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3 years ago

How do the properties in the elements in the same column of the periodic table compare?

Physics

2 answers:

butalik [34]3 years ago

8 0

Same oxidation number.

irga5000 [103]3 years ago

7 0

The columns of the periodic table, also referred to as "groups" contain elements with similar reactive properties, due to these elements having a similar configuration of electrons in their outer shell.

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A tray of electronic components contains 15 components, 4 of which are defective. If 4 components are selected, what is the poss

dlinn [17]

Answer:

a) 0.0007326

b) 0.03223

c) 0.2418

d) 0.2418

Explanation:

To find different probabilities for the selection of components among eleven good and four defective components, we will use the Combination.

a) $C(4,4) = 1; C(15,4) = 1365$

$P = \frac{C(4,4)}{C(15,4)} = \frac{1}{1365} = 0.0007326$

b) $C(4,3) = 4; C(11,1) = 11$

$P = \frac{C(4,3)*C(11,1)}{C(15,4)} = \frac{4*11}{1365} = 0.03223$

c) $C(4,2) = 6; C(11,2) = 55$

$P = \frac{C(4,2)*C(11,2)}{C(15,4)} = \frac{6*55}{1365} = 0.2418$

d) $C(11,4) = 330$

$P = \frac{C(11,4)}{C(15,4)} = \frac{330}{1365} = 0.2418$

8 0

3 years ago

Question 4 A car of mass 820 kg has a maximum power or 30 kW and moves against a constant resistance of motion to 910 N. Calcula

Naddika [18.5K]

Explanation:

power = force × velocity

velocity=power/force

=(30×1000)/910

=32.97m/s

7 0

3 years ago

A runner of mass 56.0kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its cent

SCORPION-xisa [38]

Answer: -0.84 rad/sec (clockwise)

Explanation:

Assuming no external torques act on the system (man + turntable), total angular momentum must be conserved:

L1 = L2

L1 = It ω + mm. v . r = 81.0 kg . m2 .21 rad/s – 56.0 kg. 3.1m/s . 3.1 m

L1 = -521.15 kg.m2/sec (1)

(Considering to the man as a particle that is moving opposite to the rotation of the turntable, so the sign is negative).

Once at rest, the runner is only a point mass with a given rotational inertia respect from the axis of rotation, that can be expressed as follows:

Im = m. r2 = 56.0 kg. (3.1m)2 = 538.16 kg.m2

The total angular momentum, once the runner has come to an stop, can be written as follows:

L2= (It + Im) ωf = -521.15 kg.m2/sec

L2= (81.0 kg.m2 + 538.16 kg.m2) ωf = -521.15 kg.m2/sec

Solving for ωf, we get:

ωf = -0.84 rad/sec (clockwise)

5 0

3 years ago

If the frequency of this beam is increased while the intensity is held constant, does the number of electrons ejected per second

Ivan

Answer:

if the intensity of photons is constant then number of ejected electrons will remain same

Explanation:

As per photoelectric effect we know that when light of sufficient frequency fall on the surface of metal then electrons get ejected out of the surface with certain kinetic energy

Here the energy of photons is used to eject out the electrons from metal surface and to give the kinetic energy to the ejected electrons

so we have

$h\nu = W + KE$

here W = work function of metal which shows the energy required to eject out electrons from metal surface

KE = kinetic energy of ejected electrons

now if we increase the frequency of the photons that incident on the metal surface then in that case the incident energy will increase

So the electrons will eject out with more kinetic energy while if the number of photon is constant or the intensity of photons is constant then number of ejected electrons will remain same

8 0

3 years ago

A disk, with a radius of 0.25 m, is to be rotated like a merry-go-round through 1000 rad, starting from rest, gaining angular sp

aniked [119]

Answer:

(a) The least time required for the rotation is 50 seconds

(b) The corresponding value of α₁ is 1.6 rad/s²

Explanation: Please see the attachments below

6 0

3 years ago