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3 years ago

What is the largest possible magnitude of the acceleration of the electron due to the magnetic field?

1 answer:

6 0

Thank you for posting your question here at brainly. But your question seems incomplete. I will assume you based the situation below:

An electrons moves at 2.0x10^6 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.4x10^-2 T.

The largest possible magnitude of the acceleration of the electron due to the magnetic field is = 2.6 × 10 ¹⁶ m / s ²

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True or False: Chemical energy stored in food cannot be transformed into mechanical energy

Novosadov [1.4K]

The answer to this question is false.

7 0

3 years ago

Read 2 more answers

Two blocks, joined by a string, have masses of 6.0 and 9.0 kg. They rest on a frictionless, horizontal surface. A second string,

Lynna [10]

Answer:

T= 27 N

Explanation:

Assuming that the string joining both masses is massless and inextensible, both masses accelerate at the same rate.

So, we can treat to both masses as a single system, and apply Newton's 2nd Law to both masses.

In this way, we can get the value of the acceleration without taking into account the tension in the string, as it is an internal force (actually a action-reaction pair).

Newton's 2nd law is a vector equation, so we can decompose the forces along perpendicular axis in order to convert it in two algebraic equations.

We can choose one axis as parallel to the horizontal surface (we call it x-axis, being the positive direction the one of the movement of the blocks due to the horizontal force applied to the 6.0 kg block), and the other, perpendicular to it, so it is vertical (we call y-axis, being the upward direction the positive one).

Taking into account the forces acting on both masses, we can write both equations as follows:

Fy = N- (m₁+m₂)*g = 0 (as there is no movement in the vertical direction)

Fx = Fh = (m₁ + m₂) * a ⇒ 45 N = 15.0 kg * a

⇒ a = 45 N / 15.0 kg = 3 m/s²

Now, in order to get the value of the tension T, we can choose as our system, to any mass, and apply Newton's 2nd Law again.

If we choose to the mass of 6.0 kg, in the horizontal direction, there are two forces acting on it, in opposite directions: the horizontal applied force of 45 N, and the tension in the string that join both masses.

The difference of both forces, must be equal to the mass (of this block only) times the acceleration, as follows:

F- T = m₂* a ⇒ 45 N - T = 6.0 kg * 3 m/s²

⇒ T = 45 N -18 N = 27 N

We could have arrived to the same result taking the 9.0 Kg as our system, as the only force acting in the horizontal direction is just the tension in the string that we are trying to find out, as follows:

F = m₁*a = 9.0 kg* 3 m/s² = 27 N

4 0

3 years ago

If a person consumes an extra 500 calories per day, how long would it take before he or she gained one pound of fat?

Ostrovityanka [42]

The answer is A. 10 days

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3 years ago

8. Three grams of Bismuth-218 decay to 0.375 grams in one hour. What is the half-

Evgen [1.6K]

Answer: 0.333 h

Explanation:

This problem can be solved using the Radioactive Half Life Formula:

$A=A_{o}.2^{\frac{-t}{H}}$ (1)

Where:

$A=0.375 g$ is the final amount of the material

$A_{o}=3 g$ is the initial amount of the material

$t=1 h$ is the time elapsed

$H$ is the half life of the material (the quantity we are asked to find)

Knowing this, let's substitute the values and find $h$ from (1):

$0.375 g=(3 g)2^{\frac{-1h}{H}}$ (2)

$\frac{0.375 g}{3 g}=2^{\frac{-1h}{H}}$ (3)

Applying natural logarithm in both sides:

$ln(\frac{0.375 g}{3 g})=ln(2^{\frac{-1 h}{H}})$ (4)

$-2.079=-\frac{1 h}{H}ln(2)$ (5)

Clearing $H$ :

$H=\frac{-1h}{-2.079}(0.693)$ (6)

Finally:

$h=0.333 h$ This is the half-life of the Bismuth-218 isotope

4 0

3 years ago

Equipotential surfaces a) make an angle of 45 degrees with the electric field. b) are parallel to the electric field. c) are per

Schach [20]

Answer:

Option c) are perpendicular to the electric field

Explanation:

Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90 $^{\circ}$ to the equipotential surface.

Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.

Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.

7 0

3 years ago