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ladessa [460]
3 years ago
7

A 1300 kg car starts at rest and rolls down a hill from a height of 10.0 m. It then moves across a

Physics
1 answer:
Makovka662 [10]3 years ago
5 0

Answer:

0.51 m

Explanation:

Using the principle of conservation of energy, change in potential energy equals to the change in kinetic energy of the spring.

Kinetic energy, KE=½kx²

Where k is spring constant and x is the compression of spring

Potential energy, PE=mgh

Where g is acceleration due to gravity, h is height and m is mass

Equating KE=PE

mgh=½kx²

Making x the subject of formula

x=\sqrt {\frac {2mgh}{k}}

Substituting 9.81 m/s² for g, 1300 kg for m, 10m for h and 1000000 for k then

x=\sqrt \frac {2*1300*9.81*10}{1000000}=0.50503465227646m\\x\approx 0.51 m

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A wheel is rotating freely at angular speed 660 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initi
sweet-ann [11.9K]

Answer:

(a)  110 rev/ min

(b) 5/6

Explanation:

As per the conservation of linear momentum,

L ( initial ) = L ( final )

I' ω' = ( I' + I'' ) ωf

I' is the rotational inertia of first wheel and I'' is the rotational inertia of second wheel which is at rest.

(a)

So,    ωf = I' ω' /  ( I' + I'' )

As I'' = 5I'

ωf = I' ω' /  ( I' + 5I' )

ωf = ω'/ 6

now we know ω' = 660 rev /  min

therefore    ωf = 660/6

                       = 110 rev/ min

(b)

Initial kinetic energy will be K'

K' = I'ω'² / 2

and final K.E. will be   K'' =  ( I' + I'' )ωf² / 2

                                   K'' = ( I' + 5I' ) (ω'/ 6)²/ 2

                                   K'' = 6I' ω'²/72

                                   K'' = I' ω'²/ 12

therefore the fraction lost is

                ΔK/K' = ( K' - K'' ) / K'

                           =  {( I'ω'² / 2) - (I' ω'²/ 12)} / ( I'ω'² / 2)

                            = 5/6

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3 years ago
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Answer: .36 m

Explanation:

5 0
3 years ago
An airplane flies eastward and always accelerates at a constant rate. At one position along its path it has a velocity of 34.3 m
Tomtit [17]

Explanation:

We'll need two equations.

v² = v₀² + 2a(x - x₀)

where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.

x = x₀ + ½ (v + v₀)t

where t is time.

Given:

v = 47.5 m/s

v₀ = 34.3 m/s

x - x₀ = 40100 m

Find: a and t

(47.5)² = (34.3)² + 2a(40100)

a = 0.0135 m/s²

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7 0
3 years ago
If it takes 50.0 seconds to lift 10.0 Newtons of books to a height of 7.0
Ilya [14]
Work of the force = 10 N

Time required for the work = 50 sec

Height = 7 m

We are given with the value of work and time in the question.

Substitute the values in the formula of power and then you'll get the power required.

We know that,

w = Work

p = Power

t = Time

By the formula,

Given that,
Work (w) = 7 m = 70 Joules
Time (t) = 50 sec
Substituting their values,

p = 70/50

p = 1.4 watts

Therefore, the power required is 1.4 watts.

Hope it helps!
3 0
3 years ago
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