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pashok25 [27]
4 years ago
15

An electron is moving directly toward you in a horizontal path when it suddenly enters a uniform magnetic field that is either v

ertical or horizontal.
If the electron begins to curve upward in its motion just after it enters the field, you can conclude that the direction of the magnetic field is
If the electron begins to curve upward in its motion just after it enters the field, you can conclude that the direction of the magnetic field is
upward.
downward.
to your left.
to your right.
Physics
1 answer:
blagie [28]4 years ago
3 0

Answer:

To your left

Explanation:

The direction of the force exerted on charged particle due to a magnetic field is given by the right-hand-rule, where:

- The index finger indicates the direction of motion of the electron

- the middle finger gives the direction of the magnetic field

- the thumb gives the direction of the force if the particle is positively charged - otherwise, the direction is reversed

in this case, we have an electron (so, a negatively charged particle):

- The direction of motion (index finger) is horizontal, toward you

- The electron begins to curve upward as it enters the field, so this means that the force exerted on the electrons is upward --> the thumb must point downward (because the electron is negatively charged)

- The index finger gives us the direction of the magnetic field: therefore, to your left.

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3 years ago
Students are using the experimental setup shown in the image in which the two ends of a string are attached to a car and to a ha
AlekseyPX

The Newton's second law and kinematics allows to find the result of the error when placing the masses in the car is:

  • Acceleration decreases.
  • The plot of position vs. time squared is still linear, but the slope is less.

     

Newton's second law gives a relationship between the net force, mass and acceleration of the body

      F = ma

      a = \frac{F}{m}

Where F is the force, m the mass and the acceleration

 

Kinematics studies the motion of bodies, looking for relationships between position, velocity, and acceleration.

         x = v₀ t + ½ a t²

In the experiment, the position and time of the car are measured, starting from rest in each experiment.

        x = ½ a t²

Let's substitute

      x =  ½ ( \frac{F}{m} ) t²

This is the equation that the students should graph, a graph of the position versus time squared can be made to obtain a line.

The slope of the line is the acceleration of the car, which is related to the force that is the weight of the hanging mass.

When the student makes the mistake of placing the mass on the car, the acceleration decreases therefore in a graph the position values ​​are smaller for each time.

The linearity of the graph is maintained, but when calculating the slope it gives lower values.

In conclusion using Newton's second law and kinematics we can find the result of the error when placing the masses on the car is:

  • Acceleration decreases.
  • The plot of position vs. time squared is still linear, but the slope is less.

Learn more here: brainly.com/question/11298125

7 0
3 years ago
Read 2 more answers
In a "worst-case" design scenario, a 2000-{\rm kg} elevator with broken cables is falling at 4.00{\rm m/s} when it first contact
OLga [1]

Answer:

v=3.73m/s,   a=3.35m/s^{2}

Explanation:

In order to solve this problem, we must first draw a diagram that will represent the situation (See attached picture).

So there are two ways in which we can solve this proble. One by using integrls and the other by using energy analysis. I'll do it by analyzing the energy of the system.

So first, we ned to know what the constant of the spring is, so we can find it by analyzing the energy on points A and C. So we get the following:

K_{A}+U_{Ag}=K_{C}+U_{Cs}+U_{Cg}+W_{Cf}

where K represents kinetic energy, U represents potential energy and W represents work.

We know that the kinetic energy in C will be zero because its speed is supposed to be zero. We also know the potential energy due to the gravity in C is also zero because we are at a height of 0m. So we can simplify the equation to get:

K_{A}+U_{Ag}=U_{Cs}+W_{Cf}

so now we can use the respective formulas for kinetic energy and potential energy, so we get:

\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}ky_{C}+fy_{C}

so we can now solve this for k, which is the constant of the spring.

k=\frac{mV_{A}^{2}+mgh_{A}-fy_C}{y_{C}^{2}}

and now we can substitute the respective data:

k=\frac{(2000kg)(4m/s)^{2}+(2000kg)(9.8m/s^{2})(2m)-(17000N)(2m)}{(2m)^{2}}

Which yields:

k= 9300N/m

Once we got the constant of the spring, we can now find the speed of the elevator at point B, which is one meter below the contact point, so we do the same analysis of energies, like this:

K_{A}+U_{Ag}=K_{B}+U_{Bs}+U_{Bg}+W_{Bf}

In this case ther are no zero values to eliminate so we work with all the types of energies involved in the equation, so we get:

\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}mV_{B}^{2}+\frac{1}{2}ky_{B}^{2}+mgh_{B}+fy_{B}

and we can solve for the Velocity in B, so we get:

V_{B}=\sqrt{\frac{mV_{A}^{2}-ky_{B}^{2}+2mgh_{B}-2fy_{B}}{m}}

we can now input all the data directly, so we get:

V_{B}=\sqrt{\frac{(2000kg)(4m/s)^{2}-(9300N/m)(1m)^{2}+2(2000kg)(9.8m/s^{2})(1m)-2(17000N)(1m)}{(2000kg)}}

which yields:

V_{B}=3.73m/s

which is the first answer.

Now, for the second answer we need to find the acceleration of the elevator when it reaches point B, for which we need to build a free body diagram (also included in the attached picture) and to a summation of forces:

\sum F=ma

so we get:

F_{s}+f-W=ma

Supposing the acceleration is positive when it points upwards.

So we can solve that for the acceleration, so we get:

a=\frac{F_{s}+f-W}{m}

or

a=\frac{ky_{B}+f-W}{m}

and now we substitute the data we know:

a=\frac{(9300N/m)(1m)+(17000N)-(2000kg)(9.8m/s^{2})}{2000kg}

which yields:

3.35m/s^{2}

8 0
3 years ago
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Ilia_Sergeevich [38]
The mass of the object doesn't matter. The change in its momentum is equal to the impulse that changed it ... 15 N-sec.
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