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1 year ago

what is the force per meter on a straight wire carrying 5.0 a when it is placed in a magnetic field of 0.020 t

Physics

1 answer:

Dvinal [7]1 year ago

3 0

The Force per meter on a straight wire carrying current in a magnetic field is 0.045 N/m.

Calculation:-

F/ℓ = B I sin θ

Where B – Magnetic field = 0.02 T I – Current = 5 A

Substituting the values

F/ℓ = (0.02) (5) (sin 27 deg)

F/ℓ = 0.045 N/m

A force is an influence that can alternate the motion of an item. A force can cause an item with mass to trade its pace, i.e., to boost up. force can also be described intuitively as a push or a pull. A pressure has both value and course, making it a vector quantity.

The push or pull on an item with mass causes it to change its velocity. force is an external agent capable of converting a frame's nation of relaxation or motion. It has significance and a path. A force is a push or pulls among gadgets. it is called an interplay because if one object acts on some other, its movement is matched with the aid of a reaction from the alternative object.

Learn more about force here:-brainly.com/question/12970081

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3 years ago

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A car is traveling at 39.7 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s 2 . If the coefficient of fricti

777dan777 [17]

Answer:

The minimum distance in which the car will stop is

x=167.38m

Explanation:

$39.7\frac{mi}{h}*\frac{1km}{0.621371mi}*\frac{1000m}{1km}*\frac{1h}{3600s}=17.747\frac{m}{s}$

∑F=m*a

∑F=u*m*g

The force of friction is the same value but in different direction of the force moving the car so it can stop so

$F=m*a\\a=\frac{F}{m}\\a=\frac{u*m*g}{m}\\a=u*g\\a=0.096*-9.8\frac{m}{s^{2} }$

$a=-0.9408 \frac{m}{s^{2}}$

$v_{f}^{2}=v_{o}^{2}+2*a*(x_{f}-x_{o})\\v_{f}=0 \\x_{o}=0\\0=v_{o}^{2}+2*a*x_{f}\\x_{f}=\frac{v_{o}^{2}}{2*a} \\x_{f}=\frac{(-17.747\frac{m}{s})^{2}}{2*(-0.9408)} \\x_{f}=167.38m$

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3 years ago

The distance between adjacent nodes in a standing wave pattern in a length of string is 25.0 cm:A. What is the wavelength of wav

mina [271]

A) 50 cm

B) 10000 cm/s

Explanation

Step 1

If you know the distance between nodes and antinodes then use this equation:

$\begin{gathered} \frac{\lambda}{2}=D \\ \text{where}\lambda\text{ is the wavelength} \\ D\text{ is the distance betw}een\text{ nodes} \end{gathered}$

then, let