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Gelneren [198K]
2 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]2 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: I do

Explanation:

Resistance opposes current thereby reducing the amount of current that flows through a circuit. In other words, it leads to a loss of electrical energy.

Ideally speaking, a good circuit should have no internal resistance as this would lead to more energy having to be supplied to overcome that resistance. External resistance however, is not a bad thing. For instance, oxygen being removed from lightbulbs.

7 0
3 years ago
A uniform magnetic field passes through a horizontal circular wire loop at an angle 19.5 ∘ from the vertical. The magnitude of t
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To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

\Phi = BA Cos \theta

Here,

\theta = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

\epsilon= -N \frac{d\Phi }{dt}

\epsilon = -N \frac{(BAcos\theta)}{dt}

\epsilon = -NAcos\theta \frac{dB}{dt}

\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)

\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )

At time t = 5.71s,  Induced emf is,

\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)(  (3.05T/s)-(13.9T/s)(5.71s))

\epsilon = 10.9V

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4 0
3 years ago
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Two cars collide at an intersection. One car has a mass of 1600 kg and is
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The combined momentum is 4000 kg m/s south

Explanation:

The total combined momentum of the two cars is given by the vector addition of the momenta of the two cars.

For this problem, we choose north as positive direction and south as negative direction.

The momentum of the first car travelling north is given by:

p_1 = m_1 v_1

where

m_1 = 1600 kg is the mass of the car

v_1 = +8 m/s is its velocity

Substituting,

p_1 = (1600)(8)=+12800 kg m/s

The momentum of the second car travelling south is given by:

p_2 = m_2 v_2

where

m_2 = 1400 kg is the mass of the car

v_2= -12 m/s is its velocity (negative because the car travels south)

Substituting,

p_2 = (1400)(-12)=-16800 kg m/s

And therefore, the combined momentum is

p=p_1 + p_2 = +12800 + (-16800)=-4000 kg m/s

where the negative sign means the direction of the total momentum is south.

Learn more about momentum:

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Answer:

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The recessive trait will always show up

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