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melamori03 [73]
3 years ago
6

A frictionless pendulum clock on the surface of the earth has a period of 1.00 s. On a distant planet, the length of the pendulu

m must be shortened slightly to have a period of 1.00 s. What is true about the acceleration due to gravity on the distant planet?
Physics
1 answer:
Marianna [84]3 years ago
3 0

Answer:

The acceleration due to gravity on the other planet is less than the acceleration due to gravity on Earth

Explanation:

The time period of a pendulum is given by

T=2\pi\sqrt{\frac{L}{g}}

where,

L = Length of the pendulum

g = Acceleration due to gravity

It can be seen that the time is proportional to the length and inversely proportional to the acceleration due to gravity. So, if L is reduced to keep the same time period then g must be also less.

Hence, the acceleration due to gravity on the other planet is less than the acceleration due to gravity on Earth.

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charle [14.2K]

High pressure systems entail sinking air, while lows entail rising air. High pressure systems form where air converges in the higher levels of the atmosphere. The converging air has nowhere to go and is forced to sink toward the ground. This sinking effect inhibits cloud formation and therefore precipitation.

5 0
2 years ago
A wire 6.90 m long with diameter of 2.15 mm has a resistance of 0.0320 Ω. Find the resistivity of the material of the wire. rho
spayn [35]
<h2>Answer:</h2>

1.68 x 10⁻⁸Ωm

<h2>Explanation:</h2>

The resistance (R) of a wire is related to its length(L), its material resistivity(ρ) and its crossectional area(A) as follows;

R = ρL/A               ------------------------(i)

Where;

A = πd² / 4              [where d = diameter of the wire]

From the question;

L = 6.90m

d = 2.15mm = 0.00215m

R = 0.0320Ω

First calculate the crossectional area (A) of the wire as follows;

A = πd² / 4      

[Take π = 3.142]

d = 0.00215m

∴ A = 3.142 x (0.00215)² / 4

∴ A = 0.000003631m²

Now, substitute the values of A, L, and R into equation (i) as follows;

R = ρL/A

0.0320 = ρ x 6.90 / 0.000003631

0.0320 = 1900302.95 x ρ

Solve for ρ;

=> ρ = 0.0320 / 1900302.95

=> ρ = 1.68 x 10⁻⁸Ωm

Therefore, the resistivity of the material of the wire is 1.68 x 10⁻⁸Ωm

4 0
2 years ago
If a 15 N box is lifted a distance of 3 m, how much work is done?
Naily [24]

Answer:

W=45J

Explanation:

W=Fd

W=15(3)=45

W=45J

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