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4 years ago

Why the center of Newton rings is dark?

2 answers:

3 0

The point of contact the path difference is zero but one of the interfering ray is reflected so the effective path difference becomes λ/2 thus the condition of minimum intensity is created in the center.

grandymaker [24]4 years ago

3 0

Answer:

the point of contact Thè path different is zero but

<h3>one of inter ferying ray is reflected so the effective path difference become / 2 and in this condition minimum intensity is created at center so the center of Newton ring is dark</h3>

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Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?

storchak [24]

Answer:

$\large \boxed{42\, \mu \text{C}}$$

Explanation:

The formula for the force exerted between two charges is

$F=k \dfrac{ q_1q_2}{r^2}$

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

$F=k\dfrac{q^{2}}{r^2}$

$\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}$

7 0

3 years ago

A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

4 years ago

The air pressure at the ocean surface a few miles from the shoreis most likely

frez [133]

The air pressure is most likely lower.

8 0

3 years ago

The energy of a given wave in the electromagnetic spectrum is 2.64 × 10-21 joules, and the value of Planck’s constant is 6.6 × 1

bixtya [17]

We are given
E = 2.64 × 10-21 J
h = 6.6 × 10-34 J s

The options given below are frequencies, therefore, the question must be asking about the frequency fo the given wave

The equation is
E = h f
Simply substitute and solve for f which is the frequency
f = 2.64 × 10-21 J / 6.6 × 10-34 J s
f = 4.00 × 1012 hertz

6 0

4 years ago

The revolving nosepiece of a compound microscope is used to: a. move the condenser up or down b. change the objective lens c. ad

Jlenok [28]

Answer:

Option B: change the objective lens

Explanation:

The revolving nosepiece is one of the parts of a microscope. Its responsibility is to hold the objective lenses.

6 0

3 years ago

Why the center of Newton rings is dark?​

Why the center of Newton rings is dark?