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ElenaW [278]
2 years ago
6

A small lead ball, attached to a 1.10-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled

at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is 1.3 m above the ground. The ball travels straight upward. In the absence of air resistance, to what maximum height above the ground does the ball rise
Physics
1 answer:
hjlf2 years ago
7 0

Answer:

1.84 m

Explanation:

For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.

So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,

a = g + rω²

= 9.8 m/s² + 1.10 m × (18.85 rad/s)²

= 9.8 m/s² + 390.85 m/s²

= 400.65 m/s²

Now, using v² = u² + 2a(h₂ - h₁)  where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.

v² = u² + 2a(h₂ - h₁)

So, v² - u² = 2a(h₂ - h₁)

h₂ - h₁ =  (v² - u²)/2a

h₂ =  h₁ + (v² - u²)/2a

substituting the values of the variables into the equation, we have

h₂ =  1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)

h₂ =  1.3 m + [-430.15 (m/s)²]/-801.3 m/s²

h₂ =  1.3 m + 0.54 m

h₂ =  1.84 m

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En la Tierra, una balanza muestra que tu peso es 585 N.
ra1l [238]

Answer:

a) m = 59.63 [kg]

b) Wm = 95.41 [N]

Explanation:

El peso de un cuerpo se define como el producto de la masa por la aceleración gravitacional. DE esta manera tenemos:

W = m*g

Donde:

m = masa [kg]

g = gravedad = 9.81 [m/s^2]

m = W / g

m = 585 / 9.81

m = 59.63 [kg]

Es importante aclarar que la masa se conserva independientemente de la ubicación del cuerpo en el espacio.

Por ende su masa sera la misma en la luna.

El peso en la luna se calcula como Wm y es igual a:

Wm = 59.63 * 1.6 = 95.41 [N]

5 0
3 years ago
8-30-2018 what phase change occurs as the substance at 0 degree c has its pressure dropped from 0.50 at to 0.25 atm
madam [21]

Answer:

at T = 0ºC the change of state is from the solid state to the gaseous state

Explanation:

In this exercise we are asked about the changes of state, from the data we will assume that the material is water.

Water can exist in three solid states, liquid and gas, in a graph of pressure ℗ against temperature (T) there is a point called triple at T = 0.01ºC, below this point the curve has two states at high pressure solid and low pressure gas.

As a result of the previous ones at T = 0ºC the change of state is from the solid state to the gaseous state

7 0
3 years ago
A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant
vovangra [49]

Answer:

emf induced in the loop, at the instant when 9.0s have passed = 1.576 * 10 ⁻² V.

Direction is counter clockwise.

Explanation:

See attached pictures.

6 0
3 years ago
If sulfur 34 undergoes alpha decay, what will I become?
ikadub [295]
It will decay into Silicon-30. Because alpha particles are 2 protons and 2 neutrons with an atomic mass of 4, you minus sulfur's atomic number by 2 and get silicon. And the atomic mass is 34 - 4 which equals 30.
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3 years ago
An object glides on a horizontal tabletop with a coefficient of kinetic friction of 0.5. If its initial velocity is 4.3 m/s, how
Shkiper50 [21]

Answer:

Time, t = 0.87 seconds

Explanation:

Given that,

Initial velocity of the object, u = 4.3 m/s

The coefficient of kinetic friction between horizontal tabletop and the object is 0.5

We need to find the time taken by the object for the object to come to rest i.e. final velocity will be 0.

Using first equation of motion to find it as :

v=u+at

a is the acceleration, here, a=\mu g

0=u+\mu gt

t=\dfrac{u}{\mu g}\\\\t=\dfrac{4.3}{0.5\times 9.8}\\\\t=0.87\ s

So, the time taken by the object to come at rest is 0.87 seconds. Hence, this is the required solution.

8 0
3 years ago
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