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3 years ago

14

What piece of sports equipment is aerodynamic

1 answer:

Mashcka [7]3 years ago

6 0

Aerobie. Frisbee. Discus. Javelin. I suppose an American football to some extent.
<span>Pull! Clay pigeons. Arrows. Wingsuit. Kites. Hang gliders. Sails. sailboat keels/dagger boards. Water skis. Ski jumping skis. Boomerang. </span>
<span>I'm excluding spheres and parachutes as bluff bodies even though aerodynamics often plays a big part in their motion.</span>

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What is geology >>>????

anastassius [24]

Answer:

the study of geological factors and the ways the earth

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3 years ago

Read 2 more answers

The microwaves in a certain microwave oven have a wavelength of 12.2 cm. How wide must this oven be so that it will contain five

leva [86]

Answer:

a

$l = 0.305 \ m$

b

$f = 3.0*10^{11} \ Hz$

Explanation:

From the question we are told that

The wavelength is $\lambda = 12.2 \ cm = 0.122 \ m$

The number of antinodal planes of the electric field considered is n = 5

The width is mathematically represented as

$l = \frac{ n \lambda}{2}$

$l = \frac{5 * 0.122 }{ 2}$

$l = 0.305 \ m$

Generally the frequency the errors was made is mathematically represented as

$f = \frac{c}{\lamda_k}$

Here c is the speed of light with value $c = 3.0*10^{8} \ m/s$

$\lambda_k$ is the wavelength of the microwave has to be in order for there still to be five antinodal planes of the electric field along the width of the oven, which is mathematically represented as

$\lambda_k = \frac{ \lambda * \frac{0.04}{2} }{n/2}$

$\lambda_k = \frac{0.122*0.02}{5/2}$

So

$f = \frac{3.0*10^{8}}{0.000976}$

$f = 3.0*10^{11} \ Hz$

8 0

2 years ago

(TCO 5) The relationship between Celsius (º C) and Fahrenheit (º F) degree of measuring temperature is linear. Find the linear e

gtnhenbr [62]

Answer:

$C=-\dfrac{10}{17}(F-32)$

Explanation:

Given that

32° F corresponds to 0 °C. ---Point 1

212° F corresponds to 100 °C.----Point 2

We know that if two point is given that equation of line can be found as

$y-y_1=\dfrac{y_2-y_1}{x_2-x_2}(x-x_1)$

Lets C in y- direction and F in x- direction,so we can say that

$C-C_1=\dfrac{C_2-C_1}{F_2-F_2}(F-F_1)$

$C-0=\dfrac{100-0}{32-212}(F-32)$

$C=-\dfrac{10}{17}(F-32)$

So the linear relationship is

$C=-\dfrac{10}{17}(F-32)$

8 0

3 years ago

Four point masses of 3.0 kg each are arranged in a square on masslessrods. The length of a side of the square is 0.50m. What is

Zigmanuir [339]

Answer:

Part a)

$I = 1.5 kg m^2$

Part b)

$I = 0.75 kg m^2$

Part c)

$I = 1.5 kg m^2$

Explanation:

Part a)

Moment of inertia of the system about an axis passing through B and C is given as

$I = mL^2 + mL^2 + m(0) + m(0)$

$I = 2mL^2$

$I = 2(3 kg)(0.50^2)$

$I = 1.5 kg m^2$

Part b)

Moment of inertia of the system about an axis passing through A and C is given as

$I = m(0^2) + m(\frac{L}{\sqrt2})^2 + m(0) + m(\frac{L}{\sqrt2})^2$

$I = 2m\frac{L^2}{2}$

$I = (3 kg)(0.50^2)$

$I = 0.75 kg m^2$

Part c)

Moment of inertia of the system about an axis passing through the center of the square and perpendicular to the plane of the square

$I = m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2$

$I = 4m\frac{L^2}{2}$

$I = 2(3 kg)(0.50^2)$

$I = 1.5 kg m^2$

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2 years ago

Please help with this question !!!!!

lesantik [10]

Answer:

te puedo ayudar pero corazón y estrella

y ando buscando novia por si quieres ser mia

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3 years ago