The temperature of the gas is 41.3 °C.
Answer:
The temperature of the gas is 41.3 °C.
Explanation:
So on combining the Boyle's and Charles law, we get the ideal law of gas that is PV=nRT. Here P is the pressure, V is the volume, n is the number of moles, R is gas constant and T is the temperature. The SI unit of pressure is atm. So we need to convert 1 Pa to 1 atm, that is 1 Pa = 9.86923×
atm. Thus, 171000 Pa = 1.6876 atm.
We know that the gas constant R = 0.0821 atmLMol–¹K-¹. Then the volume of the gas is given as 50 L and moles are given as 3.27 moles.
Then substituting all the values in ideal gas equation ,we get
1.6876×50=3.27×0.0821×T
Temperature = 
So the temperature is obtained to be 314.3 K. As 0°C = 273 K,
Then 314.3 K = 314.3-273 °C=41.3 °C.
Thus, the temperature is 41.3 °C.
Answer:
No; the graph fails the vertical line test.
The vertical line test is a tool used to determine if we have a function. If we can draw a single straight vertical line through more than one point on the red curve, then the graph is said to have failed the vertical line test. Consequently, this leads to the relation not being a function.
For this circle graph, we can draw a vertical line through more than one point, which is why we don't have a function here.
Put another way, there are inputs (x) that produce more than one output (y), so that's why we don't have a function.
-BBBM
Cellphones use radio frequency waves to transmit sound through the speaker and the microphone.<span> The radio frequency energy that is given off by cell phones is a type of electromagnetic energy.
I hope I helped! =D</span>
Answer:
True The grid with more slits gives more angle separation increases
True. The grating with 10 slits produces better-defined (narrower) peaks
Explanation:
Such a system can be seen as a diffraction network in this case with different number of lines per unit length, the expression for the constructive interference of a diffraction network is
d sin θ = m λ
where d is the distance between slits or lines, m the order of diffraction and λ the wavelength.
For network with 5 slits
d = 1/5 = 0.2
For the network with 10 slits
d = 1/10 = 0.1
let's calculate the separation (teat) for each one
θ = sin⁻¹ (m λ / d)
for 5 slits
θ₅ = sin⁻¹ (m λ 5)
for 10 slits
θ₁₀ = sin⁻¹ (m λ 10)
we can appreciate that for more slits the angle increases
the intensity of a series of slits is
I = I₀ sin²2 (N d/2) / sin² d/2)
when there are more slits (N) the peaks have greater intensity and are more acute (half width decreases)
let's analyze the claims
False
True The grid with more slits gives more angle separation increases
False
True The expression for the intensity of the diffraction peaks the intensity of the peaks increases with the number of slits as well as their spectral width decreases
False
Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:
The potential at a distance can be represented as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D)
₀
V(r) = ![-[\frac{qr^2}{8 \pi E_0R^3 }]](https://tex.z-dn.net/?f=-%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D)
₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:
= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]^R](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D%5ER)
V(r) = 
V(r) = 
V(r) 
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V
