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lutik1710 [3]
3 years ago
7

As a pendulum swings from its highest to lowest position, what happens to its kinetic and potential energy?

Physics
2 answers:
DIA [1.3K]3 years ago
7 0

The potential energy decreases while the kinetic energy increases.



malfutka [58]3 years ago
4 0

Answer:

the potential decreases while the kinetic energy increases

Explanation:

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8 a Name the bones that articulate (join together) in the knee joint
Marianna [84]

Answer:

The Femur and the Tibia

8 0
2 years ago
An object of mass m is hung from a spring and set into oscillation. The period of the oscillation is measured and recorded as T.
sergejj [24]

Answer:\sqrt{2}T

Explanation:

Given

object of mass m is suspended from spring and set in oscillation with time Period T

We know Time period of a mass in oscillation is given by

T=2\pi \sqrt{\frac{m}{k}}

where k=spring constant

When mass m is replaced by a mass of 2 m time period is given by

T'=2\pi \sqrt{\frac{2m}{k}}

T'=\sqrt{2}\times 2\pi \sqrt{\frac{m}{k}}

T'=\sqrt{2}T

i.e. New time period becomes \sqrt{2} times of previous one

                         

7 0
2 years ago
A 5 kg block is pushed across a table by a horizontal force of 40 N with an acceleration of 5 m/s^2. What is the frictional forc
julsineya [31]

Answer:

15

Explanation:

mass, M = 5Kg

horizontal force, F_h = 40N

acceleration, a =5 m/s^2

frictional force, F_f =?

net force = ma

net force = F_h - F_f = 40N - F_f

40  - F_f = 5 x 5

- F_f = 25 - 40

multiply both side by -1

F_f = 40 - 25 = 15

the frictional force is 15N

4 0
3 years ago
Consider a semicircular ring of radius R. Its linear mass density varies as lambda =lambda not sin theta. Locate its centre of m
bearhunter [10]

Answer:

(0, πR/4)

Explanation:

The linear mass density (mass per length) is λ = λ₀ sin θ.

A short segment of arc length is ds = R dθ.

The mass of this short length is:

dm = λ ds

dm = (λ₀ sin θ) (R dθ)

dm = R λ₀ sin θ dθ

The x coordinate of the center of mass is:

X = ∫ x dm / ∫ dm

X = ∫₀ᵖ (R cos θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

X = R ∫₀ᵖ sin θ cos θ dθ / ∫₀ᵖ sin θ dθ

X = R ∫₀ᵖ ½ sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R ∫₀ᵖ 2 sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R (-cos 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

X = ¼R (-cos 2π − (-cos 0)) / (-cos π − (-cos 0))

X = ¼R (-1 + 1) / (1 + 1)

X = 0

The y coordinate of the center of mass is:

Y = ∫ y dm / ∫ dm

Y = ∫₀ᵖ (R sin θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

Y = R ∫₀ᵖ sin² θ dθ / ∫₀ᵖ sin θ dθ

Y = R ∫₀ᵖ ½ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R ∫₀ᵖ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R (θ − ½ sin 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

Y = ½R [(π − ½ sin 2π) − (0 − ½ sin 0)] / (-cos π − (-cos 0))

Y = ½R (π − 0) / (1 + 1)

Y = ¼πR

4 0
3 years ago
An empty semi-truck sits on the scale of a weighing station. It is then loaded with 100 irritated pigeons. These two events then
mina [271]

Answer:

The reading in the scale is going to be the same, or if it experiences some change it would be minimum. The reason is because the truck will act as a big closed cage, therefore, when the pigeons fly, the air they move with their wings in order to keep flying,  exerts the same force on the closed cage, that if they were standing on the ground.

If the truck however, allows the air flow, the weight might change, because under this scenario, the air flowing could represent less force exerted on the balance.

Explanation:

4 0
3 years ago
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