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saul85 [17]
2 years ago
12

A rectangular loop of wire with width w = 5 cm, length L = 10cm, mass m = 40 g, and resistance R = 20 mΩ has an initial velocity

v0 = 1 m/s to the right. It crosses from a region with zero magnetic field to a region with B = 2T pointing out of the page. How far does the loop penetrate into the magnetic field?
Physics
1 answer:
zhuklara [117]2 years ago
5 0

Answer:

The loop penetrate 4 cm into the magnetic field.

Explanation:

Given that,

Width w= 5 cm

Length L= 10 cm

mass m = 40 g

Resistance R = 20 mΩ

Initial velocity = 1 m/s

Magnetic field = 2 T

We need to calculate the induced emf

Using formula of emf

\epsilon=v_{0}Bw

Put the value into the formula

\epsilon =1\times2\times5\times10^{-2}

\epsilon =10\times10^{-2}\ volt

We need to calculate the current

Using Lenz's formula

i=\dfrac{\epsilon}{R}

i=\dfrac{10\times10^{-2}}{20\times10^{-3}}

i=5\ A

We need to calculate the force

Using formula of force

F=i(\vec{w}\times\vec{B})

F=iwB

Put the value into the formula

F=5\times5\times10^{-2}\times2

F=0.5\ N

We need to calculate the acceleration

Using formula of acceleration

a=\dfrac{F}{m}

Put the value in to the formula

a=\dfrac{0.5}{40\times10^{-3}}

a=12.5\ m/s^2

We need to calculate the distance

Using equation of motion

v^2=u^2+2as

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{0-1^2}{2\times(-12.5)}

s=0.04\ m

s=4\ cm

Hence, The loop penetrate 4 cm into the magnetic field.

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At what angle are the electronic and the magnetic wave related in an electromagnetic signal?
VikaD [51]

Answer:

90degrees I'm pretty sure

7 0
2 years ago
An object moves from the position +16m to the position +43m in 12s. What us the total displacement
NeX [460]

First method

initial distance = 16m

final distance= 43 m

total distance covered= final -initial

                                     =43m -16m

                                     =27m

Second method

Si= 16m

Sf =43 m

t= 12 s

first we will find V

V =  (Sf-Si)/ t

V =( 43- 16)/ 12

V = 27/12  ⇒ V= 9/4

V= distance / time

distance= V×time

distance = (9/4) ×12

distance =27

3 0
3 years ago
Your 64-cm-diameter car tire is rotating at 3.3 rev/swhen suddenly you press down hard on the accelerator. After traveling 250 m
umka21 [38]

Answer:

0.76 rad/s^2

Explanation:

First, we convert the original and final velocity from rev/s to rad/s:

v_o = 3.3\frac{rev}{s} * \frac{2\pi rad}{1rev} =20.73 rad/s

v_f = 6.4\frac{rev}{s} * \frac{2\pi rad}{1rev}=40.21 rad/s

Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:

D = x*r \\x = \frac{D}{r} = \frac{250m}{0.64m/2} = 781.25 rads

Now, we just use the formula:

w_f^2-w_o^2=2\alpha*x

\alpha =\frac{w_f^2-w_o^2}{2x} = \frac{(40.21rad/s)^2-(20.73rad/s)^2}{2*781.25rad} = 0.76 rad/s^2

6 0
2 years ago
Read 2 more answers
A steel ball of mass 50 g is rolled from the left toward a ball of lead of mass 500 g. The steel ball is traveling with a veloci
alina1380 [7]

Answer: It's hard to say without characterizing the collision. But it will be either A if the collision is totally in-elastic, or B if the collision is totally elastic. It could be anywhere in between for partially elastic collisions.

Explanation:

momentum is conserved, so initial system momentum will be left to right.

The velocity of the center of mass is 50(5) / 550 = 0.4545... m/s

In an elastic collision, the lead ball will move off at twice that speed or 0.91 m/s to the right.

The steel ball will bounce back and move away at 0.91 - 5 = -4.1 m/s . The negative sign indicates the steel ball has reversed course and has negative momentum

In a totally in-elastic collision, both balls would move to the right at 0.45 m/s. The steel ball will still have positive momentum.

4 0
2 years ago
A charge of -2.65 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y =
stiks02 [169]

Answer:

A. Fnx = 5.71*10⁻⁵ N  ,  Fny= -3.67*10⁻⁵ N

B. Fn= 6.78 *10⁻⁵ N

C. α= 32.4° counterclockwise with the positive x+ axis

Explanation:

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

1cm = 10⁻²m

Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }

d_{13} =\sqrt{24.68} * 10⁻²m    = 4.9678* 10⁻²m

(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m  

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² =  ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N

F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) =  - 3.67*10⁻⁵ N

F₂₃x  = F₂₃ =  +8.8 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x =  - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N

Fn magnitude

F_{n} =\sqrt{(Fn_{x})^{2}+(Fn_{y})^{2}  }

F_{n} = \sqrt{(5.71)^{2}+(3.67)^{2}  } *10⁻⁵ N

Fn= 6.78 *10⁻⁵ N

Fn direction  (α)

\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )

\alpha =tan^{-1}( \frac{-3.67 }{5.71} )

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

4 0
3 years ago
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