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rewona [7]
3 years ago
12

Ow do quantum numbers relate to electrons?

Physics
1 answer:
strojnjashka [21]3 years ago
3 0
They enable us to dig deeper into the electron configurations by making us focus on electrons' quantum nature
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during a journey, a car travels at 40 km in 2.5 hours, next 62 km in 3 hours, then took a break for 30 minutes, again travelled
wlad13 [49]
45mph is the answer if you do the math right
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1. Tonya had a hard time deciding between the Big Burger and the Crispy Chicken sandwiches, her two favorites.
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Bro the md that she lost was the md boneless burger
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Two pendulum bobs have equal masses and lengths (8.100 m). bob a is initially held horizontally while bob b hangs vertically at
Karo-lina-s [1.5K]
Since both hv same mass and elsstic collision, so their velocity will exchange. Bob A will stop and bob B will move with speed of A just before the collision.

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5 0
3 years ago
Read 2 more answers
A car battery has a rating of 170 ampere-hours. This rating is one indication of the total charge that the battery can provide t
Pavlova-9 [17]

Answer:

612000 C

Explanation:

Current, I, is given as the rate of flow of charge, that is:

I = Δq / Δt

where q = electric charge

t = time taken

This implies that:

Δq = I * Δt

The battery rating is 170 Ampere-hours, therefore:

Δq = 170 * 1 hour

But 1 hour = 3600 seconds;

=> Δq = 170 * 3600 = 612000 C

The total charge that the battery can provide is 612000 C.

8 0
3 years ago
A boat leaves the dock at t = 0.00 s and, starting from rest, maintains a constant acceleration of (0.461 m/s2)i relative to the
liberstina [14]

Answer:

At t=4.82 s, the boat is moving at 3.464 m/s.

At t=4.82 s, the boat is 13.112 m from the dock.

Explanation:

The speed of the boat in j'th direction remains constant for all times (vj=2.16 m/s), however, the speed in i'th direction is changing due to the constant acceleration (0.461 m/s^2)i.

In order to find the velocity of the boat a t=4.82 s, first we need to compute the velocity of the boat relative to the water in the i direction (vi_b) at t=4.82 s:

vi_b = a*t = (0.461 m/s^2)*(4.82 s) = 2.222 m/s

Now, we add this velocity to the velocity of the water in the i direction:

vi = vi_b + vi_w = 2.222 m/s + 0.486 m/s = 2.708 m/s

Therefore, the speed of the boat at t = 4.82 s is: v = (vi, vj) = (2.708, 2.16) m/s. Finally, to find its speed, we just calculate the magnitude of v and obtain that the speed is: 3.464 m/s.

For the second question, first we will find the distance that the boat moved in the i'th direction and then in the j'th direction.

The speed in the i'th direction, for all times, is given by:

(0.485 + 0.461*t) and in order to find the distance advanced in the i'th direction (di) during 4.82 s, we need to integrate this velocity:

di = 0.485*t + (0.461*t^2)/2 (evaluated from t=0 to t =4.82) = 0.485*(4.82) + (0.461*(4.82)^2)/2 = 2.337 + 5.634 = 7.971 m

The speed in j'th direction, for all times, is given by:

2.16 and in order to find the distance advanced in the j'th direction (dj) during 4.82 s, we need to integrate this velocity:

dj = 2.16*t (evaluated from t=0 to t =4.82) = (2.16)*(4.82) = 10.411 m

Using Pythagoras' Theorem, we find that the the boat is at 13.112 m from the dock at t = 4.82 s.

4 0
2 years ago
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