All categories

3 years ago

The amount of diffraction depends on the size of the obstacle and the wavelength of the wave.

Physics

1 answer:

tresset_1 [31]3 years ago

8 0

I believe your answer is TRUE!
Hope this helps!:)

You might be interested in

Jason applies a force of 4.00 newtons to a sled at an angle 62.0 degrees from the ground. What is the component of force effecti

kaheart [24]

You want to draw a free body diagram of the forces on the sled in the horizontal x-direction.

If you visualize the system in an x-y coordinate plane, the force along the x-direction is the angle it makes with the x-axis multiples by the force.

The angle made with the x-axis is cosine of the angle theta.

Please see picture attached.

5 0

3 years ago

Read 2 more answers

PLEASE HELP QUICK I NEED IT TO PASS

BlackZzzverrR [31]

Answer:

Explanation:

As the contour lines have roughly the same spacing but the actual topography is much steeper, the lines on the mountainous map represent a larger vertical spacing than the lines on the gradual hills.

3 0

2 years ago

HOW MANY KILOMETERS IS EARTH FROM THE SUN?

maria [59]

Earth is 150 million kilometers away for the sun

7 0

3 years ago

Read 2 more answers

At t=0, a train approaching a station begins decelerating from a speed of 80 mi/hr according to the acceleration function a(t)=−

vredina [299]

Answer:

a) Δx = 49.23 mi , b) Δx = 5.77 mi

Explanation:

As we have an acceleration function we must use the definition of kinematics

a = dv / dt

∫dv = ∫ a dt

we integrate and evaluators

v - vo = ∫ (-1280 (1 + 8t)⁻³ dt = -1280 ∫ (1+ 8t)⁻³ dt

We change variables

1+ 8t = u

8 dt = du

v - v₀ = -1280 ∫ u⁻³ du / 8

v -v₀ = -1280 / 8 (-u⁻²/2)

v - v₀ = 80 (1+ 8t)⁻²

We evaluate between the initial t = 0 v₀ = 80 and the final instant t and v

v- 80 = 80 [(1 + 8t)⁻² - 1]

v = 80 (1+ 8t)⁻²

We repeat the process for defining speed is

v = dx / dt

dx = vdt

x-x₀ = 80 ∫ (1-8t)⁻² dt

x-x₀ = 80 ∫ u⁻² dt / 8

x-x₀ = 80 (-1 / u)

x-x₀ = -80 (1 / (1 + 8t))

We evaluate for t = 0 and x₀ and the upper point t and x

x -x₀ = -80 [1 / (1 + 8t) - 1]

We already have the function of time displacement

a) let's calculate the position at the two points and be

t = 0 h

x = x₀

t = 0.2 h

x-x₀ = -80 [1 / (1 +8 02) -1]

x-x₀ = 49.23

displacement is

Δx = x (0.2) - x (0)

Δx = 49.23 mi

b) in the interval t = 0.2 h at t = 0.4 h

t = 0.4h

x- x₀ = -80 [1 / (1+ 8 0.4) -1]

x-x₀ = 55 mi

Δx = x (0.4) - x (0.2)

Δx = 55 - 49.23

Δx = 5.77 mi

3 0

3 years ago

How many elements did mendeleev place in group 2

Pepsi [2]

Answer:5

Explanation:

8 0

3 years ago