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adell [148]
3 years ago
5

Assume the motions and currents mentioned are along the x axis and fields are in the y direction. (a) does an electric field exe

rt a force on a stationary charged object? yes no (b) does a magnetic field do so? yes no (c) does an electric field exert a force on a moving charged object? yes no (d) does a magnetic field do so? yes no (e) does an electric field exert a force on a straight current-carrying wire? yes no (f) does a magnetic field do so? yes no (g) does an electric field exert a force on a beam of moving electrons? yes no (h) does a magnetic field do so? yes no
Physics
1 answer:
matrenka [14]3 years ago
6 0
<span> (a) does an electric field exert a force on a stationary charged object? 
Yes. The force exerted by an electric field of intensity E on an object with charge q is
</span>F=qE
<span>As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

</span><span>(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with  charge q and speed v is
</span>F=qvB \sin \theta
where \theta is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

<span>(c) does an electric field exert a force on a moving charged object? 
Yes, The intensity of the electric force is still
</span>F=qE
<span>as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

</span><span>(d) does a magnetic field do so?
</span>Yes. As we said in point b, the magnetic force is
F=qvB \sin \theta
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

<span>(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.

</span><span>(f) does a magnetic field do so? 
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.

</span><span>(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
</span>F=qE
<span>So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.

</span><span>(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
</span>F=qvB \sin \theta
<span>is different from zero because v is different from zero.</span>
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Metal bar is aligned along East- West direction and it is dropped vertically down

So its velocity is along -Z direction,

Now the Earth's magnetic field is towards north so its towards +Y direction

now we have formula for force on a moving charge

F = q(v X B)

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now by the above formula

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5 0
3 years ago
The small piston of a hydraulic press has an area of 3.2 in2. If the applied force is 2.0 lb, what must the area of the large pi
BartSMP [9]

Given data:

The area of the hydraulic press is,

A_1=3.2in^2

The applied force is,

F_1=2\text{ lb}

The pressing force on the large piston is

F_2=795\text{ lb}

Given hydraulic press works on Pascal's law principle.

From the Pascal's law, the equation can be given as,

\frac{F_1}{A_1}=\frac{F_2}{A_2}

Here,

A_2

is the area of the larger piston.

Substituting the values in the above equation, we get:

\begin{gathered} \frac{2\text{ lb}}{3.2in^2}=\frac{795\text{ lb}}{A_2} \\ A_2=1272in^2 \end{gathered}

Thus, the area of the large piston is

1272in^2

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Identify which type of source is being described.
frez [133]

Answer:

Primary, secondary

Explanation:

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umka2103 [35]
D is the answer. It is a firm statement.
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A rocket sled is tested at "5 g" (5 times the acceleration due to gravity). If the sled starts from rest at position do= 0.00, h
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D=at²
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t²=441/(5*9.81)
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