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Alborosie
2 years ago
14

Determine the moment of inertia of a 14.8 kg sphere of radius 0.752 m when the axis of rotation is through its center.

Physics
1 answer:
vfiekz [6]2 years ago
6 0

Answer:

I=3.348Kgm^2

Explanation:

From the question we are told that:

Mass m=14.8kg

Radiusr=0.752m

Generally the equation for inertia is mathematically given by

I=\frac{2}{5}mR^2

I=\frac{2}{5}14.8*(0.752)^2

I=3.348Kgm^2

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Unpolarized light with an average intensity of 845 W/m2 moves along the x-axis when it enters a Polarizer A with a vertical tran
horsena [70]

Answer:

θ = 36.2º

Explanation:

When light passes through a polarizer it becomes polarized and if it then passes through a second polarizer, it must comply with Malus's law

         I = I₀ cos² tea

The non-polarized light between the first polarized of this leaves half the intensity, with vertical polarization

          I₁ = I₀ / 2

          I₁ = 845/2

          I₁ = 422.5 W / m²

In this case, the incident light in the second polarizer has an intensity of I₁ = 422.5 W / m² and the light that passes through the polarizer has a value of

I = 275 W / m ²

      Cos² θ = I / I₁

      Cos θ = √ I / I₁

      Cos θ = √ (275 / 422.5)

     Cos θ = 0.80678

     θ = cos⁻¹ 0.80678

     θ = 36.2º

This is the angle between the two polarizers

8 0
3 years ago
A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab
shutvik [7]

Answer:

a)  θ₁ = 0.487º , b)   t = 0.400 s ,        x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

            y = y₀ + v_{oy} t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

            x = v₀ₓ t

            t = x / v₀ₓ

We replace

             y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

             v_{oy} = v₀ sin θ

             v₀ₓ = vo cos θ

             

             y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

                5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

                1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

               Sec² θ = 1 - tan² θ

                 1 = 120 tan θ - 0.0213 (1 –tan²θ)

                 0.0213 tan²θ + 120 tanθ -1.0213 = 0

                 

We change variables

          u = tan θ

          u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

          u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

          u = [- 5633.8 ± 5633.82] / 2

           u₁ = 0.0085

           u₂= -5633.81

           u = tan θ

           θ = tan⁻¹ u

For u₁

           θ₁ = tan⁻¹ 0.0085

           θ₁ = 0.487º

For u₂

           θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

         x = v₀ₓ t

         t = x / v₀ cos θ

         

         t = 120 / (300 cos 0.487)

         t = 0.400 s

The deer must be at a distance of

           v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

           x = v t

           x = 29.33 0.4

           x = 11.73 ft

3 0
3 years ago
Luis and Aisha conducted an experiment. They exerted different forces on four objects. Their results are shown in the table.
lesya692 [45]

Answer:

Object 3 has greatest acceleration.

Explanation:

Objects               Mass                                Force

1                            10 kg                               4 N              

2                           100 grams                       20 N

3                            10 grams                         4 N

4                             1 kg                                 20 N

Acceleration of object 1,

a_1=\dfrac{F_1}{m_1}\\\\a_1=\dfrac{4}{10}\\\\a_1=0.4\ m/s^2

Acceleration of object 2,

a_2=\dfrac{F_2}{m_2}\\\\a_2=\dfrac{20}{0.1}\\\\a_2=200\ m/s^2

Acceleration of object 3,

a_3=\dfrac{F_3}{m_3}\\\\a_3=\dfrac{4}{0.01}\\\\a_3=400\ m/s^2

Acceleration of object 4,

a_4=\dfrac{F_4}{m_4}\\\\a_4=\dfrac{20}{1}\\\\a_3=20\ m/s^2

It is clear that the acceleration of object 3 is 400\ m/s^2 and it is greatest of all. So, the correct option is (3).

4 0
3 years ago
Read 2 more answers
In any problems involving circular motion, which way does the tangential speed vector point?
Anton [14]

In what may be one of the most remarkable coincidences in
all of physical science, the tangential component of circular
motion points along the tangent to the circle at every point. 

The object on a circular path is moving in that exact direction
at the instant when it is located at that point in the circle.  The
centripetal force ... pointing toward the center of the circle ...
is the force that bends the path of the object away from a straight
line, toward the next point on the circle.  If the centripetal force
were to suddenly disappear, the object would continue moving
from that point in a straight line, along the tangent and away from
the circle.

4 0
3 years ago
which pole of a compass needle points north ? please give simple answer with little explaining if any :)
Mumz [18]
I think the North Pole
If I understand this correctly
6 0
2 years ago
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