Answer:
the potential energy is 114 J.
Explanation:
Given;
total mechanical energy, E = 400 J
kinetic energy, K.E = 286 J
The potential energy is calculated as follows;
E = K.E + P.E
where;
P.E is the potential energy
P.E = E - K.E
P.E = 400 J - 286 J
P.E = 114 J
Therefore, the potential energy is 114 J.
Answer:
r₂ = 0.316 m
Explanation:
The sound level is expressed in decibels, therefore let's find the intensity for the new location
β = 10 log
let's write this expression for our case
β₁ = 10 log \frac{I_1}{I_o}
β₂ = 10 log \frac{I_2}{I_o}
β₂ -β₁ = 10 (
)
β₂ - β₁ = 10
log \frac{I_2}{I_1} =
= 3
= 10³
I₂ = 10³ I₁
having the relationship between the intensities, we can use the definition of intensity which is the power per unit area
I = P / A
P = I A
the area is of a sphere
A = 4π r²
the power of the sound does not change, so we can write it for the two points
P = I₁ A₁ = I₂ A₂
I₁ r₁² = I₂ r₂²
we substitute the ratio of intensities
I₁ r₁² = (10³ I₁ ) r₂²
r₁² = 10³ r₂²
r₂ = r₁ / √10³
we calculate
r₂ =
r₂ = 0.316 m
Answer:
4th answer
Explanation:
The gradient of a distance-time graph gives the speed.
gradient = distance / time = speed
Here, the gradient is a constant till 30s. So it has travelled at a constant speed. It means it had not accelarated till 30s. and has stopped moving at 30s.
Answer:
257.32
Explanation:
I jus worked it out on paper. Brainliest please?
Answer:
a) For y = 102 mA, R = 98.039 ohms
For y = 97 mA, R = 103.09 ohms
b) Check explanatios for b
Explanation:
Applied voltage, V = 10 V
For the first measurement, current 
According to ohm's law, V = IR
R = V/I
Here, 

For the second measurement, current 


b) ![y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}](https://tex.z-dn.net/?f=y%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dy_%7B1%7D%20%26y_%7B2%7D%20%5Cend%7Barray%7D%5Cright%5D%20%5E%7BT%7D)
![y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]](https://tex.z-dn.net/?f=y%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dy_%7B1%7D%20%5C%5Cy_%7B2%7D%20%5Cend%7Barray%7D%5Cright%5D)
![y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]](https://tex.z-dn.net/?f=y%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D102%2A10%5E%7B-3%7D%20%5C%5C97%2A10%5E%7B-3%7D%20%20%5Cend%7Barray%7D%5Cright%5D)
A linear equation is of the form y = Gx
The nominal value of the resistance = 100 ohms
![x = \left[\begin{array}{ccc}100\end{array}\right]](https://tex.z-dn.net/?f=x%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D100%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D102%2A10%5E%7B-3%7D%20%5C%5C97%2A10%5E%7B-3%7D%20%20%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DG_%7B1%7D%20%5C%5CG_%7B2%7D%20%20%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D100%5Cend%7Barray%7D%5Cright%5D%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DG_%7B1%7D%20%5C%5CG_%7B2%7D%20%20%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D102%2A10%5E%7B-5%7D%20%5C%5C97%2A10%5E%7B-5%7D%20%20%5Cend%7Barray%7D%5Cright%5D)