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3 years ago

15

Specialization can be both limiting and beneficial. How would a plant species benefit from a pollinator that only visits the flo

wers of that plant species?

1 answer:

Anastasy [175]3 years ago

4 0

Answer:

Extraordinarily compared to the plants that aren't

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An unknown radioactive element decays with a half life of 0.210 days. What is the decay constant, expressed in 1/microseconds, f

sergeinik [125]

The amount of a radioactive element over time can be written as:

$N=N_0e^{-\lambda t}$

So, considering that we know the time of its half life, we know how long it takes to get to half the original amount. This is (usint the amount of days in microseconds):

$0.5=1*e^{-\lambda *0.21*24*3600*10^6}\Rightarrow0.5=e^{-1.8144*10^{10}*\lambda}$ $ln(0.5)=-1.8144*10^{10}*\lambda\Rightarrow\lambda=\frac{ln(0.5)}{-1.8144*10^{10}}=3.82*10^{-11}$

Thus, lambda=3.82*10^(-11)/microseconds

3 0

2 years ago

the air in a tire is initially at 380 kpa, 20 C, when the tire volume is 0.120 m^3. as the tire is warmed by the sun, the pressu

castortr0y [4]

Answer:

final temperature is 364.32 K

mass of air is 0.5423 kg

Explanation:

given data

pressure p1 = 380 kPa

volume v1 = 0.120 m³

temperature = 20°C = 20 + 273 = 293 K

pressure p2 = 3450 kPa

volume v2 = 5% increase

to find out

final temperature and the mass

solution

we consider here ideal gas

so equation is

p1v1 / t1 = p2v2/t2

put here all value and find t2

and v2 is = 0.120 ( 1 + 0.05) = 0.126 m³

so

t2 = p2v2/ p1v1 × t1

t2 = (450 × 0.126) / (380 × 0.120) × 293

t2 = 364.32 K

so final temperature is 364.32 K

and

mass = p2v2 / Rt2

here R gas constant is 0.287 kJ/kg.K

so

mass = (450 × 0.126) / (0.287 × 364.32 )

mass = 0.5423

so mass of air is 0.5423 kg

7 0

3 years ago

How do earthquakes affect boulders

MaRussiya [10]

Weathering and rock slides

5 0

4 years ago

A luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 25 degrees above the horizontal by a force F of magnitude 145 N

Nonamiya [84]

Answer:

A) 667 J

B) 381.4 J

C) 0 J

D) 245.4 J

E) 40.2J

F) 2 m/s

Explanation:

Let g = 9.81 m/s2

A) The work done on the suitcase is the product of the force applied and the distance travelled:

w = Fs = 145 * 4.6 = 667 J

B) The work done by gravitational force the dot product between the gravity vector and the distance vector

$W_g = \vec{P}\vec{s} = mgs sin\alpha = 20*9.81*4.6*sin25^o = 381.4 J$

C) As the normal force vector is perpendicular to the distance vector, the work done by the normal force is 0

D) The work done on the suitcase by friction force is the product of the force applied and the distance travelled, whereas friction force is the product of normal force and coefficient

$W_f = F_fs = \mu N s = \mu s mgcos\alpha = 0.3* 4.6 * 20*9.81*cos25^o = 245.4 J$

E) The total workdone on the suite case would be the pulling work subtracted by gravity work and friction work

$W = w – W_g – W_f = 667 – 381.4 – 245.4 = 40.2 J$

F) As the suit case has 0 kinetic and potential energy at the bottom, and the total work done is converted to kinetic energy at 4.6 m along the ramp, we can conclude that:

$E_k = W = 40.2 j$

$mv^2/2 = 40.2$

$20v^2/2 = 40.2$

$10v^2 = 40.2$

$v^2 = 4.02$

$v = \sqrt{4.02} = 2 m/s$

3 0

3 years ago

Match the following waves on the EM Spectrum with the correct letter;

vodka [1.7K]

Answer:

(D) is visible light

Explanation:

(F) is xray and (B) is radio

5 0

3 years ago