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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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Nesterboy [21]

Answer:

we have formula of frequency :

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for wavelength we swipe it with frequency as follows

λ=c/f

λ=300,000,000/101,700,000

λ=2.949

8 0
3 years ago
Find the density of seawater at a depth where the pressure is 680 atm if the density at the surface is 1030 kg/m3. Seawater has
Pie

Answer:

1060.41kg/m^3

Explanation:

Bulk modulus is defined as the relative change in the volume of a body produced by a unit of compressive acting uniformly over its surface:

B=\rho _o \frac{\bigtriangleup P }{\bigtriangleup \rho}

Hence the density of the seawater at a depth of 680atm is calculated as:-

\rho=\rho_o +\bigtriangleup \rho=\rho_o(1+\frac{\bigtriangleup P}{B})\\\\=1030 \times (1+ \frac{(680-1)\times10^5}{2.3\times 10^9})\\=1060.41kg/m^3

4 0
3 years ago
A force does 30000 J of work along a distance of 9.5m. Find the applied force.
Elina [12.6K]

Answer: F=3158N

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W = F x d

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