If a rock weighing 3.5 newtons is resting on cliff's ledge 12 meters from the canyon floor, what is its potential energy?

A 0.150 kg ball on the end of a 1.10 m long cord (negligible mass)is swung in a vertical circle..

Aneli [31]

<span> For any body to move in a circle it requires the centripetal force (mv^2)/r. In this case a ball is moving in a vertical circle swung by a mass less cord. At the top of its arc if we draw its free body diagram and equate the forces in radial direction to the centripetal force we get it as T +mg =(mv^2)/r T is tension in cord m is mass of ball r is length of cord (radius of the vertical circle) To get the minimum value of velocity the LHS should be minimum. This is possible when T = 0. So minimum speed of ball v at top =sqrtr(rg)=sqrt(1.1*9.81) = 3.285 m/s In the second case the speed of ball at top = (2*3.285) =6.57 m/s Let us take the lowest point of the vertical circle as reference for potential energy and apllying the conservation of energy equation between top & bottom we get velocity at bottom as 9.3m/s. Now by drawing the free body diagram of the ball at the bottom and equating the net radial force to the centripetal force T-mg=(mv^2)/r We get tension in cord T=13.27 N</span>

3 0

3 years ago

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Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to

lbvjy [14]

Answer:

Explanation:

We shall consider direction towards left as positive Let the required velocity be v and let v makes an angle φ

Applying law of conservation of momentum along direction of original motion

m₁ v₁ - m₂ v₂ = m₂v₃ - m₁ v₄

0.132 x 1.25 - .143 x 1.14 = 1.03 cos43 x .143 - v cos θ

v cos θ = .8

Applying law of conservation of momentum along direction perpendicular to direction of original motion

1.03 sin 43 x .143 = .132 x v sinθ

v sinθ = .76

squaring and adding

v² = .76 ² + .8²

v = 1.1 m /s

Tan θ = .76 / .8

θ = 44°

6 0

3 years ago

A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr

valina [46]

The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
$T=2 t = 2 \cdot 0.100 s = 0.200 s$
Which means that the frequency is
$f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz$
and the angular frequency is
$\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s$

In a spring-mass system, the maximum velocity of the object is given by
$v_{max} = A \omega$
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
$v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s$

6 0

3 years ago

The Earth's _____

fenix001 [56]

Answer:

rotation

Explanation:

............................

4 0

2 years ago

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There are many interesting applications of our energy density model to the flow of blood in the human circulatory system. One in

qaws [65]

Answer:

Pressure increases due to enlargement

Explanation:

Energy density is just a fancy name for pressure

Pressure is same at the bottom of the cups (same level-Pascal's law)

thus, Air pressure 1 + h1d1g = Air pressure 2 + h2d1g

= Air pressure 3 + (h2-h1)d2g +h1d1g

from the first 2, we get that since h2>h1, AP2<AP1

from the next 2, we get that since d2<d1, AP3>AP2

from first and third, we get that AP1>AP3

thus, finally AP1>AP3>AP2

for fluids flowing in tubes (blood vessel in this case)

P+0.5dv^2 + gh is constant (also called the bernoulli equation

for the same blood vessel, the heights remain same i.e h1=h2

for same flow rate, inc in area decreases the speed at which the blood flows as vA must remain same

hence, P increases due to enlargement

5 0

3 years ago

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