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Lelu [443]
3 years ago
5

What is an ellipse? ​

Physics
2 answers:
Rzqust [24]3 years ago
6 0

Answer:An ellipse is a closed curve consisting of points whose distances from each of two fixed points (foci) all add up to the same value .

Explanation:

storchak [24]3 years ago
3 0

Answer:

A closed curve consisting of points whose

distances from each of two fixed points (foci) all add up to the same value is an ellipse. The midpoint between the foci is the center. One property of an ellipse is that the reflection off its boundary of a line from one focus will pass through the other.

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lara [203]
He created an atomic model and Bohr diagram, if that could be it.
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3 years ago
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Rank in order, from largest to smallest, the magnitudes of the electric field at the black dot. A. 2, 1, 3, 4 B. 1, 4, 2, 3 C. 3
sweet [91]

Given that,

Rank in order from largest to smallest the magnitude of the electric field at block dot.

Electric field :

Electric field is proportional to the charge divided by square of distance.

In mathematically,

E\propto\dfrac{q}{r^2}

Where, q = charge

r = distance

If the charge is greater then electric field will be greater.

If the distance is greater then electric field will be smaller.

We need to find the electric field at black dot

According to figure,

(I). The electric field at black dot due to positive charge point q to left. the distance is r.

The electric field will be

E=\dfrac{kq}{r^2}

The electric field will be largest.

(II). The electric field at black dot due to positive charge point 2q to left. The distance is 2r.

Then, the electric field will be

E=\dfrac{k2q}{(2r)^2}

E=\dfrac{kq}{2r^2}

The electric field will be smallest.

(III).  The electric field at black dot due to positive charge point 2q to left. The distance is r.

Then, the electric field will be

E=\dfrac{k2q}{(r)^2}

The electric field will be very largest.

(IV). The electric field at black dot due to positive charge point q to left. The distance is 2r.

Then, the electric field will be

E=\dfrac{kq}{(2r)^2}

E=\dfrac{kq}{4r^2}

The electric field will be very smallest.

So, The electric field from largest to smallest will be

E_{3}>E_{1}>E_{2}>E_{4}

Hence, The ranking will be 3, 1, 2, 4.

(D) is correct option.

4 0
3 years ago
How will the motion of an object that is moving to the right change, if it is pushed in the opposite direction with a greater fo
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4 0
3 years ago
1. una masa oscila a la frecuencia de 3Hz y con una amplitud de 6cm. ¿cuáles serán sus posiciones en los tiempos t=0 y t= 3.22 s
alisha [4.7K]

Answer:

yes

Explanation:dccdcc

3 0
3 years ago
The box is pushed to the right with a force of 40 Newtons and it just begins to move what is the maximum static frictional force
Vesnalui [34]

Answer:

The maximum static frictional force is 40N.

Explanation:

When an object of mass M is on a surface with a coefficient of static friction μ, there is a minimum force that you need to apply to the object in order to "break" the coefficient of static friction and be able to move the object (Called the threshold of motion, once the object is moving we have a coefficient of kinetic friction, which is smaller than the one for static friction).

This coefficient defines the maximum static friction force that we can have.

So if we apply a small force and we start to increase it, the static frictional force will be equal to our force until it reaches its maximum, and then we can move the object and now we will have frictional force.

In this case, we know that we apply a force of 40N and the object just starts to move.

Then we can assume that we are just at the point of transition between static frictional force and kinetic frictional force (the threshold of motion), thus, 40 N is the maximum of the static frictional force.

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3 years ago
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