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tankabanditka [31]
3 years ago
13

Which device would you use to most easily measure the illuminance on a surface?

Physics
2 answers:
lubasha [3.4K]3 years ago
7 0
A light meter, because illu is a root for light
vladimir2022 [97]3 years ago
3 0

Which device would you use to most easily measure the illuminance on a surface?

A. Diaphragm

B. Photometer

C. Polarizer

D. Light meter

THE CORRECT ANSWER IS D.

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3.
weeeeeb [17]

Answer:

I.72m/s²

II.8m/s²

Explanation:

acceleration equal velocity² divided by length

8 0
3 years ago
Difference between rest and motion?
Yuki888 [10]

Answer:

Rest and motion are relative terms. In simple terms, an object that changes its position is said to be in motion while the opposite action causes an object to be at rest.

Explanation:

6 0
2 years ago
A negative charge is at rest at the origin of an axis system. Location x is at coordinate point (2m,3m) while location y is at (
Sergeu [11.5K]
The magnitude of the E-field decreases as the square of the distance from the charge, just like gravity.

Location ' x ' is  √(2² + 3²) = √13 m  from the charge.

Location ' y ' is √ [ (-3)² + (-2)² ] = √13 m from the charge.

The magnitude of the E-field is the same at both locations.

The direction is also the same at both locations ... it points toward the origin.


5 0
3 years ago
Which type of force can act through empty space?
katrin2010 [14]

D. Magnetism this is shown by the planet's magnetic pull on moons and space debris.

8 0
3 years ago
Derive the formula for the moment of inertia of a uniform, flat, rectangular plate of dimensions l and w, about an axis through
Ad libitum [116K]

Answer:

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in (Figure). We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

λ = m/l (orm) = λl

If we take the differential of each side of this equation, we find

d m = d ( λ l ) = λ ( d l )

since  

λ

is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact,  

d l = d x

in this situation. We can therefore write  

d m = λ ( d x )

, giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain

I=∫r2dm=∫x2dm=∫x2λdx.

The last step is to be careful about our limits of integration. The rod extends from x=−L/2x=−L/2 to x=L/2x=L/2, since the axis is in the middle of the rod at x=0x=0. This gives us

I=L/2∫−L/2x2λdx=λx33|L/2−L/2=λ(13)[(L2)3−(−L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.

4 0
3 years ago
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