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3 years ago

9

A 20g book laying on a 40cm high table. Just before a cat knocks it off the table what is the books gravitational potential ener

gy?

1 answer:

disa [49]3 years ago

3 0

Answer:

The book gravitational potential energy is 0.0784 Joules.

Explanation:

Given:

Mass of book, m = 20 g = 20 ÷ 1000 = 0.02 kg

Table height, h = 40 cm =40 ÷ 100 = 0.4 m

To Find:

Gravitational Potential Energy = ?

Solution:

Gravitational Potential Energy :

Gravitational potential energy is energy an object possesses because of its position in a gravitational field.

Formula is given by

$\textrm{Gravitational Potential Energy}= m\times g\times h$

Where,

m = mass

g = acceleration due to gravity = 9.8 m/s²

h = height

Substituting the values we get

$\textrm{Gravitational Potential Energy}= 0.02\times 9.8\times 0.4\\\\\textrm{Gravitational Potential Energy}=0.0784 Joules$

The book gravitational potential energy is 0.0784 Joules.

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A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

So

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

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Answer:

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Explanation:

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Explanation:

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Jim takes 45 seconds to walk 180 m north to a store . What is James velocity in meters per second?

solmaris [256]

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Explanation:

Given parameters:

Time taken = 45s

Displacement = 180m

Unknown

Velocity = ?

Solution:

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Velocity = 4m/s due north of the store

Learn more:

Velocity brainly.com/question/10883914

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