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dybincka [34]
3 years ago
12

A 9.00-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wi

re and rail is 0.320 Ω . Pulling the wire at a steady speed of 4.00 m/s causes 4.30 W of power to be dissipated in the circuit.
Physics
1 answer:
m_a_m_a [10]3 years ago
5 0

Answer:

<em>0.45 N</em>

Explanation:

<em>Let Recall that,</em>

<em> The power formula is:  </em>

<em>   P = E²/R  </em>

Let A = the magnetic field  

<em>Let L = length of wire  = 9.00cm = 0.09 m  </em>

let  R = resistance of wire  = 0.320 Ω

let v =  velocity of the wire = 4 m/s  

<em>Let E = across the wire voltage </em>

Let P = the power of the wire = 4.3 W  

To Solve  for E:  

<em>The formula of E = √PR  </em>

The Voltage from  a magnetic field is given as,

E = vAL  

We therefore Use E = E

√PR = vAL  

to solve  for A,

A= √PR/vL  

BA= √4.3(0.32)/(4)(.09)  -=0.173

A = 0.173 wA/m²

Let F  be  the pulling force  

Let I  be the current in the wire  

P = I²R  

<em>I = √P/R  </em>

F = IAL  

F = √P/RAL  

F = √4.3/.32(0.173)(.09)  

<em>F = 0.45N</em>

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guajiro [1.7K]

Answer:

Vf = 60 [m/s]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f}= v_{i}+(a*t)

where:

Vf = final velocity [m/s]

Vi = initial velocity = 0

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4 0
4 years ago
A 15.5 kg block is pulled by two forces. The first is 11.8 N at a 53.7 angle and the second is 22.9 at a -15.8 angle. What is th
Ainat [17]

Answer:

1.88 m/s^2 at 6.5^{\circ}

Explanation:

We need to calculate the components of the resultant force on both the x (horizontal) and y (vertical) direction.

Components of the first force F1:

F_{1x} =(11.8) cos (53.7^{\circ})=7.0 N\\F_{1y} = (11.8) sin (53.7^{\circ})=9.5 N

Components of the second force F2:

F_{2x} =(22.9) cos (-15.8^{\circ})=22.0 N\\F_{2y} = (22.9) sin (-15.8^{\circ})=-6.2 N

So the components of the resultant force are

R_x = F_{1x}+F_{2x}=7.0+22.0 = 29.0 N\\R_y = F_{1y}+F_{2y} = 9.5+(-6.2)=3.3 N

So the magnitude of the resultant force is

F=\sqrt{(29.0)^2+(3.3)^2}=29.2 N

And the direction is

\theta = tan^{-1} (\frac{R_y}{R_x})=tan^{-1} (\frac{3.3}{29.0})=6.5^{\circ}

The magnitude of the acceleration can be found by using Newton's second law:

a=\frac{F}{m}=\frac{29.2 N}{15.5 kg}=1.88 m/s^2

while the direction is the same as the resultant force, 6.5^{\circ}.

5 0
3 years ago
Calculate the speed of light in glycerol (n = 1.47). Use the GRASS method, write your solution on paper and insert an image of i
Degger [83]

The velocity of light in glycerol is 2.04 × 10⁸ m/s.

To find the answer, we need to know about the expression of velocity of light in any medium.

<h3>What's the expression of the speed of light in any medium?</h3>
  • Mathematically, the speed of light in any medium= C/n
  • C = speed of light in vaccum, n = refractive index of the medium

<h3>What's the speed of light in glycerol having refractive index of 1.47?</h3>

Speed of light in glycerol= 3×10⁸/1.47 = 2.04 × 10⁸ m/s

Thus, we can conclude that the velocity of light in glycerol is 2.04 × 10⁸ m/s.

Learn more about the speed of light in any medium here:

brainly.com/question/18650050

#SPJ1

6 0
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