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oee [108]
3 years ago
14

What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is w

= 1.75 m wide and h = 8.00 m below the top of the cliff?
Physics
1 answer:
Yanka [14]3 years ago
8 0
<span>1.37 m/s Assuming her initial velocity is totally horizontal and her vertical velocity is only affected by gravity, let's first calculate how much time she has until she reaches the ledge 8.00 m below her. d = 1/2AT^2 8.00m = 1/2 * 9.8 m/s^2 * T^2 Solve for T 8.00 m = 4.9 m/s^2 * T^2 Divide both sides by 4.9 m/s^2 1.632653061 s^2 = T^2 Take square root of both sides 1.277753 s = T So we now know that she has 1.277753 seconds in which to reach a horizontal distance of 1.75 m. So how fast does she need to be going? 1.75 m / 1.277753 s = 1.369592 m/s Since we only have 3 significant figures in our data, round the result to 3 figures giving 1.37 m/s</span>
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TLDR: It will reach a maximum when the angle between the area vector and the magnetic field vector are perpendicular to one another.

This is an example that requires you to investigate the properties that occur in electric generators; for example, hydroelectric dams produce electricity by forcing a coil to rotate in the presence of a magnetic field, generating a current.

To solve this, we need to understand the principles of electromotive forces and Lenz’ Law; changing the magnetic field conditions around anything with this potential causes an induced current in the wire that resists this change. This principle is known as Lenz’ Law, and can be described using equations that are specific to certain situations. For this, we need the two that are useful here:

e = -N•dI/dt; dI = ABcos(theta)

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Because the number of loops remains constant and the speed of the coils rotation isn’t up for us to decide, the only thing that can increase or decrease the emf is the change in magnetic flux, represented by ABcos(theta). The magnetic field and the size of the loop are also constant, so all we can control is the angle between the two. To generate the largest emf, we need cos(theta) to be as large as possible. To do this, we can search a graph of cos(theta) for the highest point. This occurs when theta equals 90 degrees, or a right angle. Therefore, the electromotive potential will reach a maximum when the angle between the area vector and the magnetic field vector are perpendicular to one another.

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A team of dogs accelerates a 290kg dogsled from 0 to 6.0m/s in 3.0 s. Assume that the acceleration is constant.Part AWhat is the
Mademuasel [1]

Answer:

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(b) From second equation of motion we know that

s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 2\times 3^2=9m

We know that work done is given by

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So P=\frac{W}{t}=\frac{5220}{1.5}=3466.66Watt

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