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3 years ago

12

When an electric current is flowing through a wire, the force deflecting the charged particles is the greatest when the wire is

what to the magnetic field?

2 answers:

Makovka662 [10]3 years ago

5 0

The force is greatest when the wire is perpendicular to the magnetic field.

In fact, the expression for the magnetic force is
$F=ILB \sin \theta$
where I is the current, L the length of the wire, B the magnetic field intensity and $\theta$ is the angle between the directions of I and B.

We can see that F is maximum when $\sin \theta=1$ , so when $\theta = 90^{\circ}$ , so when the wire and the magnetic field are perpendicular to each other.

zzz [600]3 years ago

3 0

Answer:

C.) perpendicular

Explanation:

A particle with an electric charge experiences the maximum deflecting force when it is positioned perpendicular to the magnetic field.

You might be interested in

The 8.00-cm long second hand on a watch rotates smoothly.

dolphi86 [110]

The watch hand covers an angular displacement of 2π radians in 60 seconds.

ω = 2π/60
ω = 0.1 rad/s

v = ωr
v = 0.1 x 0.08
v = 8 x 10⁻³ m/s

4 0

3 years ago

Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, th

alex41 [277]

Answer: 321 J

Explanation:

Given

Mass of the box $m=3\ kg$

Force applied is $F=25\ N$

Displacement of the box is $s=15\ m$

Velocity acquired by the box is $v=6\ m/s$

acceleration associated with it is $a=\dfrac{F}{m}$

$\Rightarrow a=\dfrac{25}{3}\ m/s^2$

Work done by force is $W=F\cdot s$

$W=25\times 15\\W=375\ J$

change in kinetic energy is $\Delta K$

$\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J$

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy

$\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J$

Therefore, the magnitude of work done by friction is $321\ J$

3 0

3 years ago

You and a friend each carry a 15 kg suitcase up two flights of stairs, walking at a constant speed. Take each suitcase to be the

AlekseyPX

Answer:

Both of you did the same work but you expended more power.

Explanation:

<em>Work done</em> by an object is calculated by force applied multiplied by the distance.

W=F*d

From the figure given below let us calculate force applied bith you and yopur friend.

Let us take the stairs in positive x direction,

Work done by you W₁ ,

The force applied Fₓ = F - mgsinθ =maₓ

here aₓ = 0, because both of you move with constant speed

F - mgsinθ = 0

F= mgsinθ

The work done by you on the suitcase is

W = F L cos0° , where L is he length of the staircase.

W = FL = mgsinθL , by substituting value of F

Work done by you is W₁ = mgLsinθ

Similarly work done by your friend is W₂ = mgLsinθ.

Because both of you carry suitcase of same weight and in staircase is in same angle the force applied is same .

Therefore <em>work done by both of you is same</em> . Both of you did equal work.

The power , is defined as amount of energy converted or transfered per second or rate at which work is done .

P = $\frac{W}{t}$ = $\frac{FL}{t}$

Power spend by you P₁ = mgLsinθ/t

P₁ = 15*9.8*Lsinθ/30

P₁ = 4.9L sinθ eqn 1

Power spend by your friend is P₂ = mgLsinθ/t

P₂ =15*9.8*Lsinθ/60

P₂ = 2.45Lsinθ eqn 2

Dividing eqn 1 and eqn 2

P₁ = 2P₂

You have spend more power than your friend .

Hence Both of you did equal work but you spend more power.

7 0

3 years ago

Physical data is often used in the court system. In fact, police officers use radar to determine your speed when you are driving

Mashcka [7]

The driver is telling the truth, the radar gun must have been set incorrectly to record relative velocity.

The given parameters:

<em>Speed of the driver observed by the stationary police officer, Vo = 44.7 m/s</em>
<em>Speed of the driver, V = 26.8 m/s.</em>
<em>Speed limit = 60 mph</em>

The speed limit of the driver in meter per second is calculated as follows;

$= 60 \ \times \frac{miles}{hour} \times \frac{1609.34 \ m}{1 \ mile} \times \frac{1 \ hr}{3600 \ s}\\\\= 26.82 \ m/s$

From the speed limit, it is obvious that the driver's speed is within the limit. Thus, we can conclude that the driver is telling the truth, the radar gun must have been set incorrectly to record relative velocity.

Learn more about relative velocity here: brainly.com/question/17228388

6 0

2 years ago

A tennis ball is thrown vertically upward with an initial velocity of +6.2 m/s. What will the ball’s velocity be when it returns

anastassius [24]

Answer:

$v_{f}=-6.2 m/s$

Explanation:

The ball will rise decreasing its speed until it reaches the highest point where its speed will be zero. From this point the tennis ball will begin to fall again, in the free fall the tennis ball will gain speed but now in the opposite direction. When it returns to the same point where it was launched, its speed will be the same as the one that was launched but with the opposite sign.

$v_{f}=-6.2 m/s$

We can check this using the equation:

$v_{f}^2=v_{i}^2+2gh$

where $v_{i}=+6.2 m/s$

ang h is the height, but because the ball returns to the same point where it started, h =0

then

$v_{f}^2=v_{i}^2$

$v_{f}=v_{i}$

the initial and final velocity will be the same in number, but we know that the ball is going in the opposite direction, so the final velocity must have the opposite sign from the initial velocity

so if $v_{i}=+6.2 m/s$ ,

$v_{f}=-6.2 m/s$

4 0

3 years ago