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2 years ago

An atom or ion has the abbreviated electron configuration [kr]. Select the species that it could not be

Physics

1 answer:

Lera25 [3.4K]2 years ago

3 0

An atom or ion which does not have the same electronic configuration as the species [kr] is K+

The complete question is given below:

An atom or ion has the abbreviated electron configuration (Kr). Select the species that it could not A. Br" B. K+ C. Sr24 D. Rbt E. Se-

An atom can be defined as the smallest particle of an element which can take part in a chemical reaction.

Some elements are

Monoatomic; eg: C
Diatomic; eg: O2
Triatomic and others
Polyatomic

So therefore, an atom or ion which does not have the same electronic configuration as the species [kr] is K+

Learn more about atoms or ions of elements:

brainly.com/question/6258301

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A boat travels west at 20km/h. The journey lasts 3hours. How far has the boat travelled? *

fomenos

Answer:

B)60km[W]

Explanation:

The boat travels 20km/h. So every hour the boat goes 20 miles. So if one hour equals 20km. Then 3 hours will be 3*20km which equals 60km. The boat is also going west. So you should consider putting that in your answer as well. So the answer would be B)60km[W].

Hope that helps!

3 0

3 years ago

Read 2 more answers

A cannonball is fired at a 45.0° angle and an initial velocity of 670 m/s. Assume no air resistance. How long until the cannonba

Naddik [55]

Answer:

96.7 s

Explanation:

Time of flight in projectile can be calculated thus:

T = 2 × u × sin ϴ/ g

Where;

T = time of flight (s)

u = initial velocity (m/s)

ϴ = Angle of projectile (°)

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

Based on the provided information; u = 670m/s, ϴ = 45°

Hence, using T = 2.u.sin ϴ/ g

T = 2 × 670 × sin 45° ÷ 9.8

T = 1340 × 0.7071 ÷ 9.8

T = 947.52 ÷ 9.8

T = 96.68

T = 96.7s

5 0

3 years ago

A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of

Lynna [10]

We have the relation

$\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}$

where $v_{A \mid B}$ denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

$v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}$

$v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}$

Let's assume the river flows South-to-North, so that

$\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath$

and let $-90^\circ < \theta < 90^\circ$ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

$\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)$

Then the velocity of the boat relative to the Earth is

$\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath$

The crossing is 153.0 m wide, so that for some time $t$ we have

$153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s$

which is minimized when $\theta=0^\circ$ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

$\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}$