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Doss [256]
3 years ago
10

A 4-N object object swings on the end of a string as a simple pendulum. At the bottom of the swing, the tension in the string is

9 N. What is the magnitude of the centripetal acceleration of the object at the bottom of the swing? Give the answer as a function of the acceleration due to gravity, g.
Physics
1 answer:
Amanda [17]3 years ago
6 0

Answer:

Explanation:

Given mg = 4N .

m = 4 / g

At the bottom of the swing let centripetal acceleration be a

T - mg = ma

9 - 4 = ma

5 = 4 a  / g

a =  5g / 4

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Decomposers is the correct answer. ( I got your back bro)
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Which statements correctly characterize a lunar eclipse? Select all that apply.
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A) it occurs when earth is between the sun and the moon

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What is the likely identity of a metal if a sample has a mass of 63.5 g when measured in air and an apparent mass of 60.2 g when
Gre4nikov [31]

Answer:

Gold

Explanation:

Given:

Mass of sample = 63.5 g

Mass of water = 60.2 g

Find:

Object

Computation:

Mass of water displaced = 63.5 g - 60.2 g

Mass of water displaced = 3.3 g

So, volume in water = 3.3 cm³

Density = Mass / Volume

Density = 63.5 g / 3.3

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7 0
2 years ago
Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?
Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

f = focal length

Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0
3 years ago
while flying a plane parallel to the ground a pilot releases a fuel tank in order to reduce the planes mass. what is the tanks f
stepan [7]

La velocidad vertical del tanque después de caer 10 m es 14 m/seg .

La velocidad vertical del tanque se calcula mediante la aplicación de la fórmula de velocidad , la componente vertical Vfy, del movimiento horizontal como se muestra a continuación :

Vfy=?

  h = 10 m

                              Fórmula de Velocidad vertical Vfy:

                           Vfy²  = 2*g*h

                            Vfy= √(2*9.8m/seg2* 10m )

                            Vfy= 14 m/seg

7 0
3 years ago
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