In the formula 4h²o how many total hydrogen atoms are there

A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge

Serga [27]

Answer:

Explanation:

Let the plastic rod extends from - L to + L .

consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .

It will create a field at point P on y -axis . Distance of point P

= √ x² + .15²

electric field at P due to small charged length

dE = k λ dx x / (x² + .15² )

Its component along Y - axis

= dE cosθ where θ is angle between direction of field dE and y axis

= dE x .15 / √ x² + .15²

= k λ dx .15 / (x² + .15² )³/²

If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

E = ∫ k λ .15 / (x² + .15² )³/² dx

= k λ x L / .15 √( L² / 4 + .15² )

6 0

3 years ago

If two objects of the same size move through the air at different speeds, which encounters the greater air resistance?a. The fas

blagie [28]

Answer:

Option A is correct.

(The faster object encounters more resistance)

Explanation:

Option A is correct. (The faster object encounters more resistance)

Air resistance depends on various factors:

Speed of the object
Cross-sectional area of the object
Shape of the object

Formula:

$F=\frac{1}{2}C_d\rho A v^{2}$

As the speed of the object increases the amount of Air resistance/drag increases on the object, as the above formula shows direct relation between Air resistance/drag and velocity i.e F ∝ v^2.

6 0

3 years ago

Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in

Serggg [28]

Answer:

(a) The force between them quadruples

Explanation:

According to coulomb's law, initial force between the two charged objects is given as;

$F_1=\frac{Kq_1q_2}{r^2}$

where;

k is coulomb's constant

q₁ is the charge on the first object

q₂ is the charge on the second object

r is the distance between the two objects

When the charges on both objects are doubled, then;

q₁ = 2q₁

q₂ = 2q₂

Force between the two charged objects will become

$F_2 = \frac{K2q_12q_2}{r^2} = \frac{4Kq_1q_2}{r^2} = 4(\frac{Kq_1q_2}{r^2}) = 4F_1$

Therefore, the force between them quadruples

4 0

3 years ago

How can a moving coil galvanometer can be made into a dc ammeter?

Dennis_Churaev [7]

I am absolutely sure that the way how can a moving coil galvanometer can be made into a dc ammeter is of course by connecting a. low resistance across the meter. You should remember that you must connect <span>a shunt resistor straight across the galvanometer. Do hope this answer will help you! Regards.</span>

7 0

3 years ago

Two small metal spheres are 25 cm apaft.The spheres have equal amount of negative charge and repel each other with a force of 0.

Mars2501 [29]

Answer:

$0.5\times 10^{-6}C$

Explanation:

According to coulombs law force between two charges is given by $F=\frac{1}{4\pi \epsilon _0}\frac{Q_1Q_2}{R^2}$ here R is the distance between both the charges which is given as 25 cm

We have given force F =0.036 N

So $0.036=\frac{1}{4\pi \times 8.85\times 10^{-12}}\frac{Q^2}{(0.25^2)}$ As $\epsilon _0$ is constant which value is $8.85\times 10^{-12}$

$Q^2=0.250\times 10^{-12}$

$Q=0.5\times 10^{-6}C$

8 0

4 years ago