What are the four kinds of air uplift?

A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at

Elenna [48]

The tension in the cable is 23.2 N

<h3>What is the tension in the string?</h3>

The tension in the cable can be resolved into horizontal and vertical forces Tcosθ and Tsinθ respectively.

Tcosθ, is acting perpendicularly, Tcosθ = 0

Taking moments about the pivot:

Tsinθ * 2.2 = 4 * 9.8 * 0.7

Solving for θ;

θ = tan⁻¹(1.4/2.2) = 32.5°

T = 27.44/(sin 32.5 * 2.2)

T = 23.2 N

In conclusion, the tension in the cable is determined by taking moments about the pivot.

Learn more about moments of forces at: brainly.com/question/23826701

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3 0

2 years ago

Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o

love history [14]

Answer:

13.51 nm

Explanation:

To solve this problem, we are going to use angle approximation that sin θ ≈ tan θ ≈ θ where our θ is in radians

y/L=tan θ ≈ θ

and ∆θ ≈∆y/L

Where ∆y= wavelength distance= 2.92 mm =0.00292m

L=screen distance= 2.40 m

=0.00292m/2.40m

=0.001217 rad

The grating spacing is d = (90000 lines/m)^−1

=1.11 × 10−5 m.

the small-angle

approx. Using difraction formula with m = 1 gives:

mλ = d sin θ ≈ dθ →

∆λ ≈ d∆θ = =1.11 × 10^-5 m×0.001217 rad

=0.000000001351m

= 13.51 nm

6 0

3 years ago

You have a length of tubing that is closed at one end. You cut the tubing into two pieces of unequal length, giving you a tube o

umka21 [38]

Answer:

The funda mental frequency of the original tube is 182Hz.

Explanation:

See the attachment for the calculation steps.

In order to calculate the fundamental frequency of the original closed tube we need to find the length of the tube which is equal to the sum of the lengths of the two new tubes.

For closed tubes

f = nv/4L (n = 1, 3, 5,...n)

f = nv/2L (n = 1, 2, 3,...n)

The details of calculation can be found below in the attachment.

4 0

3 years ago

The potential-energy function u(x) is zero in the interval 0≤x≤l and has the constant value u0 everywhere outside this interval.

VMariaS [17]

Look first for the relation between deBroglie wavelength (λ) and kinetic energy (K):
K = ½mv²
v = √(2K/m)
λ = h/(mv)
= h/(m√(2K/m))
= h/√(2Km)

So λ is proportional to 1/√K.
in the potential well the potential energy is zero, so completely the electron's energy is in the shape of kinetic energy:
K = 6U₀

Outer the potential well the potential energy is U₀, so
K = 5U₀
(because kinetic and potential energies add up to 6U₀)

Therefore, the ratio of the de Broglie wavelength of the electron in the region x>L (outside the well) to the wavelength for 0<x<L (inside the well) is:
1/√(5U₀) : 1/√(6U₀)
= √6 : √5

5 0

3 years ago

If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h

raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory ( $u_x$ ) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

$u^2=u_x^2+u_y^2$

Where, $u_x\ and\ u_y$ are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

$(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1$

Now, we know that, for a projectile motion, the maximum height is given as:

$H=\frac{u_y^2}{2g}$

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

$H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m$

Therefore, the maximum height reached is 4.63 m.

3 0

4 years ago