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3 years ago

Which of the following statements best support the claim that human actions decreased the whooping crane population? (1

Physics

1 answer:

Anna35 [415]3 years ago

8 0

Answer:

Explanation:

A is incorrect because it would actually increase the population...

B is incorrect because other birds do not matter, it is only whooping cranes...

C is incorrect because it would also increase the population.

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Add a vector whose magnitude is 13 with angle 27 degrees to one whose magnitude is 11 with angle 45 degrees? Put the length firs

mafiozo [28]

Answer:

Magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$

Explanation:

Magnitude of first vector = $|A| = 13\ \text{units}$

Angle = $\theta_1=27^{\circ}$

Magnitude of second vector = $|B| = 11\ \text{units}$

Angle = $\theta_2=45^{\circ}$

x component of first vector

$A_{x}=|A|\cos\theta_1\\\Rightarrow A_x=13\cos27^{\circ}\\\Rightarrow A_x=11.6\ \text{units}$

y component of first vector

$A_{y}=|A|\sin\theta_1\\\Rightarrow A_y=13\sin27^{\circ}\\\Rightarrow A_y=5.9\ \text{units}$

x component of second vector

$B_{x}=|B|\cos\theta_2\\\Rightarrow B_x=11\cos45^{\circ}\\\Rightarrow B_x=7.8\ \text{units}$

y component of first vector

$B_{y}=|B|\sin\theta_2\\\Rightarrow B_y=11\sin45^{\circ}\\\Rightarrow A_y=7.8\ \text{units}$

Adding the magnitudes

$C_x=A_x+B_x=11.6+7.8\\\Rightarrow C_x=19.4\ \text{units}$

$C_y=A_y+B_y=5.9+7.8\\\Rightarrow C_y=13.7\ \text{units}$

Magnitude of the sum of the vectors would be

$|C|=\sqrt{C_x^2+C_y^2}\\\Rightarrow |C|=\sqrt{19.4^2+13.7^2}=23.75\ \text{units}$

The direction would be

$\theta=\tan^{-1}\dfrac{C_y}{C_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{13.7}{19.4}\\\Rightarrow \theta=35.23^{\circ}$

The magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$

4 0

3 years ago

Complete the passage.

nataly862011 [7]

Answer:

Answer: Sound waves and some earthquake waves are longitudinal waves. Ocean, light and other earthquake waves are transverse waves.

Explanation:

There are 2 types of waves:

1. Longitudinal waves: These waves are defined as the waves in which the particles of the medium move in the direction of the wave. This requires a medium to travel. For Example: Sound Waves.

2. Transverse wave: These waves are defined as the waves in which the particles of the medium travel perpendicularly to the direction of the wave. This does not require a medium to travel. These can travel in vacuum also. For Example: Light waves.

Hence, Sound waves and some earthquake waves are longitudinal waves. Ocean, light and other earthquake waves are transverse waves

8 0

3 years ago

At a certain location the horizontal component of the earth's magnetic field is

Katena32 [7]

At a certain location, the horizontal component of the earth’s magnetic field is, due north<span>. A proton moves eastward with just the right speed, so the magnetic force on it balances its weight.

</span>

6 0

4 years ago

A positive charge of 6.0 x 10-4 C is in an electric field that exerts a force of 4.5 x 10 -4 on it. What is the strength of the

Yuki888 [10]

Electric field is defined as force per unit charge.

So it is given by

$F = q E$

now we can find electric field by

$E = \frac{F}{q}$

$E = \frac{4.5*10^{-4}}{6 * 10^{-4}}$

$E = 0.75 N/C$

So field strength is 0.75 N/C.

6 0

4 years ago

The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?

Tju [1.3M]

Answer:

The second distance of the sound from the source is 431.78 m..

Explanation:

Given;

first distance of the sound from the source, r₁ = 1.48 m

first sound intensity level, I₁ = 120 dB

second sound intensity level, I₂ = 70.7 dB

second distance of the sound from the source, r₂ = ?

The intensity of sound in W/m² is given as;

$dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2$

$For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2$

The second distance, r₂, can be determined from sound intensity formula given as;

$I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 = \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 = \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m$

Therefore, the second distance of the sound from the source is 431.78 m.

7 0

3 years ago