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Pavlova-9 [17]
3 years ago
15

A 4.00-m long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end

is allowed to fall. What is the angular acceleration as it is released
Physics
1 answer:
Natalka [10]3 years ago
5 0

Answer:

The angular acceleration α = 14.7 rad/s²

Explanation:

The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod  and L = length of rod = 4.00 m. α = angular acceleration of rod

Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.

So Iα = Wr

Substituting the value of the variables, we have

mL²α/12 = mgL/2

Simplifying by dividing through by mL, we have

mL²α/12mL = mgL/2mL

Lα/12 = g/2

multiplying both sides by 12, we have

Lα/12 × 12 = g/2 × 12

αL = 6g

α = 6g/L

α = 6 × 9.8 m/s² ÷ 4.00 m

α = 58.8 m/s² ÷ 4.00 m

α = 14.7 rad/s²

So, the angular acceleration α = 14.7 rad/s²

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The emf induced in a coil that is rotating in a magnetic field will be at a maximum at which moment?
adelina 88 [10]
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)

where “e” describes the electromotive force, “N” describes the number of loops in the coil, “dI” describes the change in magnetic flux, “dt” describes the change in time, “A” describes the area vector of the coil (this points perpendicular to the loops, intersecting it in open space), “B” describes the magnetic field vector, and theta describes the angle between the area and mag vectors.

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.

Hope this helps!
6 0
4 years ago
Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic
Nuetrik [128]

Answer:

(a) A = 3.90 \AA

(b) A = 4.50 \AA

(c) A = 5.51 \AA

(d) A = 9.02 \AA

Solution:

As per the question:

Radius of atom, r = 1.95 \AA = 1.95\times 10^{- 10} m

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = 2\times 1.95 = 3.90 \AA

(b) For body centered cubic lattice:

A = \frac{4}{\sqrt{3}}r

A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA

(c) For face centered cubic lattice:

A = 2{\sqrt{2}}r

A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA

(d) For diamond lattice:

A = 2\times \frac{4}{\sqrt{3}}r

A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA

6 0
3 years ago
A rock climber hangs freely from a nylon rope that is 15 m long and has a diameter of 8.3 mm. If the rope stretches 5.1 cm, what
irinina [24]

Answer:

Mass of the climber = 69.38 kg

Explanation:

Change in length

        \Delta L=\frac{PL}{AE}

Load, P = m x 9.81 = 9.81m

Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

       A=\frac{\pi (8.3\times 10^{-3})^2}{4}=5.41\times 10^{-5}m^2

Length, L = 15 m

ΔL = 5.1 cm = 0.051 m

Substituting

       0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg  

Mass of the climber = 69.38 kg

6 0
3 years ago
Siven an element's atomic number and mass number, how can you tell the number of protons and neutrons in its nucleus?
sukhopar [10]
It would be the first option.

Explanation-
The number of protons is equal to the atomic number the number of neutrons is the mass minus the atomic number.
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