Using a simple pendulum, we can measure. (a) Acceleration along a circle (b) Acceleration due to gravity (c) Frequency (d) None

Physics

1 answer:

zalisa [80]3 years ago

3 0

Answer:

Option B is the correct answer.

Explanation:

Period of simple pendulum is given by the expression $T=2\pi \sqrt{\frac{l}{g}}$ , where l is the length of pendulum and g is the acceleration due to gravity value.

So if we have a simple pendulum with length l we can find its period. Using the above formula we can calculate acceleration due to gravity value of that place.

So using simple pendulum we can measure acceleration due to gravity value.

Option B is the correct answer.

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Evaluate the gravitational potential energy between two 5.00-kg spherical steel balls separated by a center-tocenter distance of

navik [9.2K]

Answer:

U = 8.30×10-⁹J

Explanation:

m1 = m2 = 5.00kg masses of the spheres

d = 15.0cm = 15×10-²m

r = 5.10cm = 5.10×10-²m

R = d + r = 15×10-² + 5.10×10-²

R = 20.10 ×10-²m = 0.201m

G = 6.67×10-¹¹Nm²/kg²

U = Gm1×m2/R = potential energybetween the spheres

U = 6.67×10-¹¹×5.00×5.00/0.201

U = 8.30×10-⁹J

7 0

3 years ago

A physics student mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them

Tatiana [17]

Answer:

A) q₂ = 75.98 cm, B) q₂' = 115.38 cm, C)

Explanation:

A) This is an exercise in geometric optics, as the two lenses are separated by a greater distance than their focal lengths from each lens, they must be worked as independent lenses.

Lens 1. More to the left

let's use the constructor equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively,

We must assume a distance to the object to perform the calculation, suppose that the object is 50 cm from lens 1 that is further to the left of the system.

$\frac{1}{q_1} = \frac{1}{f} - \frac{1}{p}$