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4 years ago

What are the 3 ways to calculate the net force ?

1 answer:

4 0

<span><span>An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force.</span>
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Imagine a third particle, which we will call a cyberon. It has three times the mass of an electron (3m). It has a positive charg

GalinKa [24]

Answer:

Explanation:

Let the potential difference between plate be V .

Electrical energy lost for cyberon = V x 3e

Kinetic energy lost by cyberon = kinetic energy gain by it

V x 3e = 1/2 x 3m x vc²

For electron , the above equation becomes

V x e = 1/2 x m x ve²

Dividing the two equations

1 = vc² / ve²

vc = ve .

vc / ve = 1 .

6 0

3 years ago

Two footballs (A & B) are thrown with the same amount of force. Football A has a greater mass. Describe the acceleration for

Sauron [17]

Newton's second law gives the relationship between force applied to an object, its mass and its acceleration:
$F=ma$
where F is the force, m the mass and a the acceleration.

A force F is applied to football A, whose mass is $m_A$ , and so the acceleration of this football will be given by (re-arranging the previous equation)
$a_A = \frac{F}{m_A}$

Similarly, the acceleration of football B will be
$a_B= \frac{F}{m_B}$
where $m_B$ is the mass of football B, and where the force F applied to the two footballs is the same.

Since football A has greater mass than football B, $m_A \ \textgreater \ m_B$ , if we compare the two previous formula we see that the acceleration of football B is greater than the acceleration of football A:
$a_B \ \textgreater \ a_A$
Therefore, if the same force is applied to the two footballs, football B will accelerate more than football A.

8 0

4 years ago

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

g100num [7]

To solve this problem we will apply trigonometric and optical concepts that allow us to obtain the minimum distance required. The resolution of the eye is given under the following condition,

$\theta = \frac{1.22\lambda}{D}$

Here,

$\lambda = Wavelength$

$D = Diameter$

With the values we have that the diameter will be,

$\theta = \frac{1.22(534nm)}{5.37mm}$

$\theta = 1.213*10^{-4}$

The relation between the distance of the lights and the distance from the eye to the lamp is given under the function,

$sin\theta = \frac{d}{L}$

For small angles $sin\theta = \theta$ , then

$\theta = \frac{d}{L}$

Here,

d = Distance between lights

L = Distance from eye to lamp

$1.213*10^{-4} = \frac{0.673m}{L}$

$L = \frac{0.673m}{1.213*10^{-4}}$

$L = 5548.22m$

Therefore the distance will be 5.5km

5 0

3 years ago

If the length of a building is 2 1 2 times the width and each dimension is increased by 7 ft, then the perimeter is 266 ft. Find

N76 [4]

Answer:

The original dimensions of the building is 95 ft × 38 ft.

Explanation:

Let the original length be 'l' and original width be 'w'.

Given:

Original length (l) = $2\frac{1}{2}\times original\ width$

Original width = 'w'.

So, $l=2\frac{1}{2}w=\frac{5}{2}w$

Now, as per question:

Length and width is increased by 7 ft.

So, new length (l') = $l+7=\frac{5w}{2}+7$

New width (w') = $w+7$

New perimeter (P) = 266 ft

Perimeter is given as:

$P=2(l' +w')\\\\266=2(\frac{5w}{2}+w)\\\\\frac{266}{2}=\frac{5w+2w}{2}\\\\266=7w\\\\w=\frac{266}{7}=38\ ft$

Therefore, original width = 38 ft.

Original length is, $l=\frac{5\times 38}{2}=\frac{190}{2}=95\ ft$

Hence, the original dimensions of the building is 95 ft × 38 ft.

7 0

4 years ago

If the speed of light is 1.5times that of the speed of light in a given material what is the index of refraction for that materi

jok3333 [9.3K]

The answer is 1.5....
The refractive index n = (Speed of light, c) /( Speed of light, v)
So c= n x v
comparing to c= 1.5 x v in this situation we see that n= 1.5

8 0

3 years ago