Improved posture is a direct result of maintianing good balance. The other answers lack relevence as answers to the question.
Answer:
1) k = 52 N/m
2) E = 1.0 J
3) ω = 8.1 rad/s
4) v = 1.4 m/s
Though asked for a velocity, we can only supply magnitude (speed) because we don't have enough information to determine direction.
If it happens to be the first time it is at y = - 10 cm after release, the velocity is upward.
Explanation:
Assuming the initial setup is after all transients are eliminated.
kx = mg
k = mg/x = 0.8(9.8) / 0.15
k = 52.26666.... ≈ 52 N/m
E = ½kA² = ½(52)(0.20²) = 1.045333... ≈ 1.0 J
ω = √(k/m) = √(52 / 0.8) = 8.0829... ≈ 8.1 rad/s
½mv² = ½kA² - ½kx²
v = √(k(A² - x²)/m) = √(52(0.20² - 0.10²)/0.8) = 1.39999... ≈ 1.4 m/s
Answer:
To find the slope, we need to have co-ordinates of atleast 2 points lying on the line.
In this case, the 2 points are
(-3,1) and (2,-2)
To find the gradient we put in the following rule :
Rise / Run = (y2 - y1) / (x2-x1) = (-2-1) / (2+3) = -3 /5
Explanation:
Therefore, the volume of the gas at temperature 100°C is 4.62L
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<u>Explanation:</u>
Given-
Volume, V1 = 4L
Temperature, T1 = 50°C = 50 + 273K = 323K
Temperature, T2 = 100°C = 100 + 273K = 373K
Volume, V2 = ?
Using Charles law:
V1/T1 = V2/T2
4/323 = V2/373
V2 = 4.62L
Therefore, the volume of the gas at temperature 100°C is 4.62L
Answer:
Angle of reflection of light is 34 degree
Explanation:
As per law of reflection of light we know that
angle of incidence of light = angle of reflection of light
So here we know that
angle of incidence on the surface of oil is given as
so we know that
so here we can say that reflection angle of light will be same as angle of incidence
