You are skydiving and accelerating at 9.8 meters per second

A charged particle of mass 0.0050 kg is subjected to a 5.0 T magnetic field which acts at a right angle to its motion. If the pa

Irina18 [472]

Answer:

0.01 C

Explanation:

Applying,

F = qvBsinФ................ Equation 1

Where F = Force on the charged particle, q = charge on the particle, v = velocity, B = magnetic field, Ф = angle

Since the charged particle noves in a circle,

F = mv²/r................. Equation 2

Where m = mass of the particle, v = velocity of the particle, r = radius of the circle

Substitute equation 2 into equation 1

mv²/r = qvBsinФ

make q the subject of the equation

q = mv/(rBsinФ)............. Equation 3

Given: m = 0.005 kg, v = 2 m/s, r = 0.2 m, B = 5 T, Ф = 90° (Act at right angle)

Substitute these values into equation 3

q = (0.005×2)/(0.2×5×sin90°)

q = 0.01/(1)

q = 0.01 C

5 0

3 years ago

How should the magnetic field lines be drawn for the magnets shown below?

sergeinik [125]

Option B is the correct answer that show how magnetic field lines should be drawn for the magnets shown in the figure.

<h3>What is Magnetic Line of Force ?</h3>

The Magnetic Line of Force of a magnet is defined as the line along which a free N - pole would tend to move if placed in the field of a line such that the tangent to it at any point gives the direction of the field at that point.

When the two unlike poles are placed to each other, there will be attraction. And when the two like poles are placed to each other, there will be repulsion. The reason is that the line of force tend to move from the north pole to the south pole.

From the given diagram, the two magnets are of the same south pole. They are of like pole and there will be repulsion between the two magnets.

Therefore, Option B is the correct answer that show how magnetic field lines should be drawn for the magnets shown in the figure.

Learn more about Magnetic Field Lines here: brainly.com/question/17011493

#SPJ1

5 0

2 years ago

Which equation is correct according to Ohm’s law?

Korvikt [17]

You didn't give us anything other than the question. No options are provided so I cannot answer. Nobody can.

4 0

3 years ago

Read 2 more answers

6A certain load and set of slings create a 20-degree angle between the load and each sling leg. Using a spreader for the same li

Otrada [13]

Solution:

The angle between the sling and the load is $20^{\circ}$

So the tension in each sling can be calculated as

$Sin \theta = Mg => T = \frac{Mg}{2Sin\theta}$

$Sin \theta=> \frac{Mg}{2Sin 20^{\circ}}$

Where

M is the mass of the load

The Horizontal reaction on the sling will be inward.

After using the spreader, the new angle between sling and load is $60^{\circ}$ , the tension in the sling will be

$T= \frac{Mg}{2 Sin 60^{\circ}}$ = $\frac{Mg}{2 Sin 20^{\circ}}$

The tension will be same as before in the sling move away through the spreader at an angle more than 90 degree the horizontal force will act opposite and will be outward

5 0

3 years ago

As an aid in working this problem, consult Concept Simulation 11.1. The volume flow rate in an artery supplying the brain is 3.5

klemol [59]

Answer

given,

flow from the artery = 3.5 x 10⁻⁶ m³/s

Radius of artery = 5.80 x 10⁻³m

area = π R²

= π x (5.8 x 10⁻³)²

= 1.06 x 10⁻⁴ m²

$v = \dfrac{Q}{A}$

$v = \dfrac{3.5 \times 10^{-6}}{1.06 \times 10^{-4}}$

v = 0.033 m/s

b) new velocity of flow

Radius = R' = R/4

A V = A' V'

R² V = R'² V'

R^2 V = (\dfrac{R}{4})^2 V'

V' = 16 V

V' = 16 x 0.033

V' =0.528 m/s

5 0

3 years ago