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3 years ago

12

Why do the noble gases and some other elements on the right side of the periodic table occur as gases instead of solids as do th

e majority of the elements elements?

1 answer:

mash [69]3 years ago

8 0

Explanation:

The left side of the periodic table has elements that have less number of electrons in the valence shell.

These elements loose electrons easily.These elements appear as metals or metalloids in nature.These are hard solids.Their inter molecular forces are very strong.

The right side of the periodic table has elements that have more number of electrons in the valence shell.

These elements gain electrons easily.These elements appear as non metals most of which are gases.Their inter molecular forces are weak.

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A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick

rewona [7]

Answer:

The new time period is $T_2 = 3.8 \ s$

Explanation:

From the question we are told that

The period of oscillation is $T = 5 \ s$

The new length is $l_2 = 0.76 \ m$

Let assume the original length was $l_1 = 1 m$

Generally the time period is mathematically represented as

$T = 2 \pi \sqrt{ \frac{ I }{ mgh } }$

Now I is the moment of inertia of the stick which is mathematically represented as

$I = \frac{m * l^2 }{12 }$

So

$T = 2 \pi \sqrt{ \frac{ m * l^2 }{12 * mgh } }$

Looking at the above equation we see that

$T \ \ \ \alpha \ \ \ l$

=> $\frac{ T_2 }{T_1} = \frac{l_2}{l_1}$

=> $\frac{ T_2}{5} = \frac{0.76}{1}$

=> $T_2 = 3.8 \ s$

3 0

3 years ago

What is the main function of a telescope?

Gwar [14]

Answer:

Explanation:The main purpose of astronomical telescope is to make objects from outer space appear as bright, contrasty and large as possible. That defines its three main function: light gathering, resolution and magnification. These are the measure of its efficiency.

3 0

3 years ago

an object of mass 6000 kg rests on the flatbed of a truck. it is held in place by metal brackets that can exert a maximum horizo

Otrada [13]

Answer:

minimum stopping distance will be d = 75 m

Explanation:

Maximum force exerted by the bracket is given as

F = 9000 N

now we know that mass of the object is

m = 6000 kg

so the maximum acceleration that truck can have is given as

$F = ma$

$9000 = 6000 a$

$a = 1.5 m/s^2$

now for finding minimum stopping distance of the truck

$v_f^2 - v_i^2 = 2a d$

$0^2 - 15^2 = 2(-1.5) d$

$d = 75 m$

4 0

3 years ago

A nearsighted person wants to see an apple that is 7 meters away but can only clearly see objects that are at most 62 cm away fr

natulia [17]

Answer:

The power of the corrective lenses is 3.162 D.

Explanation:

Given that,

Object distance = 70 cm

Image distance = 62 cm

Distance between eyes and glasses = 2.5 cm

Eyeglasses made of diverging corrective lenses can help her to see the apple clearly

So now ,

Object distance from glass =70-2.5 = 67.5 cm

Image distance from glass = 62-2.5 = 59.5 cm

We need to calculate the focal length

Using formula for focal length

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

$\dfrac{1}{f}=-\dfrac{1}{59.5}-\dfrac{1}{67.5}$

$\dfrac{1}{f}=-\dfrac{508}{16065}\ cm$

We need to calculate the power of lens

Using formula of power

$P=\dfrac{100}{f}$

$P=-\dfrac{508}{16065}\times100$

$P=-3.162 \ D$

Negative sign shows the lens is diverging.

Hence, The power of the corrective lenses is 3.162 D.

6 0

3 years ago

A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42◦ from t

Semmy [17]

Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

P.E = mgh joule

Considering h = 0 for the vertical position

The h at ∅ = 42° is h = L (1 - cos∅)

P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

P.E = 25 x 9.8 x 2.2 (1 - cos42°)

= 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

∴ K.E = P.E

1/2 mv² = 138.44

1/2 x 25 x v² 138.44

v² = 11.0752

v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

Tension on string, T = Force acting on the swing, F

$W=L\int\limits^0_\phi{F} \, d \phi$

$=L\int\limits^0_\phi{mg.sin \phi} \, d \phi$

= $-Lmg[cos\phi]_{42}^{0}$

= - 2.2 x 25 x 9.8 [cos0 - cos 42°]

= - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

3 0

3 years ago