Light travels _____________ in a material with a higher index of refraction.
2 answers:
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Answer:
a1 = 3.56 m/s²
Explanation:
We are given;
Mass of book on horizontal surface; m1 = 3 kg
Mass of hanging book; m2 = 4 kg
Diameter of pulley; D = 0.15 m
Radius of pulley; r = D/2 = 0.15/2 = 0.075 m
Change in displacement; Δx = Δy = 1 m
Time; t = 0.75
I've drawn a free body diagram to depict this question.
Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;
ΣF_x = T1 = m1 × a1
a1 is acceleration and can be calculated from Newton's 2nd equation of motion.
s = ut + ½at²
our s is now Δx and a1 is a.
Thus;
Δx = ut + ½a1(t²)
u is initial velocity and equal to zero because the 3 kg book was at rest initially.
Thus, plugging in the relevant values;
1 = 0 + ½a1(0.75²)
Multiply through by 2;
2 = 0.75²a1
a1 = 2/0.75²
a1 = 3.56 m/s²
Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.