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Gnoma [55]
3 years ago
14

Famed stunt pilot, Cleonvia Thread, pulls out rapidly from a dive. He is traveling at 222 mi/h at the bottom of his trajectory,

and at that instant is traveling on a curve of radius 820 ft. r v What acceleration does he
Physics
1 answer:
Simora [160]3 years ago
8 0

Answer:

0.39 m/s²

Explanation:

From the question,

a = v²/r.................... Equation 1

Where v = velocity, r = radius.

Given: v = 222 mi/h = (222×0.44704) m/s = 9.83 m/s, r = 820 ft = (820×0.3048) m = 249.94 m.

Substitute thses values into equation 1

a = 9.83²/249.94

a = 96.63/249.94

a = 0.39 m/s²

Hence the  acceleration is 0.39 m/s²

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3 years ago
you stretch a spring by a distance of 0.3 m. the spring has a spring constant of 440 n/m. when you release the spring, it snaps
Alina [70]

Answer:

19.8 J

Explanation:

According to the law of conservation of energy, the total mechanical energy of the spring (sum of kinetic energy and elastic potential energy) must be conserved:

K_i + U_i = K_f + U_f (1)

where we have

K_i is the initial kinetic energy of the spring, which is zero because the spring starts from rest (2)

U_i is the elastic potential energy of the spring when it is fully stretched

K_f is the kinetic energy of the spring when it reaches the natural length

U_f is the elastic potential energy of the spring when it reaches its natural length, which is zero because the stretch in this case is zero (3)

So

U_i = \frac{1}{2}k(\Delta x_i)^2

where

k = 440 N/m is the spring constant

\Delta x_i = 0.3 m is the initial stretching of the spring

Substituting,

U_i = \frac{1}{2}(440)(0.3)^2=19.8 J

And so using eq.(1) and keeping in mind (2) and (3) we find

K_f= U_i = 19.8 J

8 0
4 years ago
What is the energy Q released when 131 53Idecays and 131 54Xe is formed? The atomic mass of 131 53I is 130.906118 u and the atom
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Answer:0.967meV

Explanation:

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1 u = 931.494meV

multiply 0.001038 by 931.494

=0.001038 X 931.494

0.967 meV

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