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lyudmila [28]
2 years ago
9

Which of the following most directly limits the number of organisms in a community?

Physics
1 answer:
astraxan [27]2 years ago
7 0

Answer:

It can either be food, mating, or competition with other organisms for resources.

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A watt is a measure of power (the rate of energy change) equal to 1 j>s. (a) calculate the number of joules in a kilowatt- ho
GalinKa [24]
Before solving this, you must know the definition of these units of measurement. Watt is a measurement of power which is the amount of energy per unit time in seconds. Energy, on the other hand, is expressed through the SI unit Joules. Thus, power is the amount of energy in Joules per second.

From here, you can use the dimensional analysis technique. Also, you should know that 1 kilowatt is equal to 1000 watts, and 24 hours is equal to 86,400 seconds. Then,

100 = Energy/86,400
Energy = 8,640,000 Joules

4 0
3 years ago
Which planet has a moon called titan?
babunello [35]
Titan is the largest moon in Saturn.
8 0
2 years ago
Read 2 more answers
An electron travels with speed 6.0×10^6 m/s between the two parallel charged plates shown in the figure. The plates are separate
svlad2 [7]

Answer:

B= 3.33 m T

Explanation:

Given that

Speed ,C= 6 x 10⁶  m/s

d= 1 cm = 0.01 m

V= 200 V

The electric field E given as

V= E .d

E=Electric field

d=Distance

V=Voltage

200 = 0.01 x E

E=20000 V/m

The relationship between magnetic and electric field given as

E= C x B

20000 = 6 x 10⁶   x B

B =3333.333 x 10⁻⁶ T

B= 3.33 x 10⁻³ T

B= 3.33 m T

Therefore the magnetic filed will be 3.33 m T.

5 0
3 years ago
A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar
denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of \beta

Using formula of \beta

\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}

Put the value into the formula

\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}

\beta=8.86\times10^{9}

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}

Put the value into the formula

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}

T=5.45\times10^{-4}

(b). We need to calculate the tunnel probability for width 1.00 nm

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}

T=7.74\times10^{-8}

Hence, The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

6 0
3 years ago
Janet wants to find the spring constant of a given spring, so she hangs the spring vertically and attaches a 0.50 kg mass to the
frez [133]
Given: Mass m = 0.50 Kg;     Force = Weight = mg   F = (0.50 Kg)(9.8 m/s²)

                                               F = 4.9 N      

Displacement  x = 3.0 cm   convert to meter   x = 0.03 m

Required: Spring constant  k = "

Formula:  F = kx

                k = F/x

                k = 4.9 N/0.03 m

                k = 163.33 N/m


                    

                                                                      
3 0
3 years ago
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