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Ksenya-84 [330]
2 years ago
10

Taylor places a nail on a bar magnet. The nail sticks to the magnet when lifted up off the table. She touches a paperclip to the

nail and it sticks to the nail. Explain what happened to the magnetic domains of the nail before and after touching it to the bar magnet.
Physics
2 answers:
diamong [38]2 years ago
8 0

Taylor places a nail on a bar magnet. The nail sticks to the magnet when lifted up off the table. She touches a paperclip to the nail and it sticks to the nail. Explain what happened to the magnetic domains of the nail before and after touching it to the bar magnet.

Ratling [72]2 years ago
5 0
Prior to touching the bar magnet, the magnetic domains in the nail were pointing in random directions. When Taylor touched the nail to the bar magnet the magnetic fields of the magnetic domains aligned and it became a temporary magnet.
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A rifle, which has a mass of 5.50 kg., is used to fire a bullet, which has a massof m = 65.0 grams., at a "ballistics pendulum".
Alex787 [66]

Answer:

Part a)

U = 13 J

Part b)

v = 2.28 m/s

Part c)

v = 177.66 m/s

Part d)

W = 1012.7 J

Part e)

v = 2.1 m/s

Part f)

E = 1037.2 J

Explanation:

Part a)

As we know that the maximum angle deflected by the pendulum is

\theta = 38^o

so the maximum height reached by the pendulum is given as

h = L(1 - cos\theta)

so we will have

h = L(1 - cos38)

h = 1.25(1 - cos38)

h = 0.265 m

now gravitational potential energy of the pendulum is given as

U = mgh

U = 5(9.81)(0.265)

U = 13 J

Part b)

As we know that there is no energy loss while moving upwards after being stuck

so here we can use mechanical energy conservation law

so we have

mgh = \frac{1}{2}mv^2

v = \sqrt{2gh}

v = \sqrt{2(9.81)(0.265)}

v = 2.28 m/s

Part c)

now by momentum conservation we can say

mv = (M + m) v_f

0.065 v = (5 + 0.065)2.28

v = 177.66 m/s

Part d)

Work done by the bullet is equal to the change in kinetic energy of the system

so we have

W = \frac{1}{2}mv^2 - \frac{1}{2}(m + M)v_f^2

W = \frac{1}{2}(0.065)(177.66)^2 - \frac{1}{2}(5 + 0.065)2.28^2

W = 1012.7 J

Part e)

recoil speed of the gun can be calculated by momentum conservation

so we will have

0 = mv_1 + Mv_2

0 = 0.065(177.6) + 5.50 v

v = 2.1 m/s

Part f)

Total energy released in the process of shooting of gun

E = \frac{1}{2}Mv^2 + \frac{1}{2}mv_1^2

E = \frac{1}{2}(5.50)(2.1^2) + \frac{1}{2}(0.065)(177.6^2)

E = 1037.2 J

6 0
2 years ago
Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin
Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

\mu_o=4\pi \times 10^{-7}\ T-m/A

Resistivity, \rho=1.72\times 10^{-8}\ \Omega-m

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

R=\rho \dfrac{l}{A}

R=\rho \dfrac{l}{\pi r^2}

r=\sqrt{\dfrac{\rho l}{R\pi}}

r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}

r = 0.00199 m

or

r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m

The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

I=\dfrac{2\pi rB}{\mu_o}

I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0
2 years ago
Frequency is deoted as hertz; hertz is a measurement of the _________ _____ __________ that a wave is occurring.
nlexa [21]
Hertz is a measurement of the frequency that a wave is occurring.
4 0
2 years ago
A roadway for stunt drivers is designed for racecars moving at a speed of 40 m/s. A curved section of the roadway is a circular
Fynjy0 [20]

Answer:

Bank angle = 35.34o

Explanation:

Since the road is frictionless,

Tan (bank angle) = V^2/r*g

Where V = speed of the racing car in m/s, r = radius of the arc in metres and g = acceleration due to gravity in m/s^2

Tan ( bank angle) = 40^2/(230*9.81)

Tan (bank angle) = 0.7091

Bank angle = tan inverse (0.7091)

Bank angle = 35.34o

3 0
2 years ago
U-238 has protons and146 neutrons. A particular isotope of plutonium has 94 protons, neutrons, and a mass number of 241. Thorium
enyata [817]

#1

^{238}U

so mass number = 238

mass number = protons + neutrons

given that

neutrons = 146

238 = protons + 146

protons = 92

#2

^{241}Pu

so mass number = 241

mass number = protons + neutrons

given that

Protons = 94

241 = 94 + neutrons

neutrons = 147

#3

^ATh

A = mass number

Protons = 90

Neutrons = 137

A = protons + Neutrons

A = 90 + 137 = 227

5 0
2 years ago
Read 2 more answers
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