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Setler79 [48]
3 years ago
6

Yolanda is studying two waves. The first wave has an

Physics
2 answers:
jeka943 years ago
3 0

Answer:

she can use constructive interference to generate a wave with ab amplitude of 3.5

Korvikt [17]3 years ago
3 0

Answer:

Got a 100%

Explanation:

You might be interested in
Question 4 of 10
ch4aika [34]

Habitat fragmentation is a cost of urban development.

Option: A

Explanation:

Though from the view point or perspective of up gradation and development urban development is much needed but in cost of habitat fragmentation which feels very bitter. As habitat fragmentation leads to the loss of habitat, disruption of ecological cycle and environmental equilibrium.

Actually in the name of urban development we the human use our bread giver environment in a wrong way which causes natural disasters in long run. Animals become endangered , vulnerable and extinct with passage of time. Because they forced to enter into human settlements.

4 0
3 years ago
Find the speed of a wave with wave length 650 M and a frequency 35HZ<br><br> help.
hram777 [196]

Answer:

option A

Explanation:

wave speed= 650×35= 22750 m/s

hope it helps !

7 0
3 years ago
Read 2 more answers
Frank has a paperclip. It has a mass of 12g and a volume of 3cm3. What is its density?​
Mkey [24]

Answer:

D = 4 g/cm³

Explanation:

Density = Mass / Volume

Step 1: Define

D = x

M = 12 g

V = 3 cm³

Step 2: Substitute

D = 12g/3 cm³

Step 3: Simplify

D = 4g / cm³

4 0
3 years ago
Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.
lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = (6\times0.1)kg/min=0.6kg/min

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = \frac{dx}{dt}

So, the rate of change of salt in the tank can be given by the following equation,

1000\frac{dx}{dt} =0.6-6x

or, \int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt

or, T = 693.147 min      (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

3 0
4 years ago
Mass Center Determine the coordinates (x, y) of the center of mass of the area in blue in the figure below. Answers: x=(3)/(8)a
Naya [18.7K]

Explanation:

The x and y coordinates of the center of mass are:

xcm = ∫ x dm / m = ∫ x ρ dA / ∫ ρ dA

ycm = ∫ y dm / m = ∫ y ρ dA / ∫ ρ dA

Assuming uniform density, the center of mass is also the center of area.

xcm = ∫ x dA / ∫ dA = ∫ x y dx / A

ycm = ∫ y dA / ∫ dA = ∫ ½ y² dx / A

First, let's find the area:

A = ∫ y dx

A = ∫₀ᵃ (-h/a² x² + h) dx

A = -⅓ h/a² x³ + hx |₀ᵃ

A = -⅓ h/a² (a)³ + h(a)

A = ⅔ ha

Now, let's find the x coordinate of the center of mass:

xcm = ∫ x y dx / A

xcm = ∫₀ᵃ x (-h/a² x² + h) dx / (⅔ ha)

xcm = ∫₀ᵃ (-h/a² x³ + hx) dx / (⅔ ha)

xcm = (-¼ h/a² x⁴ + ½ hx²) |₀ᵃ / (⅔ ha)

xcm = (-¼ h/a² (a)⁴ + ½ h(a)²) / (⅔ ha)

xcm = (¼ ha²) / (⅔ ha)

xcm = ⅜ a

Next, we find the y coordinate of the center of mass:

ycm = ∫ y² dx / A

ycm = ∫₀ᵃ ½ (-h/a² x² + h)² dx / (⅔ ha)

ycm = ∫₀ᵃ ½ (h²/a⁴ x⁴ − 2h²/a² x² + h²) dx / (⅔ ha)

ycm = ½ (⅕ h²/a⁴ x⁵ − ⅔ h²/a² x³ + h² x) |₀ᵃ / (⅔ ha)

ycm = ½ (⅕ h²/a⁴ (a)⁵ − ⅔ h²/a² (a)³ + h² (a)) / (⅔ ha)

ycm = ½ (⁸/₁₅ h²a) / (⅔ ha)

ycm = ⅖ h

6 0
3 years ago
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