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3 years ago

Will mark brainliest HELP ASAP Atomic structure extension

Physics

1 answer:

Olenka [21]3 years ago

6 0

Answer:

Explanation:

atomic structure extension is very difficult to read but when done right you should get 23 yw

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Kilo- - 1,000 deka- - 10 How much larger is a kilo- than a deka-?

Ilya [14]

$lets's \ take \ an \ example \ one \ gram: \\ \\ 0,001kg \\ \\ 0,01hg \\ \\ 0,1dag \\ \\ \boxed{1g} \\ \\ 10dg \\ \\ 100cg \\ \\ 1000mg \\ \\ the \ units \ are \ rising \ and \ decreasing \ from \ 10 \ to \ 10 \ results \ \\ \\ a \ kilogram \ is \ 100 \ times \ larger \ than \ a \ dekagram$

6 0

3 years ago

A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice

Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of $\sqrt{3} v_{e}$

Explanation:

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , $v_{i}=2 v_{e}$ . Here the radius is R, so find the escape velocity as follows,

$\frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0$

$\frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}$

$v_{e}^{2}=\frac{2 G M}{R}$

$v_{e}=\sqrt{\frac{2 G M}{R}}$

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as, $U_{i}=-\frac{G M m}{R}$

R tend to infinity when far away from the planet, so $v_{f}=0$

Then, kinetic energy at initial would be,

$k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}$

Similarly, kinetic energy at final would be,

$k_{f}=\frac{1}{2} m v_{f}^{2}$

Here, $v_{f}=\text { final velocity }$

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

$U_{i}+k_{i}=k_{f}+v_{f}$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}$

'm' and $\frac{1}{2}$ as common on both sides, so gets cancelled, we get as

$4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}$

We know, $v_{e}=\sqrt{\frac{2 G M}{R}}$ , it can be wriiten as $\left(v_{e}\right)^{2}=\frac{2 G M}{R}$ , we get

$4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}$

$v_{f}^{2}=3\left(v_{e}\right)^{2}$

Taking squares out, we get,

$v_{f}=\sqrt{3} v_{e}$

4 0

3 years ago

Answer fast, please.

Tpy6a [65]

D bomb b j de violence

5 0

3 years ago

A man with a weight of 550 N climbs a ladder to a height of 3.5m. How much work did he do?

Natasha2012 [34]

Answer: 1925 joules

Explanation:

weight of man = 550 N

height climbed = 3.5m

Work = ?

Since, work is done when the man pulled his weight over the height; work is expressed as the product of weight and height climbed

i.e Workdone = 550N x 3.5m

= 1925 joules

Thus, the man did a work of 1925 joules while climbing

7 0

4 years ago

1. A liquid of mass 250g is heated with an electric heater. Its temperature rises from 30°C to 80°C, the specific heat capacity

statuscvo [17]

Answer:

1) 50 seconds 2) 100°C

Explanation:

(Follows formula of Power=Energy/Time)

1) 500W x X = 2000J/kg°C x .25kg x 50°C

X = 50 seconds.

2) 2000W x 300s = 1000J/kg°C x 2kg x X

X = 300

Initial temperature => 400°C-300°C = 100°C

8 0

3 years ago