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Gelneren [198K]
3 years ago
15

You do 120 j of work while pulling your sister back on a swing, whose chain is 5.10 m long. you start with the swing hanging ver

tically and pull it until the chain makes an angle of 32.0° with the vertical with your sister is at rest. what is your sister's mass, assuming negligible friction?
Physics
1 answer:
Goryan [66]3 years ago
5 0
The work done to pull the sister back on the swing is equal to the increase in potential energy of the sister:
W= \Delta U = mg \Delta h (1)

where m is the sister's mass, g is the gravitational acceleration and \Delta h is the increase in altitude of the sister with respect to its initial position.

By calling \theta the angle of the chain with respect to the vertical, the increase in altitude is given by
\Delta h = L - L \cos \theta = L(1 - \cos \theta) (2)
where L is the length of the chain.

Putting (2) inside (1), we find
W= m g L (1 - \cos \theta)
from which we can find the mass of the sister:
m =  \frac{W}{g L (1 - \cos \theta)} =  \frac{120 J}{(9.81 m/s^2)(5.10 m)(1- \cos 32.0^{\circ})} =15.8 kg
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Answer:

a = 120 m/s²

Explanation:

We apply Newton's second law in the x direction:

∑Fₓ = m*a Formula (1)

Known data

Where:

∑Fₓ: Algebraic sum of forces in the x direction

F: Force in Newtons (N)

m: mass (kg)

a: acceleration of the block (m/s²)

F = 1200N

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Problem development

We replace the known data in formula (1)

1200 = 10*a

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6 0
3 years ago
In an electronic transition an atom can not emit what?
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█ Question <span>█

</span><span>In an electronic transition, an atom cannot emit what?

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When an electronic transition is occurring, an atom cannot emit ultra-violet light. 

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7 0
3 years ago
A 20 cm-radius ball is uniformly charged to 71 nC.
artcher [175]

Answer:

Part a)

\rho = 2.12\mu C/m^3

Part b)

q_1 = 1.11 nC

q_2 = 8.88 nC

q_3 = 71 nC

Part c)

E_1 = 3996 N/C

E_2 = 7992 N/C

E_3 = 15975 N/C

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

V = \frac{4}{3}\pi R^3

V = \frac{4}{3}\pi(0.20)^3

V = 0.0335 m^3

now the charge density of the ball is given as

\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3

Part b)

Now the charge enclosed by the surface is given as

q = \rho V

at radius of 5 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3

q = 1.11 nC

at radius of 10 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3

q = 8.88 nC

at radius of 20 cm

q = 71 nC

Part c)

As we know that electric field is given as

E = \frac{kq}{r^2}

so we have electric field at r = 5 cm

E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}

E_1 = 3996 N/C

electric field at r = 10 cm

E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}

E_2 = 7992 N/C

electric field at r = 20 cm

E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}

E_3 = 15975 N/C

3 0
3 years ago
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Schach [20]

Answer:

Option c) are perpendicular to the electric field

Explanation:

Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90^{\circ} to the equipotential surface.

Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.

Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.

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3 years ago
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The correct answer would be  1.375 < t < 3  i hope this helps anyone


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