Vo + 2t
Vo is the initial velocity, which you can find using V = 54km/h and t = -2
Vo = V after 2 hours - a*(2hours) = 54km/h - 2(km/h^2)*2h = 54km/k - 4km/h = 50km/h
Then, the equation is: V = 50 km/h + 2t
Valid for constant acceleration.
Answer:
<em>The total potential (magnitude only) is 11045.45 V</em>
Explanation:
<u>Electric Potential
</u>
The total electric potential at location A is the sum of all four individual potentials produced by the charges, including the sign since the potential is a scalar magnitude that can be computed by

Where k is the Coulomb's constant, q is the charge, and r is the distance from the charge. Let's find the potential of the rightmost charge:

The potential of the leftmost charge is exactly the same as the above because the charges and distances are identical

The potential of the topmost charge is almost equal to the above computed, is only different in the sign:

The bottom charge has double distance and the same charge, thus the potential's magnitude is half the others':

The total electric potential in A is


The total potential (magnitude only) is 11045.45 V