Two boxes are connected by a string. The first box has a mass of 4.2 kg and the

A tennis player swings her 1000 g racket with a speed of 12 m/s. she hits a 60 g tennis ball that was approaching her at a speed

tensa zangetsu [6.8K]

Answer:

should be the answer B

Explanation:

6 0

3 years ago

A student was walking with a mass of 40 kilograms and an acceleration of 1 m/s/s. What is the force exerted on her?

ValentinkaMS [17]

Answer:

B

Explanation:40x1=40 40=b

brainlyest plz

3 0

3 years ago

An extremely long wire laying parallel to the x -axis and passing through the origin carries a current of 250A running in the po

viktelen [127]

Answer:

$1.232\times 10^{-5}\ T.$

Explanation:

<u>Given:</u>

Current through the wire, passing through the origin, $I_1 = 250\ A.$

Current through the wire, passing through the y axis, $r_y=1.8\ m.$ , $I_2 = 50\ A.$

According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to $\mu_o$ times the net current threading the loop.

$\oint \vec B \cdot d\vec l=\mu_o I.$

In case of a circular loop, the directions of magnetic field and the line element $d\vec l$ , both are along the tangent of the loop at that point, therefore, $\vec B\cdot d\vec l = B\ dl$ .

$\oint \vec B \cdot d\vec l = \oint B\ dl = B\oint dl.\\$

$\oint dl$ is the circumference of the Amperian loop = $2\pi r$

Therefore,

$B\ 2\pi r=\mu_o I\\B=\dfrac{\mu_o I}{2\pi r}.$

It is the magnetic field due to a current carrying wire at a distance r from it.

For the first wire, passing through the origin:

Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_1 = 3.510\ m.$

The magnetic field at the given point due to this wire is given by:

$B_1 = \dfrac{\mu_o I_1}{2\pi r_1}\\=\dfrac{4\pi \times 10^{-7}\times 250}{2\pi \times 3.510}=1.42\times 10^{-5}\ T.$

For the first wire, passing through the y-axis:

Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_2 = 5.310\ m.$

The magnetic field at the given point due to this wire is given by:

$B_2 = \dfrac{\mu_o I_2}{2\pi r_2}\\=\dfrac{4\pi \times 10^{-7}\times 50}{2\pi \times 5.310}=1.88\times 10^{-6}\ T.$

The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.

Thus, the net magnetic field at $r_r=-3.510\ m$ is given by

$B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.$

6 0

3 years ago

What evidence would you observe if the law of conservation of mass is obeyed when mixing two solutions?

Marianna [84]

Answer:

You should have noticed that the number of atoms in the reactants is the same as the number of atoms in the product. The number of atoms is conserved during the reaction. However, you will also see that the number of molecules in the reactants and products are not the same.

3 0

2 years ago

You shine a flashlight on a pond. The light hits the pond at an angle of 70° to the normal of the pond. What is angle between th

Solnce55 [7]

Answer:

(d) 44.8°

Explanation:

For refraction,

Using Snell's law as:

$\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}$

Where,

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₁ is the refractive index of air which is 1

n₂ is the refractive index of water which is 4/3

Given that angle of incidence = 70°

So,

$\frac {sin\theta_2}{sin70}=\frac {1}{\frac {4}{3}}$

${sin\theta_2}=0.7048$

<u>Angle of refraction = sin⁻¹ 0.7048 = 44.8°.</u>

7 0

3 years ago