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olya-2409 [2.1K]
3 years ago
6

A plane with the speed of 210 miles per hour is flying at an angle of 40 degrees from true east and a 45 mile per hour wind is b

lowing directly from the west. What is the angle measure and the speed of the true course of the plane?
Physics
1 answer:
Yanka [14]3 years ago
3 0
Medtvtvyftfunuhyfsvghjji
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MaRussiya [10]

Answer:

ballistic stretching

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3 years ago
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Which is the correct order of the academic pathway of a pulmonologist?
saul85 [17]

Answer:

B

Explanation:

on edge 2021

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3 years ago
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Dos cargas puntuales están fijas en el eje x: q1 = 6.0µC está en el origen, O, con x1 = 0.0 cm, y q2 = –3.0 µC está situada en e
erik [133]

Answer:

E_total = 1.30 10¹⁰ C / m²

Explanation:

The intensity of the electric field is

     E = k q / r²

on a positive charge proof

The total electric field at the midpoint is

as q₁= 6 10⁻⁶ C the field is outgoing to the right

for charge q₂ = -3 10⁻⁶ C, the field is directed to the right, therefore

E_total = E₁ + E₂

E_total = k q₁ / r₁² + k q₂ / r₂²

r₁ = r₂ = r = 4 10⁻² m

E_total = k/r² (q₁ + q₂)

 we calculate

E_total = 9 10⁹ / (4 10⁻²)²   (6.0 10⁻⁶ +3.0 10⁻⁶)

E_total = 1.30 10¹⁰ C / m²

8 0
3 years ago
(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup
KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx

Where

x_{o}, x_{f} - Initial and final position, respectively, measured in meters.

F(x) - Force as a function of position, measured in newtons.

Given that F = k\cdot x and the fact that F = 25\,N when x = 0.3\,m - 0.2\,m, the spring constant (k), measured in newtons per meter, is:

k = \frac{F}{x}

k = \frac{25\,N}{0.3\,m-0.2\,m}

k = 250\,\frac{N}{m}

Now, the work function is obtained:

W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx

W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]

W = 0.313\,J

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be r(\theta) = 2\cdot \sin 5\theta. The area of the region enclosed by one loop of the curve is given by the following integral:

A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta

A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta

By using trigonometrical identities, the integral is further simplified:

A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta

A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta

A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta

A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)

A = 4\pi

The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

5 0
3 years ago
What is the density of the block if it has a mass of 45.5 grams and a volume of 2.4 cm³?
rjkz [21]

Explanation:

density= mass/volume

therefore;45.5/2.4

=19.0 g/cm³

5 0
2 years ago
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