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2 years ago

5

. A 2.0-m wire carries a current of 15 A directed along the positive x axis in a region where the magnetic field is uniform and

given by B = (30 − 40 ) mT. What is the resulting magnetic force on the wire

1 answer:

Lady bird [3.3K]2 years ago

6 0

Answer:

The resulting magnetic force on the wire is -1.2kN

Explanation:

The magnetic force on a current carrying wire of length 'L' with current 'I' in a magnetic field B is

F = I (L*B)

Finding (L * B) , where L = (2, 0, 0)m , B = (30, -40, 0)

L x B = $\left[\begin{array}{ccc}i&j&k\\2&0&0\\30&-40&0\end{array}\right]$ = (0, 0, -80)

we can now solve

F = I (L x B) = I (-80)

F = -1200 kmN

F = -1200 kN * 10⁻³

F = -1.2kN

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State one advantage and disadvantage of friction

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Answer:

Advantage: Helps us to walk without slipping.
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Disadvantage:

We slip on a wet floor.
Causes the smoothness of tires and smoothness of shoes soul.

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2 years ago

flat sheet is in the shape of a rectangle with sides of lengths 0.400 mm and 0.600 mm. The sheet is immersed in a uniform electr

Ksenya-84 [330]

Answer:

$6.29591\times 10^{-6}\ N/C^2$

Explanation:

Flux is given by

$\phi=EAcos\theta$

A = Area

$A=0.4\times 10^{-3}\times 0.6\times 10^{-3}$

E = Electric field = 76.7 N/C

Angle is given by

$\theta=90-20\\\Rightarrow \theta=70^{\circ}$

$\phi=76.7\times 0.4\times 10^{-3}\times 0.6\times 10^{-3}\times cos70\\\Rightarrow \phi=6.29591\times 10^{-6}\ N/C^2$

The flux through the sheet is $6.29591\times 10^{-6}\ N/C^2$

6 0

2 years ago

In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w

Daniel [21]

Answer:

$v_2=-133.17m/s$ , the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before ( $p_i$ ) and after ( $p_f$ ) the explosion. We will take the east direction as positive.

Before the explosion we have $p_i=m_iv_i=Mv_i$ .

After the explosion we have pieces 1 and 2, so $p_f=m_1v_1+m_2v_2$ .

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

$Mv_i=m_1v_1+m_2v_2$

Which means (since we want $v_2$ and $M=m_1+m_2$ ):

$v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}$

So for our values we have:

$v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s$

5 0

2 years ago

Two identical charges q placed 2.0 mapart exert forces of magnitude 4.0 N on each other What is the value of the charge q? a) q

katen-ka-za [31]

Answer:

c) $4.2*10^{-5}C$

Explanation:

Coulomb's law says that the force exerted between two charges is inversely proportional to the square of distance between them, and is given by the expression:

$F=\frac{kq_{1}q_{2}}{r^{2}}$

where k is a proportionality constant with the value $k=9*10^{9}\frac{Nm^{2}}{C^{2}}$

In this case $q_{1}=q_{2}=q$ , so we have:

$F=\frac{kq^{2}}{r^{2}}$

Solving the equation for q, we have:

$kq^{2}=Fr^{2}$

$q^{2}=\frac{Fr^{2}}{k}$

$q=\sqrt{\frac{Fr^{2}}{k}}$

Replacing the given values:

$q=\sqrt{\frac{4.0N*(2.0m)^{2}}{9*10^{-9}\frac{Nm^{2}}{C^{2}}}}$

$q=4.2*10^{-5}C$

3 0

3 years ago

A car is traveling north at 17.7 m/s . After 6 it’s velocity is 141 in the same direction. Find the magnitude and direction of t

Furkat [3]

By equation of motion we have v = u + at

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On substitution we will get

141 = 17.7+ 6a

So, a = (141-17.7)/6 = 20. 55 m/ $s^{2}$

Aceeleration = 20. 55 m/ $s^{2}$ along north direction.

3 0

3 years ago