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vesna_86 [32]
2 years ago
13

Which combination of units can be used to express the magnetic field?

Physics
1 answer:
Zolol [24]2 years ago
8 0

Answer:

The magnetic field unit in the International System is the tesla (T). A tesla is defined as the magnetic field that exerts a force of 1 N (newton) on a load of 1 C (coulomb) that moves at a speed of 1 m / s within the field and perpendicular to the field lines

Explanation:

Magnetic induction or magnetic flux density (B), is the magnetic flux that causes a diffusion charge in motion for each unit of normal area to the direction of the flow. It is also called the magnetic field strength.

The unit of magnetic flux density in the International System of Units is the tesla (T).

The tesla (symbol T), is the magnetic induction unit (or magnetic flux density) of the International System of Units (SI). It is defined as a uniform magnetic induction that, normally distributed over a surface of one square meter, produces through this surface a total magnetic flux of a weber.

<u>Equivalences: </u>

1 T = 1 Wb · m-2 = 1 kg · s-2 · A-1 = 1 kg · C-1 · s-1

A Tesla is also defined as the induction of a magnetic field that exerts a force of 1 N (newton) on a load of 1 C (coulomb) that moves at a speed of 1 m / s within the field and perpendicular to the lines of magnetic induction.

1 T = 1 N · s · m-1 · C-1

Basic Unit in the Cegesimal System of Units (CGS): Gauss (G)

A gauss (G) is a magnetic field unit of the Cegesimal System of Units (CGS). A gauss (G) is defined as a maxwell per square centimeter.

1 gauss = 1 maxwell / cm2

A gauss is equivalent to 10-4 tesla:

1 T = 10,000 G

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Answer:

The expression of gravitational field due to mass m_! at a distance l_a

Explanation:

We have given mass is m_1

Distance of the point where we have to find the gravitational field is l_a

Gravitational constant G

We have to find the gravitational filed

Gravitational field is given by g=\frac{Gm_1}{l_a^2}

This will be the expression of gravitational field due to mass m_! at a distance l_a

4 0
1 year ago
An electric winch is used to raise a 40-kg package and a 10-kg package vertically up the side of a building as pictured in the d
just olya [345]

Answer:

Explanation:

40 divided by 10 then which would equal 4. Add the 1.0 , 2 ,and 15 together. Then multply the 60 by 18.0 after you are done dividing the answer is 3 with a remainder of 6.

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2 years ago
It requires 0.30 kJ of work to fully drive a stake into the ground. If the average resistive force on the stake by the ground is
LenaWriter [7]

Answer:

Length of the stake will be 0.3623 m

Explanation:

We have given energy required to fully drive a stake into ground = 0.30 KJ = 300 J

Average resistive force acting on the floor is equal to F = 828 N

We have to find the length of the stake

We know that work done is given by

W = Fd, here W is work done , F is average force and d is the length of the stake

So 300 = 828×d

d = 0.3623 m

So length of the stake will be 0.3623 m

6 0
2 years ago
A particle has a velocity of v→(t)=5.0ti^+t2j^−2.0t3k^m/s.
Makovka662 [10]

Answer:

a)a=5 i+2t j - 6\ t^2k

b)a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2

Explanation:

Given that

v(t) = 5 t i + t² j - 2 t³ k

We know that acceleration a is given as

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=5 i+2t j - 6\ t^2k

a=5 i+2t j - 6\ t^2k

Therefore the acceleration function a will be

a=5 i+2t j - 6\ t^2k

The acceleration at t = 2 s

a= 5 i + 2 x 2 j - 6 x 2² k  m/s²

a=5 i + 4 j -24 k m/s²

The magnitude of the acceleration will be

a=\sqrt{5^2+4^2+24^2}\ m/s^2

a= 24.83 m/s²

The direction of the acceleration a is given as

a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2

a)a=5 i+2t j - 6\ t^2k

b)a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2

5 0
2 years ago
An electron moving at right angles to a 0.1 T magnetic field experiences an acceleration of 6 × 1015 m.s-2. What is the speed of
GaryK [48]

Explanation:

It is given that,

Magnetic field, B = 0.1 T

Acceleration, a=6\times 10^{15}\ m/s^2

Charge on electron, q=1.6\times 10^{-19}\ C    

Mass of electron, m=9.1\times 10^{-31}\ kg    

(a) The force acting on the electron when it is accelerated is, F = ma

The force acting on the electron when it is in magnetic field, F=qvB\ sin\theta

Here, \theta=90

So, ma=qvB

Where

v is the velocity of the electron

B is the magnetic field

v=\dfrac{ma}{qB}

v=\dfrac{9.1\times 10^{-31}\ kg\times 6\times 10^{15}\ m/s^2}{1.6\times 10^{-19}\ C\times 0.1\ T}

v = 341250  m/s

or

v=3.41\times 10^5\ m/s

So, the speed of the electron is 3.41\times 10^5\ m/s

(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.

7 0
2 years ago
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