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erastovalidia [21]
3 years ago
8

When the distance between two charges is halved, the electrical force between them?

Physics
1 answer:
Llana [10]3 years ago
3 0
If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:
F= k \frac{q_1 q_2}{r^2}
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
r'= \frac{r}{2}
the magnitude of the force changes as follows:
F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k  \frac{q_1 q_2}{r^2}=4 F
so, the force increases by a factor 4.
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The whooping crane (Grus americana) is the tallest bird native to North America. It almost went extinct in the 1900s; in 1938, t
grin007 [14]

Answer:

The value  is   a =  2.7183

Explanation:

From the question we are told that

   The  relationship between the number of whooping cranes and the number of decades is ln N =  2.85 + 0.039t

  The  exponential relationship is  N  =  na^{kt}

Now from the given equation we have that

       N  =   e^{2.85 + 0.039t}

       N  =  17.29 e^{0.039t}

So comparing this equation obtained an the given  exponential relationship we have that

      n = 17.29

       k  =  0.039

        a =  e =  2.7183

        a =  2.7183

 

7 0
3 years ago
In a certain process, the energy change of the system is 250 \rm kJ. The process involves 480 \rm kJ of work done by the system.
alisha [4.7K]

Answer:

\Delta Q=-230kJ

Explanation:

Using the first law of thermodynamics:

\Delta U=\Delta Q-W

Where \Delta U is the change in the internal energy of the system, in this case  \Delta U=250kJ, \Delta Q is the heat tranferred, and W is the work,  W=-480kJ with a negative sign since the work is done by the system.

From the previous equation we solve for heat, because it is the unknown variable in this problem

\Delta Q=\Delta U +W

And replacing the known values:

\Delta Q=250kJ +(-480kJ)

\Delta Q=250kJ -480kJ

\Delta Q=-230kJ

The negative sign shows us that the heat is tranferred from the system into the surroundings.

3 0
4 years ago
A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220
disa [49]
First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, \mu m g. For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
ma=-\mu m g
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
2aS=v_f^2-v_i^2
where v_f=0 is the final speed of the skier and v_i=4.75 m/s is the initial speed. Substituting numbers, we find:
S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m
5 0
3 years ago
20 points and brainliest‼️‼️‼️‼️
Anastasy [175]

Answer:

0 N

Explanation:

Applying,

F = qvBsin∅................. Equation 1

Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.

From the question,

Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°

Substitute these values into equation 1

F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)

Since sin0° = 0,

Therefore,

F = 0 N

3 0
3 years ago
"I could feel his anguish" what could be the anguish​
sergey [27]

Answer:

yeah

Explanation:

the answer is right.

3 0
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