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Licemer1 [7]
3 years ago
14

Suppose that you are on an unknown planet in a distant galaxy. Let the length of a physical pendulum be 1.02 m. In the physical

pendulum experiment, the period is determined to be 0.25 s.
What is the gravitational acceleration of this planet?
Physics
1 answer:
Natalka [10]3 years ago
6 0

Answer: acceleration due gravity = 619.5 m/s²

Explanation: The relationship between the period of a pendulum, length and acceleration due to gravity is given as

T =2π * (√l/g)

T = period = 0.25s

l = length of pendulum = 1.02m

g = acceleration due to gravity on the unknown planet

0.25 = 2π * (√1.02/g)

By squaring both sides, we have that

0.25² = 4π² * (1.02/g)

0.25² * g = 4π² * 1.02

g = 4π² * 1.02/ 0.25²

g = 40.26756/ 0.0625

g = 619.5m/s²

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A metal cube with sides of length a=1cm is moving at velocity v0→=1m/sj^ across a uniform magnetic field B0→=5Tk^. The cube is o
Nitella [24]

Answer:

the magnitude of the electric field is 1.25 N/C

Explanation:

The induced emf in the cube ε = LB.v where B = magnitude of electric field = 5 T , L = length of side of cube = 1 cm = 0.01 m and v = velocity of cube = 1 m/s

ε = LB.v = 0.01 m × 5 T × 1 m/s = 0.05 V

Also, induced emf in the cube ε = ∫E.ds around the loop of the cube where E = electric field in metal cube

ε = ∫E.ds

ε = Eds since E is always parallel to the side of the cube

= E∫ds  ∫ds = 4L since we have 4 sides

= E(4L)

= 4EL

So,4EL = 0.05 V

E = 0.05 V/4L

= 0.05 V/(4 × 0.01 m)

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= 1.25 V/m

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