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3 years ago

8

2. In the pendulum lab, the period (swing time) was O a. the independent variable O b.graphed on the vertical-axis O c.graphed o

n the horizontal-axis O d. the variable you controlled

1 answer:

Nadya [2.5K]3 years ago

7 0

B is appropriate answer

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If an object is placed at the center of carvature of a convance mirror the image formed is called

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When the object is placed between centre of curvature and principal focus of a concave mirror the image formed is beyond C as shown in the figure and it is real, inverted and magnified.

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3 years ago

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The gravitational force between two objects will be greatest in which of the following situations?

Kobotan [32]

Answer:

Explanation:

Gravitational law states that, the force of attraction or repulsion between two masses is directly proportional to the product of the two masses and inversely proportional to the square of their distance apart.

So,

Let the masses be M1 and M2,

F ∝ M1 × M2

Let the distance apart be R

F ∝ 1 / R²

Combining the two equation

F ∝ M1•M2 / R²

G is the constant of proportional and it is called gravitational constant

F = G•M1•M2 / R²

So, to increase the gravitational force, the masses to the object must be increased and the distance apart must be reduced.

So, option c is correct

C. Both objects have large masses and are close together.

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3 years ago

Find the angle ϕ between the filter's polarizing axis and the direction of polarization of light necessary to increase the ratio

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D. that looks like a rubber band.

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4 years ago

A convex security mirror has a radius of curvature of 12.0 cm. What is the magnification of a pare 3.0 m from the mirror?

Makovka662 [10]

Answer:

magnification will be -0.025

Explanation:

We have given the radius of curvature = 12 cm

And object distance = 3 m

So focal length $f=\frac{R}{2}=\frac{12}{2}=6cm$

Now for mirror we know that $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

So $\frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}$

$16.66-0.333=\frac{1}{v}$

v = 0.750 m

Now magnification of the mirror is $m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025$

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4 years ago