Answer:
Solution given:
Radius of small sphere[r]=5cm=0.05m
Radius of large sphere[R]=10cm=0.1m
capacitance of small sphere[c]=4πε0r
=[tex]4π*8.85×10^{-12}×0.05=5.56*10^{-12}F[/tex]
Charge for small sphere[Q1]=100C
Charge for small sphere[Q2]=50C
Potential difference [V1]=[tex] \frac{charge}{capacitance}=\frac{100}{5.56*10^{-12}}=1.8×10^{13}[/tex]V
.
again
capacitance of small sphere[C]=4πε0R
=[tex]4π*8.85×10^{-12}×0.1=1.11*10^{-11}F[/tex]
Potential difference [V2]=[tex] \frac{charge}{capacitance}=\frac{50}{1.11*10^{-11}}=4.5×10^{12}[/tex]V
Now
Loss of energy:
[tex] \frac{cC(V1-V2)^{2}}{2(c+C)}[/tex]
=[tex] \frac{5.56*10^{-12}*1.11*10^{-11}(1.8*10^{13}-4.5*10^{12})^{2}}{2(5.56*10^{-12}+1.11*10^{-11})}[/tex]
=25Joule
You are studying a population of flowering plants for several years. When you present your research findings you make the statement that, "Increased allocation of resources to reproduction relative to growth diminished future fecundity." Which of the following graph descriptions could accurately present your data?
a) With seeds in the current year on the y-axis and seeds in the previous year on the x-axis, you would see a line that increased from left to right
b) With survivorship on the y-axis and number of seeds produced on the x-axis, you would see a line that decreased left to right.
c) With leaf area on the y-axis and number of seeds produced on the x-axis, you would see a line that increased left to right
d) With survivorship on the y-axis and number of seeds produced on the x-axis, you would see a line that increased left to right.
e) With seeds in the current year on the y-axis and seeds in the previous year on the x-axis, you would see a line that decreased from left to right
Answer:
Option A
Explanation:
The graph for this problem must depict the following ""Increased allocation of resources to reproduction relative to growth diminished future fecundity."
Hence, the survivor ship must be on the Y axis and the resources on the X axis.
Here the resources include the number of seeds produced.
hence, the higher is the number of seeds (resource), the lower is the survivorship (future fecundity)
Hence, option A is correct
What are the biotic factors in this image?
Someone help thank you!!
Suppose that an electron and a positron collide head-on. Both have kinetic energy of 1.20 MeV and rest energy of 0.511 MeV. They produce two photons, which by conservation of momentum must have equal energy and move in opposite directions. What is the energy Ephoton of one of these photons
Answer:
E = 1.711 MeV
Explanation:
From the law of the conservation of energy:
[tex]K.E_{e}+K.E_p + E_{e}+E_{p} = 2 E[/tex]
where,
[tex]K.E_e=K.Ep=[/tex] the kinetic energy of positron and electron = 1.2 MeV
[tex]E_e=E_p =[/tex] Rest energy of the electron and the positron = 0.511 MeV
E = Energy of Photon = ?
Therefore,
[tex]1.2\ MeV + 1.2\ MeV + 0.511\ MeV + 0.511\ MeV = 2E\\\\E = \frac{3.422\ MeV}{2}\\\\[/tex]
E = 1.711 MeV
What is Hooke's law? what is meant by elastic limit?
please answer me
Answer:
Explanation:
The law is about how object will be deformed when certain amount force is applied. Elastic limit is a point where object will not return its previous shape after being deformed
he electric field is dependent on the distance between the plates. false: The voltage of a connected charged capacitor decreases when the plate area is increased. false: The voltage of a disconnected charged capacitor increases when the plates are brought closer together.
Answer:
False.
Explanation:
The voltage of a disconnected charged capacitor decreases when the plates are brought closer together because the capacitance is inversely proportional to the area. If the area between plates decreases, its capacitance increases and vice versa. There is direct relationship between voltage of a disconnected charged capacitor and plates. If the distance between plates decreases, the voltage of a disconnected charged capacitor is also decreases while on the other hand, if the distance between plates increases, the voltage of a disconnected charged capacitor is also increases.
A spinning disc with a mass of 2.5kg and a radius of 0.80m is rotating with an angular velocity of 1.5 rad/s. A ball of clay with unknown mass is dropped onto the disk and sticks to the very edge causing the angular velocity of the disk to slow to 1.13 rad/s. What is the mass of the ball of clay
Answer:
M = 1.90 Kg
Explanation:
Given data: mass = 2.5 Kg
radius R = 0.8 m
angular velocity ω = 1.5 rad/s
Angular momentum L =0.5×Iω^2
Where, I is the moment of inertia of the spinning disc.
I = 0.5MR^2
I = 0.5×2.5×0.8^2
I = 0.8 Kg/m^2
Then L = 0.5×0.8×1.5^2 = 0.8×2.25 = 0.9 Kg-m^2/sec
Let unknown mass be M
New mass of disc = (2.5+M) Kg, R = 0.8 m
New I = 0.5(2.5+M)(0.8)^2
Since, angular momentum is conserved
Angular momentum before = angular momentum after
0.5×0.5(2.5+M)(0.8)^2×(1.13)^2 = 0.9
Solving for M we get
0.204304(2.5+M)=0.9
M = 1.90 Kg
Please help ASAP with questions
A 10- kg ball starting from rest rolls down a 5 m tall smooth hill from one person to another person who is standing at the bottom of the hill with a big spring whose constant is 100 N/m. How far does the spring compress in order to stop the ball
Answer: 3.13 m
Explanation:
Given
mas of the ball is m=10 kg
The ball rolls down a vertical distance of 5 m
Spring constant of spring is [tex]k=100\ N/m[/tex]
Here, the potential energy of the ball converted into kinetic energy which in turn converts into elastic potential energy
[tex]\Rightarrow mgh=\frac{1}{2}kx^2\quad [\text{x=compression in the spring}]\\\\\Rightarrow 10\times 9.8\times 5=\frac{1}{2}\cdot 100\cdot x^2\\\Rightarrow x=\sqrt{9.8}\\\Rightarrow x=3.13\ m[/tex]
Thus, the spring compresses by 3.13 m.