A balloon has a volume of 3.0 l at 25Â°c. what is the approximate volume of the balloon at 50Â°c?

Question 5

bezimeni [28]

Answer:

ffcvghnvb vyhgyhvthbgvgybn ytvg dfvthgbhtfgybhvtgbyhnt vfyhn fgb fvb

Explanation:

4 0

3 years ago

Emilio tries to jump to a nearby dock from a canoe that is floating in the water.Instead of a canoe that is floating in the wate

elixir [45]

Oooo that ones hard. ummm... idk i think we should just leave it to the experts ya know.

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

this expression holds for the points located at

-r₁ <x₁ <r₁

3) Point between the two spheres

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

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

E_ {total} = $+ k \frac{Q}{(d-x_1)^2}$ + k Q / (d-x1) 2

point range

-r₂ <x₂ <r₂

5) Right side point

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

E_ {total} = $- k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )$ - k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

8 0

3 years ago

A person travelled 350 m east from his home and returns back home an hour has displacement of_?

Svetradugi [14.3K]

Answer:

vector of zero magnitude

Explanation:

The displacement is a vector magnitude, therefore, in addition to being a module, it has direction and sense.

In this case it moved 350 m and then returned the same 350 m, so the total displacement is zero.

If we draw the vector, one has a directional direction to the right and the other direction to the left, therefore when adding the two vectors gives a vector of zero magnitude

7 0

3 years ago

Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 k

Harrizon [31]

Explanation:

Given that,

(a) Speed, $v=6.66\times 10^6\ m/s$