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svet-max [94.6K]
4 years ago
12

Which trends is indirectly proportional to effective nuclear charge

Physics
1 answer:
Minchanka [31]4 years ago
4 0

Answer:

Atomic size

Explanation:

In the periodic table , atomic size is indirectly proportional to the effective nuclear charge .the atomic size reduces from left to right across the table. This is because electrons are added to the same shell.

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If the pressure of a substance is increased during a boiling process, will the temperature also increase, or will it remain cons
7nadin3 [17]

Answer:

on increasing pressure, temperature will also increase.

Explanation:

Considering the ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Thus, at constant volume and number of moles, Pressure of the gas is directly proportional to the temperature of  the gas.

P ∝ T

Also,  

Also, using Gay-Lussac's law,

\frac {P_1}{T_1}=\frac {P_2}{T_2}

Thus, on increasing pressure, temperature will also increase.

4 0
3 years ago
Two trains on separate tracks move toward each other. Train 1 has a speed of 145 km/h; train 2, a speed of 72.0 km/h. Train 2 bl
tekilochka [14]

Answer:

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

Explanation:

Given:

Two trains on separate tracks move toward each other

For Train 1 Velocity of the observer,

v_{o}=145\ km/h=145\times \dfrac{1000}{3600}=40.28\ m/s

For Train 2 Velocity of the Source,

v_{s}=90\ km/h=90\times \dfrac{1000}{3600}=25\ m/s

Frequency of Source,

f_{s}=500\ Hz

To Find:

Frequency of Observer,

f_{o}=?  (frequency heard by the engineer on train 1)

Solution:

Here we can use the Doppler effect equation to calculate both the velocity of the source v_{s} and observer v_{o}, the original frequency of the sound waves f_{s} and the observed frequency of the sound waves f_{o},

The Equation is

f_{o}=f_{s}(\dfrac{v+v_{o}}{v -v_{s}})

Where,

v = velocity of sound in air = 343 m/s

Substituting the values we get

f_{o}=500(\dfrac{343+40.28}{343 -25})=500\times 1.205=602.64\approx 603\ Hz

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

7 0
4 years ago
Apakah itu hukum hooke.....can i knw detail n simple?
butalik [34]
Hooke's law is stated as: F = -kx

Where:

F = Force to compress or extend a spring (unit N)
k = Spring constant
x = displacement of spring

Or to help you in your native language:

<span>F = adalah gaya 
</span><span>k = adalah konstante pegas 
</span><span>x = adalah jarak pergerakan pegas dari posisi normalnya</span>
4 0
3 years ago
3. Suppose that an airplane flying 60 m/s, at a height of 300m, dropped a sack of flour (pere the effect
Arlecino [84]

469.24m. An airplane flying 60m/s at a height of 300m dropped a sack of flour that stack the ground 469.24m from the point of release.

This is a example of horizontal parabolic projectile motion,and we represents this motion in the coordinate axis, which means that the velocity has components in x axis and y axis.

The equation of components on the x axis.

v_{0}x=\frac{x}{t}, where x is the distance and Vox the initial velocity before the drop

The equation of components on the y axis.

y = v_{0}yt+\frac{gt^{2} }{2}, where y is the height, and the velocity in y component before the drop is 0, reducing the equation to y = \frac{gt^{2} }{2}

Clear t from both the equation of components on the x axis and the y axis:

t=\frac{x}{v_{0} x} and t=\sqrt{\frac{2h}{g}}

Equating both equations and clearing the distance x:

\frac{x}{v_{0} x}=\sqrt{\frac{2h}{g}}\\x={v_{0} x}\sqrt{\frac{2h}{g}}

Substituting the values:

x=60\frac{m}{s} \sqrt{\frac{2(300m)}{9.81\frac{m}{s^{2} } }}=469.24m

3 0
3 years ago
While driving north at 25 m/s during a rainstorm you notice that the rain makes an angle of 38 degrees with the vertical. while
ikadub [295]

While returning back in opposite direction driver see that rain drops are falling vertically down

so horizontal speed of rain with respect to driver must be Zero

while moving towards the north direction driver see that the rain drops makes an angle 38 degree with the vertical

tan\theta = \frac{v_y}{v_x}

tan38 = \frac{v_y}{25 + 25}

v_y = 39 m/s

so the speed of rain is

v = \sqrt{25^2 + 39^2}

v = 46.3 m/s

also the angle is given as

\theta = tan^{-1}\frac{39}{25}

\theta = 57.3 degree

8 0
4 years ago
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