** Missing info: Lines per mm = 500 **
Ans: The wavelength is = λ = 1414.21 nm
Explanation:
The formula for diffraction grading is:
dsinθ = mλ --- (1)
Where
d = 1/lines-per-meter = (1/500)*10^-3 = 2 * 10^-6
m = order = 1
λ = wavelength
θ = 45°
Plug in the values in (1):
(1) => 2*10^-6*sin(45°) = (1)λ
=> λ = 1414.21 nm
The velocity when function p(t)=11 is 8 .
According to the question
The position of a car at time t represented by function :
Now,
When function p(t) = 11 , t will be
11 = t²+2t-4
0 = t² + 2t - 15
or
t² +2t-15 = 0
t² +(5-3)t-15 = 0
t² +5t-3t-15 = 0
t(t+5)-3(t+5) = 0
(t-3)(t+5) = 0
t = 3 , -5
as t cannot be -ve as given ( t≥0)
so,
t = 3
Now,
the velocity when p(t)=11
As we know velocity =
therefore to get the value of velocity from function p(t)
we have to differentiate the function with respect to time
v(t) = 2t + 2
where v(t) = velocity at that time
as t = 3 for p(t)=11
so ,
v(t) = 2t + 2
v(t) = 2*3 + 2
v(t) = 8
Hence, the velocity when function p(t)=11 is 8 .
To know more about function here:
brainly.com/question/12431044
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Positive : The development and adoption of technology have helped societies raise productivity, inclusivity of services and improve overall well-being.
Negative: The negative impact of tech on society involves mass-made products, with most often blamed: social media.
Answer:
A. Weight
Explanation:
On Earth a gallon of milk has the same mass as a gallon of milk in space, but it has a different weight
Explanation:
Given data:
Area A = 10 cm×2 cm = 20×10⁻⁴ m²
Distance d between the plates = 1 mm = 1×10⁻³m
Voltage of the battery is emf = 100 V
Resistance = 1025 ohm
Solution:
In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by

Taking natural log on both sides,

(1)
Now we can calculate the capacitance by using the area of the plates.
C = ε₀A/d
= 
= 18×10⁻¹²F
Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)

= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)
= 15×10⁻⁹s
= 15 ns