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aleksandr82 [10.1K]

2 years ago

5

The greater the amplitude of a wave a. The quieter the sound b. The higher the frequency of the sound c. The dimmer the light in

tensity d. The greater the amount of energy in the waveImmersive Reader (1 Point) a b c d

1 answer:

Stolb23 [73]2 years ago

8 0

Answer:

The greater the amplitude the greater the energy.

(Think of a water wave - which carries greater energy a 1 ft wave or

a 10 ft wave)

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Think about the various energy sources such as the sun, fossil fuels, water, wind, and nuclear power. What is the energy source

lakkis [162]

Answer:

<em>The energy source that powers my home is gotten from burning of fossil fuels.</em>

<em>No! I do not like this energy source.</em>

<em>I personally would prefer solar source of electricity</em>

<em></em>

Explanation:

<em>Fossil fuels are fuels gotten from the decomposition of dead organisms over time, under intense heat and pressure</em>. They are usually found buried beneath the earth's crust where they have been formed and trapped.

Most electricity generating stations generate electricity by burning fossil fuels like natural gas and gasoline to generate electricity. <em>The problem with fossil fuels are the various part that they play in increasing carbon footprints in the atmosphere. The excess carbon in the atmosphere has been a major contributor to the global warming of planet Earth. </em>

My preference for solar energy source is first due to its abundance unlike the fossil fuels that are already diminishing in storage beneath the earth's crust. Also, <em>solar energy is a clean source of energy that does not leave any damage on earth from its use</em>. It also promises to be a cheap source of power in the future with advances in solar technologies.

8 0

2 years ago

T=2pi square root 1/g solve for g.<br> Explanation would be really helpful.

Natalija [7]

I added individual steps for clarity. Note that g must be positive if the solution is to be real.

$T=2\pi \sqrt{\frac{1}{g}}=2\pi g^{-\frac{1}{2}}\\g^{-\frac{1}{2}} = \frac{T}{2\pi}\\(g^{-\frac{1}{2}})^{-2} = (\frac{T}{2\pi})^{-2}\\g = \frac{4\pi^2}{T^2}\,\,\,, g>0}$

Let me know if you have any questions.

7 0

3 years ago

What is the impulse needed to stop a 45-kg boy who is running at 6 m/s in 3 seconds?

Tju [1.3M]

Impulse = change of momentum
Impulse = 45 x 6 = 270 Ns

6 0

2 years ago

Quel est le type de rayonnement produit par la radioactivité

Dmitriy789 [7]

Il existe troi types de rayons produits lors de la désintégration des éléments radioactifs:

-- "particules alpha" . . . noyaux d'hélium, composés chacun de 2 protons et 2 neutrons

-- "rayons bêta" ou "particules bêta" . . . flux d'électrons

-- "rayons gamma" . . . rayonnement électromagnétique avec les longueurs d'onde les plus courtes connues et l'énergie la plus élevée

8 0

3 years ago

An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 sec

aivan3 [116]

Answer:

244.64m

Explanation:

First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:

$x = V*t = -8\frac{m}{s} *3s = -24m$

After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

$v^2 = v_0^2 + 2a*d$

Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

$d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m$

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

$t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s$

Then, the time of the second phase will be:

$t_2 = 9s - t_1 = 9s - 1.143s = 7.857s$

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

$x = \frac{1}{2}a*t^2 + v_0*t + x_0$

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

$x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m$

So, the total distance covered by this object in meters will be the sum of all the distances we found:

$x_total = 24m + 4.57m + 216.07m = 244.64m$

8 0

2 years ago