PLEASE HELP can not find the answer

When blue light of wavelength 460 nm falls on a single slit, the first dark bands on either side of center are separated by 45.0

lyudmila [28]

Answer:

width of slit =1.23× 10⁻⁶ m

Explanation:

we know the condition of diffraction minima,

d sin θ = n λ

λ = wavelength θ = angle between the central maxima and 1st minima

d = slit width

for first minima n = 1

now,

d = $\dfrac{n \lambda}{sin \theta}$

$\theta = \dfrac{45^0}{2} = 22.5^0$

d = $\dfrac{1\times 460 \times 10^{-9}}{sin 22.5^0}$

d = 1228 × 10⁻⁹ m = 1.228× 10⁻⁶ m

d = 1.23× 10⁻⁶ m

width of slit =1.23× 10⁻⁶ m

6 0

3 years ago

How did science of cartography came in India

Kobotan [32]

Cartography (the making of maps and charts) is a science because it is a body of knowledge which can be used, built on, and can produce testable hypotheses.

The cartography of India begins with early charts for navigation and constructional plans for buildings. Indian traditions influenced Tibetan and Islamic traditions, and in turn, were influenced by the British cartographers who solidified modern concepts into India's map making

8 0

3 years ago

A 4.4-µF capacitor is initially connected to a 5.1-V battery. Once the capacitor is fully charged the battery is removed and a 2

Grace [21]

Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S

2) t = 1 mu S

3) t = 50 mu S

1) $22.4 \mu C$

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

$C=\frac{Q_0}{V_0}$

where

C is the capacitance

Q0 is the initial charge stored

V0 is the initial potential difference across the capacitor

When the capacitor is connected to the battery, we have:

$C=4.4\mu F = 4.4\cdot 10^{-6}F$

$V_0 = 5.1 V$

Solving for $Q_0$ ,

$Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C$

So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.

2) $20.0\mu C$

To find the charge on the capacitor at any other time t, we use the equation:

$Q(t) = Q_0 e^{-\frac{t}{RC}}$

where

$Q_0 = 22.4 \mu C$

t is the time

$R=2.0 \Omega$ is the resistance

$C=4.4\mu F$ is the capacitance

Therefore, at time $t=1 \mu s$ , we have:

$Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C$

3) $0.08 \mu C$

As before, we use again the equation:

$Q(t) = Q_0 e^{-\frac{t}{RC}}$

However, here the time to consider is

$t=50 \mu C$

Substituting into the formula,

$Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C$

4 0

3 years ago

During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel? Use the formula =1/2 * t^2 to s

Mila [183]

Answer:

70560 m

Explanation:

The formula to calculate the distance travelled during a free fall motion is

$d=-\frac{1}{2}gt^2$

where

d is the distance travelled

g = -9.8 m/s^2 is the acceleration due to gravity

t is the time

In this situation,

t = 120 s

Therefore the distance travelled after 120 s is

$d=-\frac{1}{2}(-9.8)(120)^2=70560 m$

6 0

4 years ago

PLEASE HELP A student climbs to the top of a press box where the cameras are. He wonders how many meters he is off of the ground

Feliz [49]

The height of the rail on top of the press box where the ball was dropped from is 11.025 m.

The given parameters:

Time of motion of the ball, t = 1.5 s
Let the height of the rail = h

<h3>Maximum height of fall;</h3>

The maximum height through which the ball was dropped from is calculated by applying second equation of motion;

$h = v_0_y t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\h = \frac{1}{2} (9.8) (1.5)^2\\\\h = 11.025 \ m$

Thus, the height of the rail on top of the press box where the ball was dropped from is 11.025 m.

Learn more about height of projectiles here: brainly.com/question/10008919

3 0

2 years ago