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qaws [65]
3 years ago
5

Analyze why a body undergoing uniform motion eventually comes to rest even in the

Physics
1 answer:
djverab [1.8K]3 years ago
8 0

Answer:

It happens due to force of friction

Explanation:

If a body is performing a uniform motion and no external unbalanced force appears to apply on it, then  the body will come to rest after certain time. The reason behind this is the force of friction that is applied in opposite direction of the motion. So, when there is no apparent unbalanced force it means that the only force acting on the body is the force of friction. This force of friction tends to stop the motion after some period of time, because it is acting in the direction opposite to that of motion.

Hence, the reason behind a body undergoing uniform motion eventually stops is <u>Force of Friction.</u>

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Ede4ka [16]
The answer is a 20 or 140 u choose because u subtract or add these numbers and sorry I can’t speak sri lanka or Indian but I know how to read them
8 0
3 years ago
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Which of the following takes place in the combustion chamber of a gas turbine engine?A. A glow plug is used to add enough heat t
erica [24]

Answer:

Fuel oil is mixed with a proper portion of compressed air

Explanation:

A gas turbine has three main part, which are

  • combustion chamber
  • air compressor
  • power turbine

The combustion chamber is responsible for mixing fuel with a proper portion of compressed air.

The air compressor supplies air in sufficient quantity to satisfy the requirements of the combustion chamber

The power turbine produces the power to drive the air compressor.

6 0
3 years ago
A thin hoop is hung on a wall, supported by a horizontal nail. The hoop's mass is M=2.0 kg and its radius is R=0.6 m. What is th
boyakko [2]

Answer:

Explanation:

Given that,

Mass of the thin hoop

M = 2kg

Radius of the hoop

R = 0.6m

Moment of inertial of a hoop is

I = MR²

I = 2 × 0.6²

I = 0.72 kgm²

Period of a physical pendulum of small amplitude is given by

T = 2π √(I / Mgd)

Where,

T is the period in seconds

I is the moment of inertia in kgm²

I = 0.72 kgm²

M is the mass of the hoop

M = 2kg

g is the acceleration due to gravity

g = 9.8m/s²

d is the distance from rotational axis to center of of gravity

Therefore, d = r = 0.6m

Then, applying the formula

T = 2π √ (I / MgR)

T = 2π √ (0.72 / (2 × 9.8× 0.6)

T = 2π √ ( 0.72 / 11.76)

T = 2π √0.06122

T = 2π × 0.2474

T = 1.5547 seconds

T ≈ 1.55 seconds to 2d•p

Then, the period of oscillation is 1.55seconds

6 0
3 years ago
The host says during the video that her interests during middle school were?
Jobisdone [24]

Answer:

huh? What video?

Explanation:

7 0
3 years ago
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
Sergeeva-Olga [200]

Answer:

33.23 m

Explanation:

At the point where both objects will meet, the vertical height will be equal.

From the equations of motion, the vertical height of the body falling at any time is given as

(y - y₀) = ut + ½gt²

y = vertical height at any time T

y₀ = initial height of the object = 81.5 m

u = initial velocity = 0 m/s (body was dropped)

g = -9.8 m/s²

(y - 81.5) = 0 - 4.9T²

y = 81.5 - 4.9T² (eqn 1)

For an object thrown up, the vertical height of the body at any time, t, is given as

(y - y₀) = ut + ½gt²

y = vertical height of the object at any time t

y₀ = initial height of the object = 0 m

u = initial velocity = 40 m/s

g = -9.8 m/s²

y = 40t - 4.9t² (eqn 2)

At the point where the two objects meet, we equate eqn 1 and eqn 2

y = y

81.5 - 4.9T² = 40t - 4.9t²

But T = (t + 2.2) (Since object 2 was dropped 2.2 s after object 1)

81.5 - 4.9(t + 2.2)² = 40t - 4.9t²

81.5 - 4.9(t² + 4.4t + 4.84) = 40t - 4.9t²

81.5 - 4.9t² - 21.56t - 23.716 = 40t - 4.9t²

81.5 - 21.56t - 23.716 - 40t = 0

57.784 = 61.56t

t = (57.784/61.56) = 0.93866 = 0.94 s

Therefore, the vertical height at t = 0.93866 s is

y = (40×0.93866) - 4.9(0.93866²) = 33.23 m

Hope this Helps!!!

5 0
3 years ago
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