Please help!How is constant or uniform acceleration used to explain free fall?

1 answer:

4 0

Free fall is a special case of motion with constant acceleration, because acceleration due to gravity is always constant and downward. For example, when a ball is thrown up in the air, the ball's velocity is initially upward.

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What is the best use of an atomic model to explain the charge of the particles in Thomson’s beams?

kodGreya [7K]

Answer:

The interpretation of the subject in question is characterized in the discussion section below.

Explanation:

The atom seems to be the smallest particle of negatively charged electrons as well as positive. The negative particulates are comparatively small, as well as distant from both the considerably high beneficial particles. When the negative objects traveled away from the desired ones those who formed an intangible beam which was electrically charged.
Thomson would use a closed glass globe with just a single positive and another negative electrode with extraordinarily low present pressure. He was forced to submit those other gases to quite a voltage level, and also that the rise in popularity of emission levels, which have been called cathode rays, must have been observed.
As the negatives shifted away from those in the positives, an imaginary electrically conductive beam was formed. Not only that but the negative particles have been completely circumvented due to the extreme distance seen between positively and negatively particle size.

5 0

3 years ago

Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax

kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its x-component is

$F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its x-component is

$F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N$

So, the x-component of the resultant force is

$F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N$

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its y-component is

$F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its y-component is

$F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N$