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kvasek [131]
2 years ago
9

A plane has a speed of 275 km/hr. The wind is at a 70 degree angle to the direction of the plane with a speed of 75 km/hr. What

is the combined speed?
Physics
1 answer:
8090 [49]2 years ago
6 0

The combined speed of the plane and the wind is 249.35 km/h.

<h3>Relative speed</h3>

The relative speed of the plane due to speed of the wind is calculated as follows;

Let the direction of the plane be horizontal direction.

<h3>Speed of the wind</h3>

The speed of the wind is calculated as follows;

V_b = V \times cos\theta\\\\V_b = 75 \times cos(70)\\\\V_b = 25.65 \ km/hr

<h3>Combined speed</h3>

The combined speed of the plane and the wind is calculated as follows;

V = 275 km/h - 25.65 km/h

V = 249.35 km/h

Thus, the combined speed of the plane and the wind is 249.35 km/h.

Learn more about relative speed here: brainly.com/question/17228388

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PLS HELP!!!- The moon rotates around the Earth every 28 days. What is the frequency of the moon’s revolution?
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The equation that describes a transverse wave on a string is y = (0.0120 m)sin[(927 rad/s)t - (3.00 rad/m)x] where y is the disp
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Answer:

Speed, v = 312.34 m/s

Explanation:

The equation that describes a transverse wave on the string is given by :

y=0.0120\ msin[(927\ rad/s)t-(3\ rad/m)x]..............(1)

Where

y = displacement of a string particle

x = position of the particle on the string

The wave is travelling in the +x direction. We have to find the speed of the wave.

The general equation of traverse wave is given by :

y=A\ sin(kx-\omega t)................(2)

On comparing equation (1) and (2) we get,

k = 3 rad/m

Since, k=\dfrac{2\pi}{\lambda}

\lambda=\dfrac{2\pi}{3} ..............(3)

Also, \omega=927\ rad/s

Since, \omega=2\pi \nu

\nu=\dfrac{927}{2\pi}...............(4)

Speed of the wave is the product of frequency and wavelength i.e.

v=\nu\times \lambda

Using equation (3) and (4), the speed of the wave can be calculated as :

v=\dfrac{927}{2\pi}\times \dfrac{2\pi}{3}

v = 312.34 m/s

Hence, the speed of the transverse wave is 312.34 m/s

5 0
3 years ago
A 165 pound football player assumed a 4 point stance at the line. If his base of support was 30” wide, 38” deep, with the center
Hatshy [7]

Answer:

(1) The front attack is = 1528.56 inch-pounds (2) From the side is = 1237.5 inch- pounds. (3) from the rear is = 1528.32 inch pounds

Explanation:

Solution

Given

The weight of a football player is = 65 pound

Instance = 4 point

Now,

Let us consider that the weight on both hands are the same and weight on both knees are also same.

Let say, weight on each hand  be declared as m1, and weight of each leg be m2

Then,

m₁+ m₁  + m₂ +m₂ = 165 pounds

m₁ + m₂ = 82.5 ---------( equation 1)

On the coordinate plane let us assume that the center of gravity G is at the origin (0,0)

so,

2m₁x₁ + 2m₂x₂ /2m₁ + 2m₂ = 0

m₁ * 16 - m₂ * 22 = 0

m₂ = 8 /π m₁------ (2)

Now, let substitute 8 /π m for m₂ in equation 1

m₁ + 8 /π m₁ = 82.5

where m₁ = 47.76 pounds and m₂ = 34.74 pounds

Now,

(a) The front attack

The weight in front that is the one arm is displaced and about the center of gravity

so,

the inch of pound needed is denoted as:

Mfront = 2m₂ * x₂ = 2 * 34.74 *22

The Mfront becomes = 1528.56 inch-pounds

(2) From the side

The weight on one leg and one hand is displaced about the center of gravity G

Mside = 2/2 * y (m₁ +m₂) = 15¹¹ * (82.5)

so,

Mside = 15 * 82.5

Mside = 1237.5 inch- pounds

(c) For the rear attack

Now,

For the real attack, the weight in rear end on the knees is displaced about the center of gravity

Mrear =  2m₁ * x₂ = 2 * 47.76 * 16

Therefore the Mrear = 1528.32 inch pounds

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