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siniylev [52]
3 years ago
10

A boulder is raised above the ground, so that its potential energy relative to the ground is 200 J. Then it is dropped. Estimate

what its kinetic energy will be just before hitting the ground.
Physics
1 answer:
babymother [125]3 years ago
6 0

Answer:

200 J

Explanation:

In this problem, I assume there is no air resistance, so the  system is isolated (=no external forces).

For an isolated system, the total mechanical energy is constant, and it is given by:

E=KE+PE

where

KE is the kinetic energy

PE is the potential energy

The kinetic energy is the energy due to the motion of the object,  while the potential energy is the energy due to the position of the object relative to the ground.

At the beginning, when the boulder is raised above the ground, its height above the ground is maximum, while its  speed is zero; it means that all its mechanical energy is just potential energy, and it is:

E=PE_{max}=200 J

As the boulder falls  down, its altitude decreases, so its potential energy decreases, while the speed increases, and the kinetic energy increases. Therefore, potential energy is converted into kinetic energy.

Eventually, just before the boulder hits the ground, the height of the object is zero, and the speed is maximum; this means that all the energy has now converted into kinetic energy, and we have

E=KE_{max}=200 J

Therefore, the kinetic energy just before hitting the ground is 200 J.

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2×3.14√(1.0m/(9.8〖ms〗^(-1) )=)
il63 [147K]

This is the period in a simple harmonic motion which is 2 seconds in this question.

<h3>What is Period ?</h3>

The period of an oscillatory object can be defined as the total time taken  by a vibrating body to make one complete revolution about a reference point.

We are given the below question

2×3.14√(1.0m/(9.8〖ms〗^(2) )= T

This question can as well be expressed as

2π√(L/g) which is equal to period T.

In a nut shell, Period T = 2×3.14√(1.0m/9.8)

T = 6.28√0.102

T = 6.28 × 0.32

T = 2.006 s

Therefore, the period T of the oscillation is 2 seconds approximately.

Learn more about Period here: brainly.com/question/12588483

#SPJ1

8 0
1 year ago
Water enters a student's house 10.0 m above the ground through a pipe with a cross section area of 1.00 x 10-4m2 at ground. Insi
dezoksy [38]

Answer:

(a). V₁ = 10m/s (velocity inside the house), V₂ = 5m/s (velocity at ground level)

(b). P₂ = 236500 Pa

Explanation:

This is quite straight-forward so let us begin by defining the terms given.

Given that;

The cross-section area inside the student's house A₁ = 0.50 0.50 x 10-4m2.

Let us make the velocity of water inside the house be V₁

such that the Volume of water entering the per second is = A₁V₁

Therefore, in 90sec:

45 L =  90 A₁V₁

V₁ = 45 * 10⁻³m³ / 90*0.5*10⁻⁴

V₁ = 10m/s            (velocity of water inside the house)

From the continuity equation we have that;

A₁V₁ = A₂V₂

0.5*10⁻⁴ * 10 = 1*10⁻⁴ V₂

V₂ = 5m/s               (velocity at ground level)

(b). We are told to calculate the water pressure in the pipeline at the ground level.

Using Bernoulli's equation;

P₁ + pgh₁ + 1/2PV₁²  (inside)      =       P₂ + pgh₂ + 1/2PV₂²   (ground level)

1.01*10⁵ + 1000*9.8*10 + 1/2*1000*(10)² = P₂ + 0 + 1/2*1000*(5)²

P₂ (pressure) = 1.01*10⁵Pa

Therefore we have;

101000 + 98000 + 50000 = P₂ + 12500

P₂ = 236500 Pa

cheers I hope this helped !!

3 0
3 years ago
A box with a mass of 40 kg sits at rest on a frictionless tile floor. with your foot, you apply a 20 N force in a horizontal dir
Grace [21]

Answer:

0.5 m/s²

Explanation:

according to Newton's second law, we are goven a relationship between force, mass and acceleration, with the formula:

F = m×a

F for force

m for mass

a for acceleration

we use the given data and get:

20 = 40×a

we find a=20/40=0.5m/s²

4 0
2 years ago
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How was bohr atomic model similar to Rutherford's model​
Juliette [100K]
Both believe that an atom contains negative charges and positive charges.
But both were different in the placement of charges
6 0
3 years ago
Read 2 more answers
A series circuit consists of a 100-ω resistor, a 10.0-μf capacitor, and a 0.350-h inductor. the circuit is connected to a 120-v
Tpy6a [65]
Current will be I=\dfrac{V_{rms}}{Z}=\dfrac{V_{rms}}{\sqrt{ R^{2}+(X_{C}-X_{L})^{2}}}\\where~X_{c}=\dfrac{1}{j.\omega .C}~and~X_{L}=j.\omega.L~where~\omega=2.\pi f~and~f=60Hz
now just pluf in the values and Voila..
7 0
3 years ago
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