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3 years ago

5

. A 2.00 kg ball is attached to a ceiling by a string. The distance from the ceiling to the center of the ball is 1.00 m, and th

e height of the room is 3.00 m. What is the gravitational potential energy associated with the ball relative to each of the following? (a) the ceiling (b) the floor (c) a point at the same elevation as the ball

1 answer:

cupoosta [38]3 years ago

4 0

<h2>a) Potential energy of ball relative to ceiling is 19.62 J</h2><h2>b) Potential energy of ball relative to floor is 39.24 J</h2><h2>c) Potential energy of ball relative to same elevation is 0 J</h2>

Explanation:

Mass of ball, m = 2 kg

Acceleration due to gravity, g = 9.81 m/s²

Height of ball from ground = 3 - 1 = 2 m

Potential energy, PE = mgh

Potential energy of ball, PE = 2 x 9.81 x 2 = 39.24 J

a) Potential energy of ball at ceiling = 2 x 9.81 x 3 = 58.86 J

Potential energy of ball relative to ceiling = 58.86 - 39.24 = 19.62 J

b) Potential energy of ball at floor = 2 x 9.81 x 0 = 0 J

Potential energy of ball relative to floor = 39.24 - 0 = 39.24 J

c) Potential energy of ball at same elevation = 2 x 9.81 x 2 = 39.24 J

Potential energy of ball relative to same elevation = 39.24 - 39.24 = 0 J

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Answer:

a) $E=50.53\times 10^{6}\ N/C$

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b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

a)

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

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I believe it’s density and temperature

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3 years ago

A firecracker breaks up into two pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.0 m/

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Answer:

Total momentum, p = 21.24 kg-m/s

Explanation:

Given that,

Mass of first piece, $m_1=200\ g= 0.2\ kg$

Mass of the second piece, $m_2=300\ g= 0.3\ kg$

Speed of the first piece, $v_1=82\ m/s$ (along x axis)

Speed of the second piece, $v_2=45\ m/s$ (along y axis)

To find,

The total momentum of the two pieces.

Solve,

The total momentum of two pieces is equal to the sum of momentum along x axis and along y axis.

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The net momentum is given by :

$p=\sqrt{p_x^2+p_y^2}$

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p = 21.24 kg-m/s

Therefore, the total momentum of the two pieces is 21.24 kg-m/s.

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Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

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Answer:

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b) $(x_4,y_4)=(-1.917,-1.75)m$

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The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m$

If we solve for a we got:

$(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0$

$a=-\frac{(5kg*2.3m)}{6kg}=-1.917m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m$

$(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0$

And solving for b we got:

$b=-\frac{(7kg*1.5m)}{6kg}=-1.75m$

So the coordinates for this new particle are:

$(x_4,y_4)=(-1.917,-1.75)m$

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