A piece of soft iron is placed in a solenoid increasing the magnetic field in an arrangement that can be switched on and off. Such a device is calleda. a hysteresis loop.b. an electrosolenoid.c. a permanent magnet.d. a ferromagnet.

Answers

Answer 1

Answer:

Question:

A piece of soft iron is placed in a solenoid increasing the magnetic field in an arrangement that can be switched on and off. Such a device is called

Answer

a hysteresis loop.

an electrosolenoid.

a permanent magnet.

a ferromagnet.

an electromagnet

Answer:

A piece of soft iron is placed in a solenoid increasing the magnetic field in an arrangement that can be switched on and off. Such a device is called an electromagnet.

Explanation:

When a magnet is processed by supplying electricity than it is called as electromagnet and therefore its strength can be fluctuated as dependent on amount of electricity supplied.

Here, when a iron piece is moved closer to a long coil of wire named solenoid physical interaction named electromagnetic force take place between two electrically charged particles.

This force is supplied by electromagnetic fields constructed from electric and magnetic field. This is also called as Lorentz force constituted from both electricity and magnetism. This property itself gives evidence of characteristics of material used for electromagnetism.

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A baseball leaves the bat with a speed of 40 m/s at an angle of 35 degrees. A 12m tall fence is placed 130 m from the point the ball was struck. Assuming the batter hit the ball 1m above ground level, does the ball go over the fence? If not, how does the ball hit the fence? If yes, how far beyond the fence does the ball land?

Answers

Answer:

The ball land at 3.00 m.

Explanation:

Given that,

Speed = 40 m/s

Angle = 35°

Height h = 1 m

Height of fence h'= 12 m

We need to calculate the horizontal velocity

Using formula of horizontal velocity

$V_(x)=V_(i)\cos\theta$

$V_(x)=40*\cos35$

$V_(x)=32.76\ m/s$

We need to calculate the time

Using formula of time

$t = (d)/(v)$

$t=(130)/(32.76)$

$t=3.96\ sec$

We need to calculate the vertical velocity

$v_(y)=v_(y)\sin\theta$

$v_(y)=40*\sin35$

$v_(y)=22.94\ m/s$

We need to calculate the vertical position

Using formula of distance

$y(t)=y_(0)+V_(i)t+(1)/(2)gt^2$

Put the value into the formula

$y(3.96)=1+22.94*3.96+(1)/(2)*(-9.8)*(3.96)^2$

$y(3.96)=15.00\ m$

We need to calculate the distance

$s = y-h'$

$s=15.00-12$

$s=3.00\ m$

Hence, The ball land at 3.00 m.

An unstable atomic nucleus of mass 1.82 10-26 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.18 10-27 kg, moves in the y direction with a speed of 6.00 106 m/s. Another particle, of mass 8.50 10-27 kg, moves in the x direction with a speed of 4.00 106 m/s. (a) Find the velocity of the third particle.

Answers

Answer:

Explanation:

Using Conservation of momentum (total final momentum of system is)

m1•v1f + m2•v2 f + m3•v3 f=0

and it must be zero to equal the original momentum( since the original body is at rest).

Given that

original mass M=1.82×10^-26

First disintegrate mass m1=5.18×10^-27kg

In y direction V1f=6×10^6 I'm/s

Second disintegrate mass m2=8.5×10^-27kg

In x direction V2f=4×10^6 im/s

Then the third disintegrate will be

m3=M-m1-m2

m3=1.82×10^-26-5.18×10^-27-8.5×10^-27

m3=4.52×10^-27

And the velocity is unknown

Now using the formula above

m1•v1f + m2•v2 f + m3•v3 f=0

m3•V3f= - m1•v1f - m2•v2 f

4.52E-27V3f=-5.18E-27×6E6j - 8.5E-27×4E6 i

Divide thorough by 4.52E-27

V3f= - 6.88×10^6j - 7.52×10^6i

V3f= - 7.52×10^6i - 6.88×10^6j

The final velocity of the third mass disintegrate is 6.88×10^6j - 7.52×10^6i m/s

A 2.00 kg cart on a frictionless track is pulled by force of 3.00 N. What is the acceleration of the cart?

