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3 years ago

What is the speed of a wave with a wavelength of 10 mm and a frequency of 5.0 Hz?

Physics

2 answers:

sleet_krkn [62]3 years ago

4 0

Speed = wavelength × frequency
speed = 10/1000 × 5.0
speed = 0.001 × 5.0
speed = 0.005m/s

Dmitriy789 [7]3 years ago

3 0

Answer:

Speed of the wave, v = 0.05 m/s

Explanation:

It is given that,

Wavelength of the wave, $\lambda=10\ mm=0.01\ m$

Frequency of the wave, $\nu=5\ Hz$

The speed of the wave is given by the product of frequency and the wavelength i.e.

$v=\nu\times \lambda$

$v=5\times 0.01$

$v=0.05\ m/s$

Hence, the speed of the wave is 0.05 m/s

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The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizo

Jobisdone [24]

(a) 4.0 m/s

We can solve this part just by analyzing the vertical motion of the froghopper.

The initial vertical velocity of the froghopper as it jumps from the ground is given by

$u_y = u_0 sin \theta$ (1)

where

$u_0$ is the takeoff speed

$\theta=58.0^{\circ}$ is the angle of takeoff

The maximum height reached by the froghopper is

h = 58.7 cm = 0.587 m

We know that at the point of maximum height, the vertical velocity is zero:

$v_y = 0$

Since the vertical motion is an accelerated motion with constant (de)celeration $g=-9.8 m/s^2$ , we can use the following SUVAT equation:

$v_y^2 - u_y^2 = 2gh$

Solving for $u_y$ ,

$u_y = \sqrt{v_y^2-2gh}=\sqrt{-2(-9.8)(0.587)}=3.4 m/s$

And using eq.(1), we can now find the initial takeoff speed:

$u_0 = \frac{u_y}{sin \theta}=\frac{3.4}{sin 58.0^{\circ}}=4.0 m/s$

(b) 1.47 m

For this part, we have to analyze the horizontal motion of the froghopper.

The horizontal velocity of the froghopper is

$u_x = u_0 cos \theta = (4.0) cos 58.0^{\circ} =2.1 m/s$

And this horizontal velocity is constant during the entire motion.

We now have to calculate the time the froghopper takes to reach the ground: this is equal to twice the time it takes to reach the maximum height.

The time needed to reach the maximum height can be found through the equation

$v_y = u_y + gt$

Solving for t,

$t=-\frac{u_y}{g}=-\frac{3.4}{9.8}=0.35 s$

So the time the froghopper takes to reach the ground is

$T=2t=2(0.35)=0.70 s$

And since the horizontal motion is a uniform motion, we can now find the horizontal distance covered:

$d=u_x T = (2.1)(0.70)=1.47 m$

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3 years ago

Using two boxes, design two solar ovens. Before you build, describe your designs. Discuss elements such as materials, box shape,

Marizza181 [45]

Answer:

Find two boxes just a bit smaller than the other. The long sides of the box should be less than twice as long as the short side of the boxes. The smaller box should fit inside the larger box with about 1 inch in each direction to spare. The boxes can be cut down so that they fit together properly. Leave the flaps on the boxes. Buy a small sheet of Plexiglas (tm) a little bit smaller than the width and length of the top of the box. You will also need four pieces of cardboard to use for reflectors.

Explanation:

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2 years ago

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sasho [114]

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A cylindrical tank has a radius of 53 cm and a height of 1200 mm. It weighs 9.6 kN

xxTIMURxx [149]

Answer:

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qaws [65]

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