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Monochromatic light falling on two very narrow slits 0.048mm apart. Successive fringes on a screen 5.00m away are 6.5cm apart ne

tino4ka555 [31]

Answer:

λ = 5.85 x 10⁻⁷ m = 585 nm

f = 5.13 x 10¹⁴ Hz

Explanation:

We will use Young's Double Slit Experiment's Formula here:

$Y = \frac{\lambda L}{d}\\\\\lambda = \frac{Yd}{L}$

where,

λ = wavelength = ?

Y = Fringe Spacing = 6.5 cm = 0.065 m

d = slit separation = 0.048 mm = 4.8 x 10⁻⁵ m

L = screen distance = 5 m

Therefore,

$\lambda = \frac{(0.065\ m)(4.8\ x\ 10^{-5}\ m)}{5\ m}$

λ = 5.85 x 10⁻⁷ m = 585 nm

Now, the frequency can be given as:

$f = \frac{c}{\lambda}$

where,

f = frequency = ?

c = speed of light = 3 x 10⁸ m/s

Therefore,

$f = \frac{3\ x\ 10^8\ m/s}{5.85\ x\ 10^{-7}\ m}\\\\$

f = 5.13 x 10¹⁴ Hz

5 0

2 years ago

Please help me, this is a physics test.

sweet-ann [11.9K]

Answer:

a = 2 [m/s²]

Explanation:

To be able to solve this problem we must make it clear that the starting point when the time is equal to zero, the velocity is 5 [m/s] and when three seconds have passed the velocity is 11 [m/s], this point is the final point or the final velocity.

We can use the following equation.

$v_{f}=v_{o}+a*t\\$

where:

Vf = final velocity = 11 [m/s]

Vo = initial velocity = 5 [m/s]

a = acceleration [m/s²]

t = time = 3 [s]

$11 = 5 + a*3\\6=3*a\\a= 2[m/s^{2} ]$

4 0

2 years ago

Help with physics projectile motion

BlackZzzverrR [31]

Answer:

10.4 m/s

Explanation:

First, find the time it takes for the projectile to fall 6 m.

Given:

y₀ = 6 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 1.11 s

Now find the horizontal position of the target after that time:

Given:

x₀ = 6 m

v₀ = 5 m/s

a = 0 m/s²

t = 1.11 s

Find: x

x = x₀ + v₀ t + ½ at²

x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²

x = 11.5 m

Finally, find the launch velocity needed to travel that distance in that time.

Given:

x₀ = 0 m

x = 11.5 m

t = 1.11 s

a = 0 m/s²

Find: v₀

(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²

v₀ = 10.4 m/s

3 0

3 years ago

A steel rod is pulled in tension with a stress that is less than the yield strength. The modulus of elasticity may be calculated

pshichka [43]

Answer:

B. Axial stress divided by axial strain

Explanation:

Elasticity:

It is the tendency of an object to deform along the axis when an opposing force is applied without facing permanent change in shape.

Plasticity:

When an object crosses the elasticity limit, it enters plasticity where the change due to stress is permanent and the object might even break.

Yield strength:

Yield strength is the point of maximum bearable stress that indicates the limit of elasticity.

Our case:

As the stress applied is less than the yield strength, the rod is still in the elasticity state and its modulus can be calculated.

Modulus of Elasticity = Stress along axis/Ratio of change in length to original length

Axial strain is basically the ratio of change in length to original length.

So, Modulus of Elasticity = Axial Stress/ Axial Strain

6 0

3 years ago

Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o

Helen [10]

Answer:

Charge Z can be placed at x = -2.7 m or at x = 0.27 m.

Explanation:

The Coulomb force between two charges, $Q_1$ and $Q_2$ , separated by a distance, $d$ , is given

$F = k\dfrac{Q_1Q_2}{r^2}$

k is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the x-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be Q. It is positive.

Let the distance from charge X be x m. Then the distance from charge Y will be (0.60 - x) m.

Force due to charge X

$F_X = k\dfrac{18Q}{x^2}$

Force due to charge Y

$F_Y = k\dfrac{-27Q}{(0.60-x)^2}$

Since both forces are equal and opposite,

$F_X = -F_Y$

$k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}$

$\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}$

$2(0.60-x)^2 = 3x^2$

$2(0.36-1.20x+x^2) = 3x^2$

$0.72-2.40x+2x^2 = 3x^2$

$x^2+2.40x-0.72 = 0$

Applying the quadratic formula,

$x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}$

$x = -2.7$ or $x = 0.27$

Charge Z can be placed at x = -2.7 m or at x = 0.27 m

3 0

3 years ago