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Helen [10]
3 years ago
12

A steel ball with mass m is suspended from the ceiling at the bottom end of a light, 17.0-m-long rope. The ball swings back and

forth like a pendulum. When the ball is at its lowest point and the rope is vertical, the tension in the rope is three times the weight of the ball, so T=3mg.
1. What is the speed of the ball as it swings through this point? Express your answer with the appropriate units.
2. What is the speed of the ball if T=mg at this point, where the rope is vertical? Express your answer with the appropriate units.
Physics
1 answer:
iogann1982 [59]3 years ago
6 0

Answer:

1. 18.25 m/s

2. 0 m/s

Explanation:

1.So the centripetal acceleration of the ball at this lowest point must be, taking gravity into account

a_c = \frac{T - mg}{m} = \frac{3mg - mg}{m} = 2g

The speed at this point would then be

v^2 = a_c r = 2gr = 2*9.8*17 = 333.2

v = \sqrt{333.2} = 18.25 m/s

2. Similarly, if T = mg, then the centripetal acceleration must be

a_c = \frac{T - mg}{m} = \frac{mg - mg}{m} = 0

As the ball has no centripetal acceleration, its speed must also be 0 as well.

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3 0
3 years ago
A test charge of -1.4 x 10-7 coulombs experiences a force of 5.4 x 10-1 newtons. Calculate the magnitude of the electric field c
VARVARA [1.3K]

Answer:

3.86×10⁶ Newton/coulombs

Explaination:

Applying,

E = F/q....................... Equation 1

Where E = Electric Field, F  = Force, q = charge.

From the question,

Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs

Substitute these values into equation 1

E = 5.4×10⁻¹/ -1.4×10⁻⁷

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Hence the magnitude of the electric field created by the

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Read 2 more answers
Uma partícula com carga “Q”, no vácuo, gera um potencial elétrico de 600 volts a uma distância de 6,0 m, determine o valor dessa
garik1379 [7]

Answer:

q =  400 nC

the correct answer is b

Explanation:

The expression for the electric potential of a point charge is

            V = k q / r

they ask us for the electrical charge

          q = V r / k

let's calculate

         Q = 600 6.0 / 9 10⁹

         Q = 4 10⁻⁷ C

let's reduce to nC

          Q = 4 10⁻⁷ C (10⁹ nC / 1C)

          q = 4 10² nC = 400 nC

the correct answer is b

Traslate

La expresión para el potencial eléctrico de una carga puntual es

            V = k q/r

nos piden la carga eléctrica

          q= V r /k

          calculemos

         Q= 600  6,0 / 9 10⁹

         Q=  4 10⁻⁷ C

reduzcamos a nC

          Q = 4 10⁻⁷ C(10⁹ nC/1C )  

          q = 4 10² nC = 400 nC

la respuesta correcta es b

8 0
3 years ago
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