5. A 1.2 m guitar string is under a tension of 888 N. The waves produced on the string (when

How can groundwater flow, elevation can transport contaminates

Pavel [41]

Answer: it's under explanation

Explanation:Groundwater contamination occurs when man-made products such as gasoline, oil, road salts and chemicals get into the groundwater and cause it to become unsafe and unfit for human use. Materials from the land's surface can move through the soil and end up in the groundwater.

3 0

3 years ago

What would happen to the marbles speed if we had longer tables

Zanzabum

Answer:

The marble will start to decrease in speed

Explanation:

The longer the table is the more the marbles speed would decrease because marbles can only be fast for a certain amount of time

7 0

3 years ago

Read 2 more answers

Determine tge energy in joules of a photon whose frequency is 3.55x10

JulijaS [17]

E = hf
E = 6.63×10^-34 × 3.55×10 eV
1 eV = 1.60×10^-19 J
E = 6.63×10^-34 × 3.55×10 × 1.60×10^-19
E = 3.77×10^-51 J

Hope it helped!

6 0

3 years ago

A car going at v = 29.7 m/s (67 mph) rounds a curve of radius R = 50.0 m, where the road is banked at an angle of θ = 30.0°. Wha

Nikolay [14]

Answer:

μ = 0.6

Explanation:

given,

speed of car = 29.7 m/s

Radius of curve = 50 m

θ = 30.0°

minimum static friction = ?

now,

writing all the forces acting along y-direction

N cos θ - f sinθ = mg

N cos θ -μN sinθ = mg

$N = \dfrac{m g}{cos\theta-\mu sin \theta}$

now, writing the forces acting along x- direction

N sin θ + f cos θ = F_{net}

N cos θ + μN sinθ = F_{net}

$\dfrac{m g}{cos\theta-\mu sin \theta}(cos \theta + \mu sin\theta)=F_{net}$

taking cos θ from nominator and denominator

$F_{net} =\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg$

$\dfrac{mv^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg$

$\dfrac{v^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}g$

$\mu=\dfrac{v^2 -r g tan\theta}{v^2tan\theta + r g}$

now, inserting all the given values

$\mu=\dfrac{29.7^2 -50 \times 9.8tan 30^0}{29.7^2\times tan 30^0 +50 \times 9.8}$

μ = 0.6

7 0

3 years ago

The New England Merchants Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.17 Hz, and the acc

aksik [14]

Answer:

d = 8.4 cm

Explanation:

In order to calculate the amplitude of oscillation of the top of the building, you use the following formula for the max acceleration of as simple harmonic motion:

$a_{max}=A\omega^2$ (1)

A: amplitude of the oscillation

w: angular speed of the oscillation = 2 $\pi$ f

f: frequency = 0.17Hz

The maximum acceleration of the top of the building is a 2.0% of the free-fall acceleration. Then, you have:

$a_{max}=0.02(9.8m/s^2)=0.196\frac{m}{s^2}$

Then, you solve for A in the equation (1) and replace the values of the parameters:

$A=\frac{a_{max}}{\omega^2}=\frac{a_{max}}{4\p^2i f^2}\\\\A=\frac{0.196m/s^2}{16\pi^2(0.17Hz)^2}\\\\A=0.042m=4.2cm$

The total distance, side to side, of the oscilation of the top of the building is twice the amplitude A. Then you obtain:

d = 2A = 2(4.2cm) = 8.4cm

4 0

3 years ago