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Virty [35]
4 years ago
11

Which of the following is an example of a population?

Physics
2 answers:
svlad2 [7]4 years ago
7 0
D the gray wolves in the forest.is the correct answer.

Bogdan [553]4 years ago
7 0
<span>D is the answer because it describes a single species in a specific environment.


</span>
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Does the thickness of the wire affect the strength of an electroscope
TiliK225 [7]
It probably does. I'm not 100% sure about it, but a thicker wire would increase the number of positive and negative charges in it.
4 0
3 years ago
Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900
Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron m_{e}=9.11\times10^{-31}\ kg

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}

E=0.582\ Mev

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

E=0.350\ Mev

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0
3 years ago
An electron has an initial velocity of (19.0 j + 18.0 k) km/s and a constant acceleration of (3.00 ✕ 1012 m/s2)i in a region in
asambeis [7]

Answer:

E=(-17.08 i +7.2 j -7.6 k )N/C

Explanation:

v= (19.0j+18.k)km/s

a=3.0x10^{13}m/s^2

\beta= 400x10{-6}T

Electron information needed to solve the question:

m_e=9.11x10^{-31}kg

q=-1.6x10{-19}C

F=F_E+F_B=q*(E+Vx \beta)

F=m*a

m*a=q*(E+Vx\beta)

E=\frac{m*a}{q}-(Vx\beta )

E=\frac{9.11x10{-31}kg*3.0x10^{12}m/s^2}{-1.6x10{-19}C}-[(19.0x10^3mj+18.0x10^3m)xi(400x10^{-6}T)]

E=-i17.08N/C-[7.6(-k)+7.2(j)]N/C

E=(-17.08 i +7.2 j -7.6 k )N/C

6 0
3 years ago
Microwave ovens use microwave radiation to heat food. the microwaves are absorbed by the water molecules in the food, which is t
PIT_PIT [208]
<span>step 1: energy required to heat coffee E = m Cp dT E = energy to heat coffee m = mass coffee = 225 mL x (0.997 g / mL) = 224g Cp = heat capacity of coffee = 4.184 J / gK dT = change in temp of coffee = 62.0 - 25.0 C = 37.0 C E = (224 g) x (4.184 J / gK) x (37.0 C) = 3.46x10^4 J step2: find energy of a single photon of the radiation E = hc / λ E = energy of the photon h = planck's constant = 6.626x10^-34 J s c = speed of light = 3.00x10^8 m/s λ = wavelength = 11.2 cm = 11.2 cm x (1m / 100 cm) = 0.112 m E = (6.626x10^-34 J s) x (3.00x10^8 m/s) / (0.112 m) = 1.77x10^-16 J step3: Number of photons 3.46x10^4 J x ( 1 photon / 1.77x10^-16 J) = 1.95x10^20 photons</span>
7 0
3 years ago
Read 2 more answers
A miner of mass 90kg travels down a slide calculate the potential energy of the miner when he moves15m vertically downwards
Lena [83]
Answer:

Gravitational Potential Energy = mgh

Explanation:

As the miner moves down, the GPE changes because the height changes.

Gravitational Potential Energy = mgh


8 0
3 years ago
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