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mars1129 [50]
3 years ago
8

In which biome would you most likely find plants and animals that are adapted to survive in hot humid climates

Physics
2 answers:
joja [24]3 years ago
7 0

in the desert,  I believe

Serggg [28]3 years ago
4 0

Answer: Desert biome

Explanation:

The animals and plants of desert biome are perfectly suited to live there. These plants ad animals have certain adaptation that helps them to survive in such a hot climate and at a very high temperature.

The animals there lose water very easily because of heat so they have hard cuticles on their skin to absorb water like blotting paper.

Animals living there most search food at night time to avoid heat loss.

Plants have no leaves and more cuticles to avoid transpiration.

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A train is moving at a speed of 50 km/h. How many hours will it take thetrain to travel 600 kilometers
olga_2 [115]

Answer:

12

Explanation:

600/50 is 12 so the answer is twelve

5 0
2 years ago
Do you think toxicity is a qualitative or quantitative property? explain
Tomtit [17]

Answer:

<u>Toxicity is a quantitative property</u>

Explanation:

  • Qualitative property of a object cannot be measured it can just be observed
  • Quantitative property of a substance  can be measured and be assigned a numerical value .
  • <u>The toxicity level of a substance can be measured and be assigned a numeral value </u>

<u />

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3 0
3 years ago
Which of the following metals require ultraviolet light to exhibit the photoelectric effect?The options available: a. Cs, work f
Eddi Din [679]

Answer:

b. AG, work function=4.74eV

Explanation:

Ultraviolet light starts at the end of the visible light spectrum, where violet light ends:

\lambda=380 nm =3.8\cdot 10^{-7}m (wavelength of lowest-energy ultraviolet light)

So, the lowest energy of ultraviolet light can be found by using the formula

E=\frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

Substituting,

E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.8\cdot 10^{-7} m}=5.23\cdot 10^{-19}J

And keeping in mind that

1 eV = 1.6\cdot 10^{-19}J

This energy converted into electronvolts is

E=\frac{5.23\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=3.27 eV

The work function of a metal is the minimum energy needed to extract a photoelectron from the surface of the metal. Therefore, the metals that exhibit photoelectric effect are the ones whose work function is larger than the energy we found previously, so:

b. AG, work function=4.74eV

Because for all the other metals, visible light will be enough to extract photoelectrons.

7 0
3 years ago
The uniform slender bar AB has a mass of 6.4 kg and swings in a vertical plane about the pivot at A. If θ˙ = 2.7 rad/s when θ =
dolphi86 [110]

Answer:

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Explanation:

Given data,

The mass of the bar AB, m = 6.4 kg

The angular velocity of the bar,  θ˙ = 2.7 rad/s

The angle of the bar at A, θ = 24°

Let the length of the bar be, L = l

The angular moment at point A is,

                        ∑ Mₐ = Iα

Where,     Mₐ - the moment about A

                 α  - angular acceleration

                 I - moment of inertia of the rod AB

                       -mg(\frac{lcos\theta}{2})=\frac{1}{3}(ml^{2})\alpha

                        \alpha=\frac{-3gcos\theta}{2l}

Let G be the center of gravity of the bar AB

The position vector at A with respect to the origin at G is,

                          \vec{r_{G}}=[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]

The acceleration at the center of the bar

                          \vec{a_{G}}=\vec{a_{a}}+\vec{\alpha}X\vec{r_{G}}-\omega^{2}\vec{r_{G}}

Since the point A is fixed, acceleration is 0

The acceleration with respect to the coordinate axes is,

                         (\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=0+(\frac{-3gcos\theta}{2l})\hat{k}\times[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]-\omega^{2}[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]

(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=[-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}\hat{i}+(\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4})\hat{j}]

Comparing the coefficients of i

=-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}

Comparing coefficients of j

(\vec{a_{G}})_{y}=\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4}

Net force on x direction

F_{x}=(\vec{a_{G}})_{x}

substituting the values

F_{x}=1.5(14.58L+11.96)

Similarly net force on y direction

F_{y}=(\vec{a_{G}})_{y}+mg

               = 3.2(2.97L - 157.03) + 62.72

Where L is the length of the bar AB

Therefore the net force,

F=\sqrt{F_{x}^{2}+F_{y}^{2}}

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Substituting the value of L gives the force at pin A

8 0
2 years ago
Why is the speed of different sounds in a single medium remain the same though wavelength is different?
WINSTONCH [101]

Answer:

I think it's because the refractive index of that medium is constant

6 0
2 years ago
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