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3 years ago

Blown balloon kept outside on a sunny day bursts what will happen to the ballon

1 answer:

5 0

When the balloon is kept in the sun, due to Sun's heat, the kinetic energy of gaseous particles inside the balloons also gets increased and the balloon expands. This will increase the pressure on the walls of the balloon. It continues to expand and comes to a stage when the baloon bursts.

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Which of the following neutral elements has the same electron configuration as the bromide ion, Br−?

NikAS [45]

Answer:

B) krypton

Explanation:

Electron configuration of Br: [Ar] 3d10 4s2 4p5

Since we have the bromide ion with a negative charge, Br's atom acquired an electron and therefore another electron is added to the electronic configuration .

Electron configuration of Br-: [Ar] 3d10 4s2 4p6

And now this electronic configuration coincides with that of Krypton as we see below :

Electron configuration of Kr: [Ar] 3d10 4s2 4p6

5 0

4 years ago

Define the law of conservation of mass

Leya [2.2K]

Answer:

The law of conservation of mass or principle of mass conservation states that for any system closed to all transfers of matter and energy, the mass of the system must remain constant over time, as system's mass cannot change, so quantity can neither be added nor be removed.

Explanation:

7 0

4 years ago

A balloon rises at the rate of 8 feet per second from a point on the ground 12 feet from an observer. To 2 decimal places in rad

Hitman42 [59]

Answer:

$\displaystyle \theta' =0.24\ rad/s$

Explanation:

<u>Rate Of Change</u>

Let some variable y depend on time t. we can express y as a function of t as

$y=f(t)$

The instant rate of change of y respect to t is the first derivative, i.e.

$y'=f'(t)$

The balloon, the ground and the observer form a right triangle (shown below) where the height of the balloon y, the horizontal distance x, and the angle of elevation are related with the trigonometric formula

$\displaystyle tan\theta =\frac{y}{x}$

Since x is constant, we take the derivative with respect to time by using the chain rule:

$\displaystyle sec^2\theta \ \theta' =\frac{y'}{x}$

Solving for $\theta'$

$\displaystyle \theta' =\frac{y'}{xsec^2\theta}$

Let's compute the actual angle with the initial conditions y=9 feet, x=12 feet

$\displaystyle tan\theta =\frac{y}{x}$

$\displaystyle tan\theta =\frac{9}{12}=\frac{3}{4}$

Knowing that

$\sec^2\theta=1+tan^2\theta$

$\displaystyle \sec^2\theta=1+\left(\frac{3}{4}\right)^2$

$\displaystyle \sec^2\theta=\frac{25}{16}$

The balloon is rising at y'=8 feet/sec, thus we compute the change of the angle of elevation:

$\displaystyle \theta' =\frac{8}{12\ \frac{25}{16}}$

$\displaystyle \theta' =\frac{32}{75}\ rad/s$

$\boxed{\displaystyle \theta' =0.43\ rad/s}$

6 0

4 years ago

B What is SI and CGs unit of force

choli [55]

Answer:

The relation between SI and CGS unit of force is the CGS unit of force is equal to SI unit of force. The force is the product of mass and acceleration. its SI unit is Newton and CGS unit is dyne. The relation between SI and CGS unit of force is.

HOPE THIS HELPS!!!!!

Explanation:

6 0

3 years ago

Two identical point charges are 3.00 cm apart. find the charge on each of them if the force or repulsion is 4.00 x 10^-7. (Use C

DanielleElmas [232]

Answer:

Charge on each is 2 x 10⁻¹⁰.

Explanation:

We know that Force between two point charges is given b the Coulomb's law as:

F = kq₁q₂/r^2

k = 9 x 10^9

r = 3.00 cm

= 0.03 m

q₁ = q₂

F = 4.00 x 10^-7

Rearranging the formula, we get:

F = k q²/r²

q² = Fr²/k

q² = 4 x 10⁻⁷ x 0.03²/(9x10⁹)

q² = 4 x 10⁻²⁰

q = 2 x 10⁻¹⁰

As there is force of repulsion between the charges, the charges must be both positive or both negative.

3 0

4 years ago

Blown balloon kept outside on a sunny day bursts what will happen to the ballon​

Blown balloon kept outside on a sunny day bursts what will happen to the ballon