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KATRIN_1 [288]
2 years ago
12

Determine the direction of the force (if any) that will act on the charge in each of the following situations. A positive charge

within an electric field that points to the right. negative charge within an electric field that points downward. A positive charge moving downward in a magnetic field that points out of the screen. A negative charge moving to the right in a magnetic field that points into the screen. A positive charge moving upward in an electric field that points into the screen. A negative charge moving to the right in a magnetic field that points to the right. to the right to the left into the screen zero force downward out of the screen upward
Physics
1 answer:
NeTakaya2 years ago
3 0

Answer:

a) F on the right , b)    F up , c) Force to the left , d)  Force down , e)  Force towards the screen , f) Force is zero

Explanation:

The equations for the forces are

Electric

          F = q E

Magnetic

          F = q v x B

The Bold are vectors, the charges are positive

Let's apply these equation to the proposed situations

a) as the load is positive the force goes in the direction of the field

      F on the right

b) as the load is negative the force goes in the opposite direction to the field

     F is up

c) positive charge with speed down and magnetic field out of the screen.

For this part we will use the right hand rule. The thumb is in the direction of the speed, the fingers extended in the direction of the magnetic field and the palm is in the direction of the force for a positive charge

       Force to the left

d) negative charge, speed to the right magnetic field between the screen

   The force has direction

      Force down

e) positive charge, electric field towards the screen.

     Force in the direction of the field

        Force towards the screen

.f) negative charge, speed to the right, magnetic field to the right

    The vector product is zero,

     Force is zero

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Answer:

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Explanation:

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Windmills use wind, that is renewable.

So, the answer is D.

7 0
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Read the following and identify the main parts of the experiment.
slava [35]

The independent variable refers to the type of socks (synthetic socks), the dependent variable is the number of blisters and the control is the two weeks that she switches back to the cotton socks.

<h3>What is the dependent variable?</h3>

In an experiment, the dependent variable is those being tested and changes according to the independent variable.

The control group is the set of experimental conditions that is used to compare a given outcome in an experiment.

In conclusion, The independent variable refers to the type of socks (synthetic socks), the dependent variable is the number of blisters and the control is the two weeks that she switches back to the cotton socks.

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4 0
1 year ago
I need help with the question 8B
ratelena [41]
The answer of this is C
3 0
3 years ago
2. An airplane traveling north at 220. meters per second encounters a 50.0-meters-per-second crosswind
Alex777 [14]

The resultant speed of the plane  is (3) 226 m/s

Why?

We can calculate the resultant speed of the plane by using the Pythagorean Theorem since both speeds are perpendicular (forming a right triangle).

So, calculating we have:

ResultantSpeed=\sqrt{VerticalSpeed^{2}+HorizontalSpeed^{2}}\\\\ResultantSpeed=\sqrt{(220\frac{m}{s})^{2}+50\frac{m}{s})^{2}

ResulntantSpeed=\sqrt{48400\frac{m^{2} }{s^{2} }+2500\frac{m^{2} }{s^{2} } } \\\\ResultantSpeed=\sqrt{50900\frac{m^{2} }{s^{2} }}=226\frac{m}{s}

Hence, we have that the resultant speed of the plane  is (3) 226 m/s

Have a nice day!

5 0
2 years ago
A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an
Nat2105 [25]

Answer:

a)T=0.0031416s

b)v_{max}=6.283\frac{m}{s}

c) E=0.1974J

d)F=80N

e)F=40N

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by X=Acos(\omega t +\phi)    (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)   (2)

For the acceleration

\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)   (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

A\omega^2=8x10^{3}\frac{m}{s^2}

Since we know the amplitude A=0.002m  we can solve for \omega like this:

\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}

And we with this value we can find the period with the following formula

T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens cos(\omega t +\phi)=1, and we also know that the maximum acceleration is given by::

|\frac{d^2X}{dt^2}|=A\omega^2

So then we have that:

F=ma=mA\omega^2

And since we have everything we can find the force

F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N

6) Part e

When the mass it's at the half of it's maximum displacement the term cos(\omega t +\phi)=1/2 and on this case the acceleration would be given by;

|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}

And the force would be given by:

F=ma=\frac{1}{2}mA\omega^2

And replacing we have:

F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N

8 0
2 years ago
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