I NEED HELP

Learning Goal: How do 2 ordinary waves build up a "standing" wave? A very generic formula for a traveling wave is: y1(x,t)=Asin(

zheka24 [161]

Answer:

Explanation:

=Asin(kx−ωt). This general mathematical form can represent the displacement of a string, or the strength of an electric field, or the height of the surface of water, or a large number of other physical waves!

Part C Find ye(x) and yt(t). Remember that yt(t) must be a trig function of unit amplitude. Express your answers in terms of A, k, x, ω, and t. Separate the two functions with a comma. Use parentheses around the argument of any trig functions.

Part E At the position x=0, what is the displacement of the string (assuming that the standing wave ys(x,t) is present)? Part G From

Part F we know that the string is perfectly straight at time t=π2ω. Which of the following statements does the string's being straight imply about the energy stored in the strJHJMNMMUJJHTGGHing?

a.There is no energy stored in the string: The string will remain straight for all subsequent times.

b.Energy will flow into the string, causing the standing wave to form at a later time.

c.Although the string is straight at time t=π2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string.

d.The total mechanical energy in the string oscillates but is constant if averaged over a complete cycle.

3 0

3 years ago

A ramp is needed to allow vehicles to climb a 2 foot wall. The angle of elevation in order for the vehicles to safely go up must

stepan [7]

Answer:

Yes the ramp can be safely used

Explanation:

Here, we have

Length of longest ramp = 5 ft

Height of wall = 2 ft

Therefore, the sine of the angle adjacent to the ramp which is equal to the angle of elevation is given by;

$Sin\theta = \frac{Opposite \, side \, to\, angle}{Hypothenus\, side \, of\, triangle}$

Where:

The opposite side to angle = 2 ft wall and

Hypotenuse side = Ramp = 5 ft

Therefore,

$Sin\theta = \frac{2}{5} = 0.4$ and θ = sin⁻¹0.4 = 23.55 °

The ramp can be safely used as the angle it is adjacent to is less than the specified 30°.

5 0

4 years ago

A ship's propeller of diameter 3 m makes 10.6 revolutions in 30s. What is the angular velocity of the propeller?

ycow [4]

Answer:

The angular velocity of the propeller is 2.22 rad/s.

Explanation:

The angular velocity (ω) of the propeller is:

$\omega = \frac{\Delta \theta}{\Delta t}$

Where:

θ: is the angular displacement = 10.6 revolutions

t: is the time = 30 s

$\omega = \frac{\Delta \theta}{\Delta t} = \frac{10.6 rev*\frac{2\pi rad}{1 rev}}{30 s} = 2.22 rad/s$

Therefore, the angular velocity of the propeller is 2.22 rad/s.

I hope it helps you!

5 0

3 years ago

URGENT I WILL GIVE BRAINLIEST

Reil [10]

Weight : It is the gravitational attraction force on an object near its surface.

It is always measured in the unit of Newton(N)

it is given by the product of mass and acceleration due to gravity

$W = mg$

here we know that mass of the pig is m = 140 kg

also we know that acceleration due to gravity near the surface of earth is approximately g = 9.8 m/s/s

so here we know by above formula

$W = (140 kg)(9.8 m/s^2)$

$W = 1372 N$

so correct answer will be

4 0

3 years ago

(7)Figure 4 shows three charges: Q₁, Q₂ and Q3 . Determine the net force (Fnet) acting on Q3. (Hint: Draw a free body diagram of

NISA [10]

Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is

$F = \dfrac{kq_1q_2}{r^2}$

where k ≈ 8,98 × 10⁹ kg•m³/(s²•C²) is Coulomb's constant.

8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.

8.2. First convert everything to base SI units:

0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C

0,03 µC = 3 × 10⁻⁸ C

0,04 µC = 4 × 10⁻⁸ C

300 mm = 300 × 10⁻³ m = 0,3 m

600 mm = 0,6 m

Force due to Q₁ :

$F_{Q_2/Q_1} = \dfrac{k (6 \times 10^{-16} \,\mathrm C)}{(0,3 \, \mathrm m)^2} \approx \boxed{6,0 \times 10^{-5} \,\mathrm N} = 0,06 \,\mathrm{mN}$

Force due to Q₃ :

$F_{Q_2/Q_3} = \dfrac{k (12 \times 10^{-16} \,\mathrm C)}{(0,6 \, \mathrm m)^2} \approx \boxed{3,0 \times 10^{-5} \,\mathrm N} = 0,03 \,\mathrm{mN}$

8.3. The net force on the particle at Q₂ is the vector

$\vec F = F_{Q_2/Q_1} \, \vec\imath + F_{Q_2/Q_3} \,\vec\jmath = \left(-0,06\,\vec\imath - 0,03\,\vec\jmath\right) \,\mathrm{mN}$

Its magnitude is

$\|\vec F\| = \sqrt{\left(-0,06\,\mathrm{mN}\right)^2 + \left(-0,03\,\mathrm{mN}\right)^2} \approx 0,07 \,\mathrm{mN} = \boxed{7,0 \times 10^{-5} \,\mathrm N}$

and makes an angle θ with the positive horizontal axis (pointing to the right) such that

$\tan(\theta) = \dfrac{-0,03}{-0,06} \implies \theta = \tan^{-1}\left(\dfrac12\right) - 180^\circ \approx \boxed{-153^\circ}$

where we subtract 180° because $\vec F$ terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.

8 0

2 years ago