Do it in order. from smallest to largest

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr

fgiga [73]

Answer:

The balloon is 66.62 m high

Explanation:

Combined Motion

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

$\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}$

The values are

$\displaystyle y_o=15\ m$

$\displaystyle v_o=14\ m/s$

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

$\displaystyle y=0$

$\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0$

Multiplying by 2

$\displaystyle 2y_o+2v_ot-gt^2=0$

Clearing the coefficient of $t^2$

$\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0$

Plugging in the given values, we reach to a second-degree equation

$\displaystyle t^2-2.857t-3.061=0$

The equation has two roots, but we only keep the positive root

$\displaystyle \boxed {t=3.69\ s}$

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

$\displaystyle Y_r=15+V_r.t$

$\displaystyle Y_r=15+14\times3.69$

$\displaystyle \boxed{Y_r=66.62\ m}$

3 0

3 years ago

If 1.8 1016 electrons enter a light bulb in 3 milliseconds, what is the magnitude of the electron current at that point in the c

Mice21 [21]

Answer:

I = 0.96 A

Explanation:

No of electrons, $n=1.8\times 10^{16}$

Time, t = 3 ms = $3\times 10^{-3}\ s$

We need to find the electric current. We know that electric charge per unit time is equal to the electric current.

$I=\dfrac{q}{t}$

q = ne (Quantization of electric charge)

$I=\dfrac{ne}{t}\\\\I=\dfrac{1.8\times 10^{16}\times 1.6\times 10^{-19}}{3\times 10^{-3}}\\\\I=0.96\ A$

So, the electric current is 0.96 A.

7 0

3 years ago

when an element tends to lose its valence electrons in chemical reactions , does it behave more like a metal or nonmetal

Juli2301 [7.4K]

It behaves more like a metal

Explanation:

When an element tends to lose its valence electrons in chemical reactions, they behave more like a metal.

Metals are electropositive.

Electropositivity or metallicity is the a measure of the tendency of atoms of an element to lose electrons.

This is closely related to ionization energy and the electronegativity of the element.

The lower the ionization energy of an element, the more electropositive or metallic the element is .

Metals are usually large size and prefers to be in reactions where they can easily lose their valence electrons.

When most metals lose their valence electrons, they attain stability.

Non-metals are electronegative. They prefer to gain electrons.

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4 0

3 years ago

What is 1 abiotic factors shown in this diagram?

yawa3891 [41]

Answer:

B. The water

Explanation:

Water is abiotic factor because it is non living

7 0

3 years ago

Which of the following would be most likely to contribute to molecules

lana66690 [7]

Answer:

B. Containing charged regions

Explanation:

The term i.e. intermolecular forces would be used to explain the attraction forces. Here the interaction would be done between molecules etc that acts between the acts & the other types of particles i.e. neighboring like atoms or ions

So in the given case, the option b would be contributed to the molecules that have intermolecular forces

hence, the option b is correct

8 0

2 years ago