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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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8 0
2 years ago
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A 20-kilogram child is riding on a 10-kg sled over a frictionless icy surface at 8.0 meters per second. Calculate the kinetic en
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K = 960 J

Explanation:

Given that,

Mass of a child = 20 kg

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Speed of child on sled = 8 m/s

We need to find the kinetic energy of the sled with the child.

The total mass of child and the sled = 20 kg + 10 kg

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The formula for the kinetic energy of an object is given by :

K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 30\times (8)^2\\\\K=960\ J

Hence, the kinetic energy of the sled with the child is 960 J.

6 0
2 years ago
A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the seco
Sunny_sXe [5.5K]

Answer:

(a)

x_1=17.68km\\y_1=-17.68km\\x_2=20km\\y_2=34.64km

(b) r=41.32km

\alpha =24.23^o

Explanation:

Let us take the north direction to be the positive y-axis and the east to be positive x-axis.

First day:

25.0 km southeast, which implies 45^o south of east. The y-component will be negative and the x-component will be positive.

x_1=25cos45^o=17.68km

y_1=-25sin45^o=-17.68km

Second day:

She starts off at the stopping point of last day. This time, both the y- and x-components are positive.

x_2=40cos60^o=20km

y_2=40sin60^o=34.64km

Therefore, total displacements:

x=x_1+x_2=(17.68+20)km=37.68km

y=y_1+y_2=(-17.68+34.64)km=16.96km

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6 0
3 years ago
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Answer:

Range = 22.61 m

Explanation:

We can use the formula for the Range in flat ground, given by:

Range=v_i^2\frac{sin(2\theta)}{g}

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Range=15^2\frac{sin(80^o)}{9.8} \approx 22.61\,\,m

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Answer:

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