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Anna11 [10]
3 years ago
15

A student is holding a stone at a certain height. The stone has 50 joules of potential energy and 0 joules of kinetic energy. Th

e stone is released from that height. How much energy does the stone have
right before it hits the ground?
Select one:
a) The stone will have no energy
b) The stone will have about 25 joules each of kinetic and potential energy
c) The stone will have about 50 joules of potential energy and ovogres van kinetic
energy
d) The stone will have about 50 joules of kinetic energy and ovoules of potencia
energy​
Physics
1 answer:
Hitman42 [59]3 years ago
6 0

Answer:

d) The stone will have about 50 joules of kinetic energy and 0 joules of potential energy​ .

Explanation:

Given :

Initial Potential energy , P_i=50\ J .

Initial Kinetic energy , K_i=0\ J . ( because ball is in rest )

Now , we know , kinetic energy is maximum when an object reaches ground .

Also , potential energy is zero when an object is in ground .

We know , by conservation of energy :

Initial total energy = Final total energy

P_i+K_i=P_f+K_f\\\\50+0=0+K_f\\\\K_f=50 \ J

Therefore , option d) is correct .

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3 years ago
In a double slit experiment, 450 nm light passes through two slits producing an interference pattern where the first bright frin
musickatia [10]

Answer:

1.3 x 10⁻⁴ m

Explanation:

\lambda = wavelength of the light = 450 nm = 450 x 10⁻⁹ m

n = order of the bright fringe = 1

θ = angle = 0.2°

d = separation between the slits

For bright fringe, Using the equation

d Sinθ = n \lambda

Inserting the values

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6 0
2 years ago
To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
motikmotik

Answer:

0.79 s

Explanation:

We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

\sum F_x:f_{max}=ma_x\\\sum F_y:N-mg=0

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that f_{max}=\mu N

Now, we find the acceleration:

\mu N=ma_x\\\mu mg=ma_x\\a_x=\mu g\\a_x=0.83(9.8\frac{m}{s^2})\\a_x=8.134\frac{m}{s^2}

Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

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8 0
3 years ago
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