Ask question

Ask question

All categories

2 years ago

11

A 2.3 m -long wire carries a current of 8.6 A and is immersed within a uniform magnetic field B⃗ . When this wire lies along the

+x axis, a magnetic force F⃗ =(−2.0j)N acts on the wire, and when it lies on the +y axis, the force is F⃗ =(2.0i−4.8k)N.

1 answer:

Volgvan2 years ago

5 0

Answer:

Explanation:

Given that :

The current I = 8.6 A

The length vector L = 2.3 m i

The force vector is = -2.0 j N

When L = 2.3 m i ; the

Force vector F = (2.0 i - 4.8 k) N

Compute the components of the magnetic field as follows:

$F = I(L*B)$

Replacing 8.6 A for I ; -2.0 j N for F & 2.3 m i for L

$-2.0 j N = (8.6 A) (2.3 m i *(B_xi + B_yj+B_zk)$

$-2.0 j N = 19.78B_z \ and \ 0 = 19.78B_y$

$B_z = -0.1011 T \ and \ B_y = 0$

However in y direction ; we have :

$(2.0 i - 4.8 k) = 8,6 A (2.3 mi*(B_xi+B_yj+B_zk)$

$- 2.0 = 19.78 B_z \ and \ -4.8 = 19.78 B_x$

$B_z = -0.1011 T \ and \ B_x = -0.2427T$

Hence, the component of magnetic field is as follows:

$B_x = -.02427T \ ; B_y = 0 T \ ; B_z = - 0.1011 T)$

You might be interested in

What is the centripetal force that holds planets in orbit?

Fudgin [204]

<h2>Answer: Gravity force</h2>

If we approximate the orbit of the planets around the Sun to circular orbits with a uniform circular motion, where the velocity $\vec{V}$ is a vector, whose direction is perpendicular to the radius $r$ of the trajectory; the acceleration $\vec{a}$ is directed towards the center of the circumference (that's why it's called centripetal acceleration).

Now, according to Newton's 2nd law, the force $\vec{F}$ is directly proportional and in the same direction as the acceleration:

$\vec{F}=m.\vec{a}$

Therefore the net force resulting from the movement of a planet orbiting the Sun points towards the center of the circle, this is called Centripetal Force which is a central force that in this case is equal to the gravity force.

8 0

3 years ago

Two carts, one of mass 2m and one of mass m, approach each other with the same speed, v. When the carts collide, they

Svetradugi [14.3K]

Answer: Graph C is the correct option

Explanation:

The question is incomplete, please remember to submit the whole question :)

However, the rest of the question with its corresponding figures is below:

Assume that positive momentum is to the right, which of the following best represents the momentum of the cart of mass m as a function of time before and after the collision?

The initial momentum $p_{o}$ of the cart with mass $m$ (before the collision) is:

$p_{o}=-mV$ (1) Note the negative sign indicates the direction of cart's velocity (to the left, as seen in the first image attached)

On the other hand, the final momentum $p_{f}$ of both carts (after the inelastic collision) is:

$p_{f}=(2m+m) V_{f}=3m V_{f}$ (2)

So, according to this, the correct graph tha best represents the situation is C. Since before the collision the momentum is negative, then both carts slow down after the collision ( $V_{f}$ ), and taking into account the linear momentum is directly proportional to the velocity $p_{f}$ (although is in the positive direction) is less than $p_{o}$ .

6 0

3 years ago

The acceleration of a particle traveling along a straight line isa=14s1/2m/s2, wheresis in meters. Ifv= 0,s= 1 m whent= 0, deter

Yuliya22 [10]

Answer:

0.78m/s

Explanation:

We are given that

Acceleration= $a=\frac{1}{4}s^{\frac{1}{2}}m/s^2$

v=0, s=1 when t=0

We have to find the particle's velocity at s=2m

We know that

$a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=\frac{dv}{ds}v$

$vdv=ads$

$\int_{0}^{v} vdv=\int_{1}^{s}0.25s^{\frac{1}{2}}ds$

$\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2})^{s}_{1}$

By using formula: $\int x^ndx=\frac{x^{n+1}}{n+1}+C$

$\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2}}-1$

Substitute s=2

$\frac{v^2}{2}=\frac{0.50}{3}((2^{1.5})-1)$

$\frac{v^2}{2}=\frac{0.50}{3}\times 1.83$

$v^2=2\times 0.305=0.61$

$v=\sqrt{0.61}=0.78m/s$

Hence, the velocity of particle at s=2m=0.78m/s

3 0

3 years ago

4) Block A has a mass of 3kg and velocity of 13m/s, catching up with a second block B that has a mass of 3kg and is moving with

serious [3.7K]

Answer:

Option A is the correct answer.

Explanation:

Here momentum is conserved.

That is $\left (m_Av_A+m_Bv_B \right )_{initial}=\left (m_Av_A+m_Bv_B \right )_{final}$

Substituting values

$3\times 13+3\times 5=3v_A+3\times 8\\\\3v_A=39+15-24\\\\3v_A=30\\\\v_A=10m/s$

Speed of block A after collision = 10 m/s

Option A is the correct answer.

5 0

3 years ago

Is gravity a non-contact form?

sattari [20]

No, we cannot touch gravity nor can we physically see it. We can only see how it works.

8 0

2 years ago