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4 years ago

10

William wanted to create a report on a geographical location with the greatest species diversity. Which ecosystem can consider f

or his report?

1 answer:

FrozenT [24]4 years ago

8 0

Answer:

The ecosystem that he should consider is the tropical rain forest ecosystem.

Explanation:

In that area, there are uncountable amounts of various plants and animals that have not all yet been discovered and who all live together to build the biome. This biome is indeed the most diverse one even at this point without the knowledge of all possible life forms.

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A 15lb bowling ball and a 10 pound bowling ball are

emmasim [6.3K]

Same time
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3 years ago

Earth's orbit around the Sun is slightly elliptical. Thus, Earth actually gets closer to the Sun during part of the year. What h

Fudgin [204]

Answer:

C)Earth's orbital speed is greater when it is closer to the Sun than when it is farther from the Sun.

Explanation:

We can use Kepler's 2nd law or the law of area to answer this question.

The law states that the rate of area swept out by a planet's orbit is same throughout the orbit. For the farthest point since the distance is large as compared to to the nearest point, the possibility that area swept is large. Hence, to compensate the extra swept area. the orbital speed has to decrease at the largest point.

Hence, planet's speed is greater when it is closer to sun than the speed when it is farther.

6 0

3 years ago

An electron is moving through a magnetic field whose magnitude is 9.21 × 10^-4 T. The electron experiences only a magnetic force

Marysya12 [62]

Answer:

$\theta=10.60^{\circ}$

Explanation:

Given that,

Magnetic field, $B=9.21\times 10^{-4}\ T$

Acceleration of the electron, $a=2.3\times 10^{14}\ m/s^2$

Speed of electron, $v=7.69\times 10^{6}\ m/s$

The force due to this acceleration is balanced by the magnetic force as :

$ma=qvB\ sin\theta$

$sin\theta=\dfrac{ma}{qvB}$

$sin\theta=\dfrac{9.1\times 10^{-31}\times 2.3\times 10^{14}}{1.6\times 10^{-19}\times 7.69\times 10^{6}\times 9.21\times 10^{-4}}$

$\theta=10.60^{\circ}$

So, the angle between the electron's velocity and the magnetic field is 10.6 degrees. Hence, this is the required solution,

8 0

4 years ago

How does electricity turn into light in a common lightbulb?

Anestetic [448]

Answer:

C or D

Explanation:

5 0

3 years ago

Read 2 more answers

Two strings on a musical instrument are tuned to play at 196 hz (g) and 523 hz (c). (a) what are the first two overtones for eac

Tems11 [23]

(a) first two overtones for each string:
The first string has a fundamental frequency of 196 Hz. The n-th overtone corresponds to the (n+1)-th harmonic, which can be found by using
$f_n = n f_1$
where f1 is the fundamental frequency.

So, the first overtone (2nd harmonic) of the string is
$f_2 = 2 f_1 = 2 \cdot 196 Hz = 392 Hz$
while the second overtone (3rd harmonic) is
$f_3 = 3 f_1 = 3 \cdot 196 Hz = 588 Hz$

Similarly, for the second string with fundamental frequency $f_1 = 523 Hz$ , the first overtone is
$f_2 = 2 f_1 = 2 \cdot 523 Hz = 1046 Hz$
and the second overtone is
$f_3 = 3 f_1 = 3 \cdot 523 Hz = 1569 Hz$

(b) The fundamental frequency of a string is given by
$f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }$
where L is the string length, T the tension, and $\mu = m/L$ is the mass per unit of length. This part of the problem says that the tension T and the length L of the string are the same, while the masses are different (let's calle them $m_{196}$ , the mass of the string of frequency 196 Hz, and $m_{523}$ , the mass of the string of frequency 523 Hz.
The ratio between the fundamental frequencies of the two strings is therefore:
$\frac{523 Hz}{196 Hz} = \frac{ \frac{1}{2L} \sqrt{ \frac{T}{m_{523}/L} } }{\frac{1}{2L} \sqrt{ \frac{T}{m_{196}/L} }}$
and since L and T simplify in the equation, we can find the ratio between the two masses:
$\frac{m_{196}}{m_{523}}=( \frac{523 Hz}{196 Hz} )^2 = 7.1$

(c) Now the tension T and the mass per unit of length $\mu$ is the same for the strings, while the lengths are different (let's call them $L_{196}$ and $L_{523}$ ). Let's write again the ratio between the two fundamental frequencies
$\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L_{523}} \sqrt{ \frac{T}{\mu} } }{\frac{1}{2L_{196}} \sqrt{ \frac{T}{\mu} }}$
And since T and $\mu$ simplify, we get the ratio between the two lengths:
$\frac{L_{196}}{L_{523}}= \frac{523 Hz}{196 Hz}=2.67$

(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them $T_{196}$ and $T_{523}$ . Let's write again the ratio of the frequencies:
$\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L} \sqrt{ \frac{T_{523}}{m/L} } }{\frac{1}{2L} \sqrt{ \frac{T_{196}}{m/L} }}$
Now m and L simplify, and we get the ratio between the two tensions:
$\frac{T_{196}}{T_{523}}=( \frac{196 Hz}{523 Hz} )^2=0.14$

7 0

3 years ago