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3 years ago

Which of newtons laws is illustrated by a squid moving forward by shooting water out behind it? Why?

1 answer:

5 0

I believe its newtons 3rd law for every action there is an equal but opposite reaction since the squid is moving foward by shooting the water its pushing the squid back as its reaction. Hope this helped !

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A tennis ball with a speed of 11.3 m/s is moving perpendicular to a wall. after striking the wall, the ball rebounds in the oppo

Arlecino [84]

To calculate the average acceleration of the ball we use the formula,

$a= \frac{v_{f} -v_{i}}{t}$

Here, $v_{f}$ is the final velocity of the ball and $v_{i}$ is the initial velocity of the ball t is the time in contact with the wall.

Given $v_{i} = 11.3 m/s$ towards the wall and $v_{f} = 7.2207 m/s$

away from the wall and $t=0.00803 s$ .

Substituting these values in above formula , we get

$a=\frac{(-7.2207 m/s)-11.3 m/s}{0.00803 s}$

Here $v_{f}$ is negative because ball is moving away from the wall.

$a= - 2306 .4 \ m/s^{2}$

Therefore, average acceleration of the ball is $- 2306 .4 \ m/s^{2}$ (away from the wall).

4 0

3 years ago

Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance

siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

E_ {total} = E₁ + E₂

E_{ total} = $k \frac{Q}{x_1^2} + k \frac{Q}{x_2^2}$

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

x₂ = x₁ -d

E total = $k \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}$

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

E_ {total} = $k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )$

E_ {total} = $k \frac{Q}{x_1^2}$ $( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )$

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

E_ {total} = 0

x₁² = (x₁ + d)²

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

E_ {total} = $-k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}$