Myth: An organism's kingdom only describes physical characteristics. Fact: Evidence:

1 answer:

7 0

FACT: An organism's kingdom does not only describe physical characteristics.

Organisms are classified into the different kingdoms based on physical appearance, their function, how they reproduce and how they obtain their food.

EVIDENCE:
Kingdom Plantae: They obtain their food through photosynthesis
Reproduce asexually.
Kingdom Animalia : They obtain their food through hunting or consuming other organisms.
Reproduce sexually.

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An ultrasound pulse has a wavelength of 1.0mm. Its speed in water is 1400m. What’s the frequency?

timama [110]

Answer:

$1.4\cdot 10^6 Hz$

Explanation:

The relationship between the frequency, the wavelength and the speed of a wave is given by the wave's equation:

$v=f \lambda$

where

v is the speed of the wave

f is the frequency

$\lambda$ is the wavelength

For the pulse in this problem,

$\lambda = 1.0 mm = 0.001 m\\v = 1400 m/s$

Solving for f, we find the frequency:

$f=\frac{v}{\lambda}=\frac{1400}{0.001}=1.4\cdot 10^6 Hz$

6 0

2 years ago

A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?

Tcecarenko [31]

The weight of the meterstick is:
$W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N$
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
$d_1 = 0.50 m - 0.40 m=0.10 m$
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
$M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm$

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
$(mg) d_2 = 0.20 Nm$
from which we find the value of d2:
$d_2 = \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m$

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.

4 0

2 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$