True or false cleft palates is the failure of palate to form and join correctly

If we wish to expand (x + y) 8 , what is the coefficient of x 5 y 3 ? what is the coefficient of x 3 y 5 ?

Molodets [167]

We use the binomial theorem to answer this question. Suppose we have a trinomial (a + b)ⁿ, we can determine any term to be:

[n!/(n-r)!r!] a^(r) b^(n-r)

a.) For x⁵y³, the variables are: x=a and y=b. We already know the exponents of the variables. So, we equate this with the form of the binomial theorem.

r = 5
n - r = 3
Solving for n,
n = 3 + 5 = 8

Therefore, the coefficient is equal to:
Coefficient = n!/(n-r)!r! = 8!/(8-5)!8! = 56

b.) For x³y⁵, the variables are: x=a and y=b. We already know the exponents of the variables. So, we equate this with the form of the binomial theorem.

r = 3
n - r = 5
Solving for n,
n = 5 + 3 = 8

Therefore, the coefficient is equal to:
Coefficient = n!/(n-r)!r! = 8!/(8-3)!8! = 56

7 0

3 years ago

The sources of marine debris and explain why it is a problem with global origins

Lera25 [3.4K]

Marine debris is any trash or litter that ends up in a marine environment. This harms wildlife through entanglement, ingestion, and disturbs the habitat of many species of animals. This becomes a global problem when it is a danger to human health. Nails, syringes, and glass can cause harm to people on the beach. On top of that, trash in the waterway increases the amount of chemicals destroying our water quality.
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4 0

3 years ago

The ratio between the volumes of two spheres is 27 to 8. what is the ratio between their respective radii?

kiruha [24]

Plz see the attachment...

7 0

4 years ago

A block of mass M=10 kg is on a frictionless surface as shown in the photo attached. And it's attached to a wall by two springs

nordsb [41]

a.

i. the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s

ii. the angular velocity when the two springs are in parallel is 7.07 rad/s

b.

i. the speed of the block of mass when the springs are connected in series is 11.2 A m/s

ii. the angular velocity when the two springs are in series is 11.2 rad/s

<h3>a. </h3><h3>i. How to calculate the velocity of the mass when the springs are connected in parallel?</h3>

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is k' = k + k

= 2k

= 2 × 250 N/m

= 500 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k'A² = 1/2k'x² + 1/2Mv² where

k' = equivalent spring constant in parallel = 500 N/m,
A = maximum displacement of spring,
x = equilibrium position = 0 m,
M = mass of block = 10 kg and
v = speed of block at equilibrium position

Making v subject of the formula, we have

v = √[k'(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k'(A² - x²)/M]

v = √[500 N/m(A² - (0)²)/10]

v = √[50 N/m(A² - 0)]

v = [√50]A m/s

v = [5√2] A m/s

v = 7.07 A m/s

So, the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s

<h3>ii. The angular velocity of mass when the springs are in parallel</h3>

Since velocity of spring v = ω√(A² - x²) where

ω = angular velocity of spring,
A = maximum displacement of spring and
x = equilbrium position of spring = 0 m

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 7.07 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 7.07 A m/s/√(A² - 0²)

ω = 7.07 A m/s/√(A² - 0)

ω = 7.07 A m/s/√A²

ω = 7.07 A m/s/A m

ω = 7.07 rad/s

So, the angular velocity when the two springs are in parallel is 7.07 rad/s

<h3>b. </h3><h3>i. How to calculate the velocity of the mass when the springs are connected in series?</h3>

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is 1/k" = 1/k + 1/k

= 2/k

⇒ k" = k/2

k" = 250 N/m ÷ 2

= 125 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k"A² = 1/2k"x² + 1/2Mv'² where

k" = equivalent spring constant in series = 125 N/m,
A = maximum displacement of spring,
x = equilibrium position = 0 m,
M = mass of block = 10 kg and
v' = speed of block at equilibrium position

Making v subject of the formula, we have

v = √[k"(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k"(A² - x²)/M]

v = √[125 N/m(A² - (0)²)/10]

v = √[125 N/m(A² - 0)]

v = [√125]A m/s

v = [5√5] A m/s

v = 11.2 A m/s

So, the speed of the block of mass when the springs are connected in series is 11.2 A m/s

<h3>ii. The angular velocity of the mass when the springs are in series</h3>

Since velocity of spring v = ω√(A² - x²) where

ω = angular velocity of spring,
A = maximum displacement of spring and
x = equilbrium position of spring = 0 m

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 11.2 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 11.2 A m/s/√(A² - 0²)

ω = 11.2 A m/s/√(A² - 0)

ω = 11.2 A m/s/√A²

ω = 11.2 A m/s/A m

ω = 11.2 rad/s

So, the angular velocity when the two springs are in series is 11.2 rad/s

Learn more about speed of block of mass here:

brainly.com/question/21521118

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3 0

2 years ago

What is the maximum number of lines per centimeter a diffraction grating can have and produce a complete first-order spectrum fo

Vitek1552 [10]

Answer:

1315789.47368 lines/m

Explanation:

m = Order = 1

$sin\theta$ = 1 For maximum condition

$\lambda$ = Wavelength = 760 nm maximum

From Rayleigh criteria we have the expression for the gap

$d=\dfrac{m\lambda}{sin\theta}\\\Rightarrow d=\dfrac{1\times 760}{1}\\\Rightarrow d=760\ nm$

The number of lines is the reciprocal of the slit distance

$n=\dfrac{1}{d}\\\Rightarrow n=\dfrac{1}{760\times 10^{-9}}\\\Rightarrow n=1315789.47368\ /m$

The number of lines is 1315789.47368 per meter

7 0

3 years ago