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3 years ago

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which describes the magnetic field that forms when a wire carries an electric current? a. It is in a line perpendicular to the w

ire. b. It is in a line parallel with the wire. c. It is in a cylinder around the wire. d. It is in a cylinder across the wire.

2 answers:

frez [133]3 years ago

5 0

C. the magnetic field is shaped like a cylinder through which the wire is positioned along the axis of symmetry.

True [87]3 years ago

4 0

Answer:

Option (c)

Explanation:

According to the Maxwell's right hand rule, If we hold a current carrying conductor in our right hand, such that the thumb indicates the direction of current flowing in the conductor then the curling of fingers gives the direction of magnetic field around the conductor.

So, the magnetic field around the conductor is in cylindrical form around the wire.

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10m/s2 because it travels 10 meters every second

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3 years ago

Work is done on an object when

liubo4ka [24]

Answer:

Incomplete question but the closest ans is

A) The displacement is not zero

Explanation:

For work to be done, the direction of force has to be parallel to or in the same direction as the displacement, this can then be calculated by

$W=Fs \ cos$ θ

Where θ is the angle between direction of force and direction of displacement

By this equation,

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3 years ago

3.1 * Consider a gun of mass M (when unloaded) that fires a shell of mass m with muzzle speed v. (That is, the shell's speed rel

kramer

Answer:

$v_s=\frac{v}{1+\frac{m}{M} }$

Explanation:

Let the shells speed with respect to ground be $v_s$

Let the shells speed with respect to ground be $v_g$

mass of shell, $m$

mass of gun, $M$

The relative velocity of shell with respect to a free unconstrained gun, $v$

According to the Newton's third law of motion the direction of the velocity of gun and shell will be in the opposite direction.

So, the relation between the relative velocity and their individual velocity will be:

$v=v_s+v_g$ ......................(1)

<u>And according to the conservation of momentum (as the condition is very close to the elastic collision):</u>

$M.v_g-m.v_s=0$

substitute the value of $v_g$ from equation (1)

$M\times (v-v_s)=m\times v_s$

$M.v-M.v_s=m.v_s$

$M.v=M.v_s+m.v_s$

$v_s=\frac{M.v}{(M+m)}$

$v_s=\frac{v}{(\frac{(M+m)}{M}) }$

$v_s=\frac{v}{1+\frac{m}{M} }$

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3 years ago

Positions B and D are 11.3 meters and 1.9 meters above the playing field, respectively. If the ball had a speed of 6.2 m/s in po

Andreyy89

Answer:

Speed of the ball in position D $=14.92 \frac{m}{s}$

<u>Explanation:</u>

Given:

Position of B=11.3 $\text { meters }$

Position of D=1.9 $\text { meters }$

Velocity of B=6.2 $\text { meters }$

To Find:

Velocity of D

Solution:

According to the formula, Velocity is given as

$V d=\sqrt{\left[V b^{2}+(2 \times g \times d y)\right]}$

$V b$ =Velocity of B

$V d$ =Velocity of D

g=acceleration due to gravity=9.8 m/s^2

$d y$ =Change in position of B and D

Substitute the all values in the above equation we get

$d y=11.3-1.9$

$V d=\sqrt{\left[6.2^{2}+(2 \times 9.8 \times(11.3-1.9))\right]}$

$=\sqrt{[38.44+(2 \times 9.8 \times(9.4))]}$

$=\sqrt{[38.44+(19.6 \times 9.4)]}$

$=\sqrt{38.44+186.24}$

$=\sqrt{222.68}$

$=14.92 \frac{m}{s}$

Result :

The velocity of D is $=14.92 \frac{m}{s}$

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3 years ago