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3 years ago

The region, or sphere, of influence around a magnet is called the magnetic_____.

Physics

2 answers:

Firdavs [7]3 years ago

7 0

Field* maybe... I might be wrong

vovikov84 [41]3 years ago

3 0

its called the magnetic field

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The revolution of the earth around the sun demonstrate what motion?

olga2289 [7]

Answer:

Anticlockwise directions

Please mark me Brainliest to help me

3 0

1 year ago

A car takes off from rest takes of from rest and covers a distance of 80m on a straight road in 10s.Calculate the magnitude of i

hodyreva [135]

Initial velocity (u) = 0 m/s [the car was at rest]
Distance (s) = 80 m
Time (t) = 10 s
Let the magnitude of acceleration be a.
By using the equation of motion, $s = ut + \frac{1}{2} a {t}^{2}$ we get, $80 = 0 \times 10 + \frac{1}{2} \times a \times {10}^{2} \\ = > 80 = \frac{1}{2} \times 100a \\ = > 80 = 50a \\ = > a = \frac{80}{50} \\ = > a = 1.6$

Answer:

The magnitude of its acceleration is 1.6 m/s^2.

Hope you could get an idea from here.

Doubt clarification - use comment section.

4 0

2 years ago

Help pls i need this right now

pantera1 [17]

Answer:

The x-component of $F_{3}$ is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

$\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O$ (1)

Where:

$\vec F_{1}$ , $\vec F_{2}$ , $\vec F_{3}$ - External forces exerted on the ring, measured in newtons.

$\vec O$ - Vector zero, measured in newtons.

If we know that $\vec F_{1} = (70.711,70.711)\,[N]$ , $\vec F_{2} = (-126.859, 46.173)\,[N]$ , $F_{3} = (F_{3,x},F_{3,y})$ and $\vec O = (0,0)\,[N]$ , then we construct the following system of linear equations:

$\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N$ (2)

$\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N$ (3)

The solution of this system is:

$F_{3,x} = 56.148\,N$ , $F_{3,y} = -116.884\,N$

The x-component of $F_{3}$ is 56.148 newtons.

5 0

2 years ago

A long, thin solenoid has 450 turns per meter and a radius of 1.06 . The current in the solenoid is increasing at a uniform rate

kirill115 [55]

Answer:9.34 A/s

Explanation:

Given

radius of solenoid $R=1.06 m$

Emf induced $E=8.50\times 10^{-6} V/m$

no of turns per meter n=450

we know Induced EMF is given by

$\int Edl=-\frac{\mathrm{d} \phi}{\mathrm{d} t}=-\frac{\mathrm{d} B}{\mathrm{d} t}A$

Magnetic Field is given by

$B=\mu _0ni$

thus $\frac{\mathrm{d} B}{\mathrm{d} t}=-\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}$

Area of cross-section

$A=\pi R^2$ where

solving integration we get

$E.\cdot 2\pi r=\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}\pi R^2$

where r=distance from axis

R=radius of Solenoid

$\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{Er}{\mu _0nR^2}$

$\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{8.50\times 10^{-6}\times 3.49\times 10^{-2}}{4\pi \times 10^{-7}\times 450\times 1.06^2}$

$\frac{\mathrm{d} i}{\mathrm{d} t}=9.34 A/s$

4 0

3 years ago

Evaluate A car company wants to build a wind-powered car that converts 100 percent of the mechanical energy in the wind to the m

lorasvet [3.4K]

Answer:

There's one or two reasons, depending on what is meant by "wind-powered car".

The first reason is that it's impossible for any transfer of energy to be 100% efficient. There will always be frictional losses.

Secondly, if the company means that they want to attach a wind turbine to the car so that the car is powered by the same wind that it generates, that violates the conservation of energy.

4 0

3 years ago