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3 years ago

What is the frequency of a wave if 3 crests pass in 6 seconds?

Physics

2 answers:

AlexFokin [52]3 years ago

8 0

A. 0.5 hz because frequency os the amount of waves that pass per one second.

zavuch27 [327]3 years ago

6 0

Answer:

b. 2Hz

Explanation:

Frequency is defined as the number of wavelengths passing a particular point per second.

If 3 crests pass a point in two seconds, then we know that 6 crests pass the point per second.

Therefore 2 wavelengths pass the point per second, and hence the frequency of the wave is 2Hz.

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How do the nonliving parts of earth's systems provide the basic materials to support life?

MatroZZZ [7]

The abiotic community of the ecosystem or the nonliving objects in the earth's biosphere influences the survival, growth and development of the organism. They provide basic mmaterials to support life by the biogeochemical processes which is composed of many biological and environmental cycles which profoundly has factored the intiation and continuation of life on earth. They also provide the basic materials to support life by the undergoing process which is also initiated by the living creatures or the biotic society of the ecosystem.

3 0

3 years ago

1-Autotrophic plants are also called

tresset_1 [31]

Answer:

1. Producers

2. cell (guards)

3. Rhizobium

4. Rafflesia Arnoldii

5. Xylem

3 0

3 years ago

When a certain rubber band is stretched a distance x, it exerts a restoring force of magnitude f = ax, where a is a constant. th

Veseljchak [2.6K]

Given:
F = ax
where
x = distance by which the rubber band is stretched
a = constant

The work done in stretching the rubber band from x = 0 to x = L is
$W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2} [x^{2} ]_{0}^{L} = \frac{aL^{2}}{2}$

Answer: $\frac{aL^{2}}{2}$

4 0

4 years ago

A 36N force applied vertically upwards on an object causes it to accelerate at 2.0m/s2. What is the object's mass?

Natalija [7]

Force = mass * acceleration

mass = Force/acceleration

= 36/2

mass = 18 kg

7 0

3 years ago

In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope

zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

If we have the initial velocity $v_o$ and its angle $\theta$ , we can obtain the vertical component of the velocity $v_{oy}$ using trigonometry:

$v_{oy}=v_osin\theta$

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

$v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }$

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$

In words, the projectile travels 7.49m horizontally before it lands.

8 0

4 years ago