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3 years ago

What is the direction of the magnetic field b⃗ a at point a?

Physics

1 answer:

pogonyaev3 years ago

5 0

Answer: Use the Right -hand Rule to find the direction of the magnetic field.

Explanation: A magnetic force only occur, in a magnetic field, when a particle is moving perpendicularly relative to the magnetic field. So, all three quantities are perpendicular among themselves.

The Right-Hand Rule apply for the positive charges moving in conventional current. If the moving charge is negative, the velocity must be reversed.

Right-Hand Rule determines the direction of any or all of the three and it is as follows: With you right hand, point your index finger in the direction of the velocity, your middle finger in the direction of the magnetic field and your thumb will be pointing in the direction of the magnetic force.

Another way of using the rule is when there is a current carrying wire:

The fingers of your right hand will curl into a half circle around the wire, they will point in the direction of the field. The thumb will point in the direction of the conventional current.

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Which statement about astrology is not true?

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A. planets do not affect birth defects

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3 years ago

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The force F is expressed in terms of the mass “m” and acceleration “a” according to the

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Answer:

F = [M] × [L1 T-2] = M1 L1 T-2.

Explanation:

Therefore, Force is dimensionally represented as M1 L1 T-2.

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3 years ago

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.70 mm. A 21

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Answer:

12353 V m⁻¹ = 12.4 kV m⁻¹

Explanation:

Electric field between the plates of the parallel plate capacitor depends on the potential difference across the plates and their distance of separation.Potential difference across the plates V over the distance between the plates gives the electric field between the plates. Potential difference is the amount of work done per unit charge and is given here as 21 V. Electric field is the voltage over distance.

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3 years ago

A gray kangaroo can bound across level ground with each jump carrying it 8.7 from the takeoff point. Typically the kangaroo leav

oksano4ka [1.4K]

Answer:

a) The takeoff speed is 10 m/s.

b) The maximum height above the ground is 1.2 m.

Explanation:

The position of the kangaroo and its velocity at any given time "t" can be calculated by the following equations:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v =(v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t".

x0 = initial horizontal position.

v0 = initial velocity.

α = jumping angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity vector at time "t"

a) Please see the attached figure for a better understanding of the problem. In red is depicted the position vector at the final time (r final). The components of r final are known:

r final = (8.7 m, 0 m)

Then at final time:

8.7 m = x0 + v0 · t · cos α

0 m = y0 + v0 · t · sin α + 1/2 · g · t²

(notice in the figure that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0). Then:

8.7 m = v0 · t · cos α

Solving for "v0":

8.7 m /(t · cos α) = v0

Replacing v0 in the equation of the y-component, we can obtain the final time:

0 m = 8.7 m · tan 29° - 1/2 · 9.8 m/s² · t² (remember: sin α / cos α = tan α)

- 8.7 m · tan 29° / -4.9 m/s² = t²

t = 0.99 s

Now, we can calculate the initial speed:

8.7 m /t · cos α = v0

v0 = 8.7 m / (0.99 s · cos 29°)

The takeoff speed is 10 m/s

b) When the kangaroo is at its maximum height, the velocity vector is horizontal (see figure). That means that the y-component of the velocity at that time is 0:

0 = v0 · sin α + g · t

Solving for "t":

-v0 · sin α / g = t

t = - 10 m/s · sin 29° / 9.8 m/s²

t = 0.49 s

Notice that we could have halved the final time (0.99 s, calculated above) to obtain the time at which the kangaroo is at its maximum height. That´s because the trajectory is parabolic.

Now, let´s find the height of the kangaroo at that time:

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 10 m/s · 0.49 s · sin 29° - 1/2 · 9.8 m/s² · (0.49 s)²

The maximum height above the ground is 1.2 m.