Answers

1.5 will be its acceleration

A 4.0-m-long, 500 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 70 kg construction worker stands at the far end of the beam.What is the magnitude of the torque about the bolt due to the worker and the weight of the beam?

Answers

Answer:

T=12544 N*m

Explanation:

Given

L=4.0m

ms=500kg

mw=70kg

Torque is the force in a distance the relation is proportional so the torque of weight first is:

Ts = Fs*d

Ts = ms*g*L

Ts = 500kg*9.8m/s^2*2m

Ts = 9800 N*m

now torque of the worker

Tw = Fw*d

Tw = 70kg*9.8m/s^2*4m

Tw = 2744 N*m

Torque net is

Tnet = Tw+Ts

Tnet= 2744 + 9800 =12544 N*m

Final answer:

The total torque about the bolt due to the worker and the weight of the beam is 12544 Nm. This is found by adding the torque due to the beam and the worker which can be calculated using their weights and their distance from the pivot point (bolt).

Explanation:

The key to solving this question is understanding torque, which in physics represents the rotational effect of a force. Torque is calculated using the formula τ = r x F, where τ is the torque, r is the distance from the pivot point, and F is the force applied.

In this case, there are two forces to consider: the weight of the beam and the weight of the worker. Both of these can be calculated using the formula for weight (F = m*g), where m is mass and g is gravitational acceleration, which is approximately 9.8 m/s^2 on Earth. The weight of the beam is therefore 500 kg * 9.8 m/s^2 = 4900 N, and the weight of the worker is 70 kg * 9.8 m/s^2 = 686 N.

The distance from the pivot (bolt) for the beam's weight is considered to be the midpoint of the beam, so it is 4.0 m / 2 = 2.0 m. For the worker, r equals the full length of the beam, which is 4.0 m. The total torque can be calculated by adding the torque due to the beam and the worker. Therefore, the total torque τ = (2.0 m * 4900 N) + (4.0 m * 686 N) = 9800 Nm + 2744 Nm = 12544 Nm.

Learn more about Torque here:

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The most soaring vocal melody is in Johann Sebastian Bach's Mass in B minor. In one section, the basses, tenors, altos, and sopranos carry the melody from a low D to a high A. In concert pitch, these notes are now assigned frequencies of 146.8 Hz and 880.0 Hz. (Use 343 m/s as the speed of sound, and 1.20 kg/m3 as the density of air.)a. Find the wavelength of the initial note.
b. Find the wavelength of the final note.
c. Assume the choir sings the melody with a uniform sound level of 70.0 dB. Find the pressure amplitude of the initial note.
d. Find the pressure amplitude of the final note.
e. Find the displacement amplitude of the initial note.
f. Find the displacement amplitude of the final note.

Answers

Answer:

Detailed step wise solution is attached below

Explanation:

(a) wavelength of the initial note 2.34 meters

(b) wavelength of the final note 0.389 meters

(d) pressure amplitude of the final note 0.09 Pa

(e) displacement amplitude of the initial note 4.78*10^(-7) meters

(f) displacement amplitude of the final note 3.95*10^(-8) meters

Sharece knows that wave peaks and valleys can add and subtract. What would be the net effect if she was able to cross Wave 1 (a large-amplitude wave in a valley phase) with Wave 2 (a wave with slightly smaller amplitude than Wave 2, in a peak phase)?Sharece knows that wave peaks and valleys can add and subtract. What would be the net effect if she was able to cross Wave 1 (a large-amplitude wave in a valley phase) with Wave 2 (a wave with slightly smaller amplitude than Wave 2, in a peak phase)?

Answers

Answer:

The two waves will add vectorially to produce a small amplitude wave in a valley phase.

Explanation:

The two waves will add vectorially to produce a small amplitude wave in a valley phase. This is because the amplitudes of the waves are slightly different and in opposite directions. When wave 1 cancels out all of wave 2, the resultant wave would be the slight difference between both waves, and it would be in the direction of wave 1 which is a valley phase.

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