Oxygen. Plants need water sunlight and CO2 to make their food. They get the carbon and the energy from the CO2 and the sun, and they need water as well. Then they release oxygen
Question:
A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.
a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.
Express your answer using two significant figures.
Answer:
The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity
Explanation:
A point charge ,q = is located in the center of a spherical cavity of radius , m inside an insulating spherical charged solid.
The charge density in the solid , d =
Distance from the center of the cavity,R =
Volume of shell of charge= V =
Charge on the shell ,Q =
Electric field at m due to shell
E1 =
Electric field at due to 'q' at center
E2 =
The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity
= E2- E1
Answer:
M. Magnetism is a property of individual atoms.
Explanation:
when a magnet is broken into pieces the new pieces behave like the original magnet this observation shows that magnetism is the property of individual atoms.
Answer:
d= 1.56 m
Explanation:
In order to have a constructive interference, the path difference between the sources of the sound, must be equal to an even multiple of the semi-wavelength, as follows:
⇒ d = d₂ - d₁ = 2n*(λ/2)
The minimum possible value for this distance, is when n=1, as it can be seen here:
dmin = λ
In any wave, there exists a fixed relationship between the wave speed, the frequency and the wavelength:
v = λ*f
If v = vsound = 343 m/s, and f = 220 1/s, we can solve for λ:
λ =
⇒ dmin =λ = 1.56 m
Answer:
a. 16 s b. -1.866 kJ
Explanation:
a. Since the initial rotational speed ω₀= 3313 rev/min = 3313/60 × 2π rad/s = 346.94 rad/s. Its rotational speed becomes ω₁ = 0.75ω₀ in time t = 4 s.
We find it rotational acceleration using α = (ω₁ - ω₀)/t = (0.75ω₀ - ω₀)/t = ω₀(0.75 - 1)/t = -0.25ω₀/t = (-0.25 × 346.94 rad/s)/4 s = -21.68 rad/s².
Since the turntable stops at ω = 0, the time it takes to stop is gotten from
ω = ω₀ + αt and t = (ω - ω₀)/α = (0 - 346.94 rad/s)/-21.68 rad/s² = (-346.94/-21.68) s = 16 s.
So it takes the turntable 16 s to stop.
b. The workdone by the turntable to stop W equals its rotational kinetic energy change.
So, W = 1/2Iω² - 1/2Iω₀² = 1/2 × 0.031 kgm² × 0² - 1/2 × 0.031 kgm² × (346.94 rad/s)² = 0 - 1865.7 J = -1865.7 J = -1.8657 kJ ≅ -1.866 kJ