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Sladkaya [172]
3 years ago
5

Atom in the ground state is said to be​

Physics
1 answer:
koban [17]3 years ago
3 0

Answer:

Ground-state atom

Explanation:

When an atom is not excited, it is in its ground-state, which we refer as "standard" or "normal" state.

(Hopefully that helped you!)

GOOD LUCK

Astrophysicist Dr. D

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Please help! Will give brainliest. 10 points. Show work!
Natasha_Volkova [10]

Answer:

421.83 m.

Explanation:

The following data were obtained from the question:

Height (h) = 396.9 m

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

First, we shall determine the time taken for the ball to get to the ground.

This can be calculated by doing the following:

t = √(2h/g)

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 396.9 m

Time (t) =.?

t = √(2h/g)

t = √(2 x 396.9 / 9.8)

t = √81

t = 9 secs.

Therefore, it took 9 secs fir the ball to get to the ground.

Finally, we shall determine the horizontal distance travelled by the ball as illustrated below:

Time (t) = 9 secs.

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

s = ut

s = 46.87 x 9

s = 421.83 m

Therefore, the horizontal distance travelled by the ball is 421.83 m

5 0
3 years ago
Bob is moving at 0.967c with respect to Alice. At the exact instant he passes Alice, she fires a very short laser pulse in the s
yulyashka [42]

Explanation:

Speed of Bob, v = 0.967 c

At the exact instant he passes Alice, she fires a very short laser pulse in the same direction Bob is moving.

(a) We need to find the distance measured by Alice  between Bob and the laser pulse. It is given by :

d=ct-vt

d=t(c-v)

d=5.59\times (c-0.967c)

d=5.53\times 10^7\ meters

(b) Distance measured by Bob between himself and the laser pulse is given by :

d_B=ct

d_B=3\times 10^8\times 5.59

d_B=6.67\times 10^9\ meters

Hence, this is the required solution.  

8 0
3 years ago
What is 64 nanometers to m?
defon
\sf Hello!

\sf We\: know \:that,
\sf 1\: meter = \sf 10^{9} nm

\sf Then,

\sf Distance\: in \:m = \sf Distance\: in\: nm × \dfrac{\sf 1}{\sf 10^{9}}\: \sf m

⇒ \sf Distance\: in \:m = \sf 64 × 10^{-9} \:m

⇒ \sf Distance\: in\: m = \sf 6.4 × 10^{-8} \:m

~ \sf iCarl
3 0
3 years ago
Read 2 more answers
When an airplane is flying 200 mph at 5000-ft altitude in a standard atmosphere, the air velocity at a certain point on the wing
Sedaia [141]

Answer:

P1 = 0 gage

P2 = 87.9 lb/ft³

Explanation:

Given data

Airplane flying = 200 mph = 293.33 ft/s

altitude height = 5000-ft

air velocity relative to the airplane = 273 mph = 400.4 ft/s

Solution

we know density at height 5000-ft is 2.04 × 10^{-3} slug/ft³

so here P1 + \frac{\rho v1^2}{2}  = P2 + \frac{\rho v2^2}{2}

and here

P1 = 0 gage

because P1 = atmospheric pressure

and so here put here value and we get

P1 + \frac{\rho v1^2}{2}  = P2 + \frac{\rho v2^2}{2}

0 + \frac{2.048 \times 10^{-3} \times 293.33^2}{2}  = P2 + \frac{2.048 \times 10^{-3} \times 400.4^2}{2}  

solve it we get

P2 = 87.9 lb/ft³

5 0
3 years ago
What is moderate injuries?
Reika [66]
Injuries that are warm and not cold or  hot.
3 0
3 years ago
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