Answer:
The intensity I₂ of the light beam emerging from the second polarizer is zero.
Explanation:
Given:
Intensity of first polarizer = Io/2
For the second polarizer, the intensity is equal:
Tyson throws a shot put ball weighing 7.26 kg. At a height of 2.1 m above the ground, the mechanical energy of the ball is 172.1
2 answers:
Answer:
Velocity, v = 2.50 m/s
Explanation:
Given that,
Mass of the ball, m = 7.26 kg
Height above the ground, h = 2.1 m
Mechanical energy of the ball, T = 172.1 J
To find,
The velocity of the ball at the given point.
Solution,
The sum of potential energy and the kinetic energy is called the mechanical energy of an object. It is given by :
K is the kinetic energy
P is the potential energy
On rearranging the above equation,
v = 2.50 meters
Therefore, the velocity of the ball at the given point is 2.50 m/s.
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Answer:
a) r = (0.6 i- 2039 j ^ + 0.102 k⁾ m and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s c) v_{z} = 1.02 10⁻¹m / s
Explanation:
a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration
Fe + Fm = m a
a = (Fe + Fm) / m
the electric force is
Fe = q E k ^
Fe = 4.25 10-4 60 k ^
Fe = 2.55 10-2 k ^
the magnetic force is
Fm = q v x B
Fm = 4.25 10⁻⁴
fm = 4.25 10⁻⁴ (-j ^ 30 4)
fm 0 = ^ -5,10 10⁻² j
We look for every component of acceleration
X axis
aₓ = 0
there is no force
Axis y
ay = -5.10 10²/5 10⁻⁵ j ^
ay = -1.02 107 j ^ / s2
z axis
az = 2.55 10⁻² / 5 10⁻⁵ k ^
az = 5.1 10² k ^ m / s²
Having the acceleration in each axis we can encocoar the position using kinematics
X axis
the initial velocity is vo = 30 m / s and an initial position xo = 0
x = vo t + ½ aₓ t₂2
x = 30 0.02 + 0
x = 0.6m
Axis y
acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m
y = I + go t + ½ ay t²
y = 1 + 0 + ½ (-1.02 10⁷) 0.02²
y = 1 - 2.04 10³
y = -2039 m j ^
z axis
acceleration is aza = 5.1 10² m / s², the position and initial speed are zero
z = zo + v₀ t + ½ az t²
z = 0 + 0 + ½ 5.1 10² 0.02²
z = 1.02 10⁻¹ m k ^
therefore the position of the bodies is
r = (0.6 i- 2039 j ^ + 0.102 k⁾ m
b) x axis
since there is no acceleration the speed remains constant
vₓ = 30.0 m / s
Axis y
let's use the equation v = v₀ + t
= 0 + -1.02 10⁷ 0.02
v_{y} = 2.04 10⁵ m / s
z axis
v_{z} = vo + az t
v_{z} = 0 + 5.1 10² 0.02
v_{z} = 1.02 10⁻¹m / s
The energy transferred to the spring is given by:
where
k is the spring constant
x is the elongation of the spring with respect its initial length
Let's convert the data into the SI units:
so now we can use these data inside the equation ,to find the energy transferred to the spring:
where
k is the spring constant
x is the elongation of the spring with respect its initial length
Let's convert the data into the SI units:
so now we can use these data inside the equation ,to find the energy transferred to the spring:
Answer:
The maximum speed of the car at the bottom of that drop is 26.34 m/s.
Explanation:
Given that,
The maximum vertical distance covered by the roller coaster, h = 35.4 m
We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :
v = 26.34 m/s
So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.