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3 years ago

Hi guys this is my last points and i need some answers :)

Physics

1 answer:

Zinaida [17]3 years ago

6 0

Answer:

Number 1: B Number 2: D

Explanation:

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A single-turn current loop, carrying a current of 4.03 A, is in the shape of a right triangle with sides 68.1, 151, and 166 cm.

SashulF [63]

Answer:

Part a)

$F = 0$

Part b)

$F = 0.25 N$

Part c)

$F = 0.25 N$

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

$F_{net} = 0$

Explanation:

As we know that force on a current carrying wire is given as

$\vec F = i(\vec L \times \vec B)$

now we have

Part a)

current in side 166 cm and magnetic field is parallel

so we have

$F = i(\vec L \times \vec B)$

here we know that L and B is parallel to each other so

$F = 0$

Part b)

For 68.1 cm length wire we have

$F = iLB sin\theta$

here we know that

$cos\theta = \frac{68.1}{166}$

$\theta = 65.8$

so we have

$F = (4.03)(0.681)(99.3 \times 10^{-3})sin65.8$

$F = 0.25 N$

Part c)

For 151 cm length wire we have

$F = iLB sin\phi$

here we know that

$cos\phi = \frac{151}{166}$

$\theta = 24.5$

so we have

$F = (4.03)(1.51)(99.3 \times 10^{-3})sin24.5$

$F = 0.25 N$

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

$F_{net} = 0$

4 0

2 years ago

When will electric charges flow?

Norma-Jean [14]

C. when the circuit is closed

7 0

2 years ago

Read 2 more answers

The main component of all computer memory is

never [62]

Hi!

The main component of all computer memory is RAM.

Hope this helps !

4 0

2 years ago

Read 2 more answers

A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the

irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

$x=V_{o} cos \theta t$ (1)

$y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2}$ (2)

Where:

$x$ is the horizontal distance between the cannon and the ball

$V_{o}=23 m/s$ is the cannonball initial velocity

$\theta=0\°$ since the cannonball was shoot horizontally

$t$ is the time

$y=0$ is the final height of the cannonball

$y_{o}=1.96 m$ is the initial height of the cannonball

$g=9.8 m/s^{2}$ is the acceleration due gravity

Isolating $t$ from (2):

$t=\sqrt{-\frac{2(y-y_{o})}{g}}$ (3)

$t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}}$ (4)

$t=0.63 s$ (5)

Substituting (5) in (1):

$x=(23 m/s) cos(0\°) 0.63 s$ (6)

Finally:

$x=14.49 m$

5 0

2 years ago

1. Which object is farthest from the origin at t=2sec

Stolb23 [73]

Answer:

that one i know only pe not that sorry again

6 0

2 years ago