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maria [59]
3 years ago
11

If we're interested in knowing the rate at which light energy is received by a unit of area on a particular surface, we're reall

y trying to figure out the
A. incandescence.
B. luminous flux.
C. illuminance of a surface.
D. luminous intensity.
Physics
1 answer:
frosja888 [35]3 years ago
4 0

Answer: C. illuminance of a surface.

a)  Incandescence:  The phenomenon of light emission by a body as a result of high temperature.

B. Luminous flux :  It is the quantity of the energy of the light emitted per second in all directions.

C. Illuminance of a surface :describes the quantity of light emitted by a light source or received at a surface.

D. luminous intensity : the quantity of visible light that a point source radiates in a given direction.

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Margarita [4]
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7 0
3 years ago
The wheel having a mass of 100 kg and a radius of gyration about the z axis of kz=300mm, rests on the smooth horizontal plane.a.
pickupchik [31]

Answer:

a) 20 rad/s

b) 6 m/s

Explanation:

b) Force acting on the wheel is 200 N

mass of the wheel is 100 kg

From Newton's second law of motion, F = m × a

Where F is the net force acting on the body

m is mass of the body

a is the acceleration of the body

By substituting the values we get, a = 2 m/s²

As acceleration is constant, we can use the below formula for calculating the final velocity of the object

v = u + a × t

Where v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

u = 0 (∵ it starts from rest)

By substituting the values we get

v = 0 + 2 × 3 = 6 m/s

∴ Speed of center of mass after 3 seconds = 6 m/s

a) As the wheel rotates about z-axis, radius of gyration will be the radius of wheel

∴ Radius of the wheel = 300 mm

Torque acting on the wheel about axis of rotation = 300 mm × 200 N =

60 N·m

Torque = (Moment of inertia) × (angular acceleration)

Assuming that the mass of spokes of the wheel to be negligible,

Moment of inertia of the wheel about axis of rotation = 100 × 300² × 10^{-6} = 9 kg·m²

Then,

60 = 9 × (angular acceleration)

∴ angular acceleration ≈ 6·67 rad/s²

As angular acceleration of the wheel is constant, we can use the below formula for calculation of final angular speed

w_{f} = w_{i} + α × t

Where

w_{f} is the final angular velocity

w_{i} is the initial angular velocity

α is the angular acceleration

t is the time taken

w_{i} is 0 (∵ initially it starts from rest)

By substituting the values we get

w_{f} = 6·67 × 3 = 20 rad/s

∴ Angular velocity of the wheel after three seconds = 20 rad/s

3 0
3 years ago
Two forces of 83 pounds and 56 pounds act simultaneously on an object. The resultant forms an angle of 52° with the 56-pound for
Nikitich [7]

Answer:

95.9°

Explanation:

The diagram illustrating the action of the two forces on the object is given in the attached photo.

Using sine rule a/SineA = b/SineB, we can obtain the value of B° as shown in the attached photo as follow:

a/SineA = b/SineB,

83/Sine52 = 56/SineB

Cross multiply to express in linear form

83 x SineB = 56 x Sine52

Divide both side by 83

SineB = (56 x Sine52)/83

SineB = 0.5317

B = Sine^-1(0.5317)

B = 32.1°

Now, we can obtain the angle θ, between the two forces as shown in the attached photo as follow:

52° + B° + θ = 180° ( sum of angles in a triangle)

52° + 32.1° + θ = 180°

Collect like terms

θ = 180° - 52° - 32.1°

θ = 95.9°

Therefore, the angle between the two forces is 95.9°

3 0
3 years ago
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