The word supersonic describes:
2 answers:
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<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>
<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>
where
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>
<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
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>
<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>
<span>is different from zero because v is different from zero.</span>
Yes. The force exerted by an electric field of intensity E on an object with charge q is
</span>
<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>
where
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>
<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
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>
<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>
<span>is different from zero because v is different from zero.</span>
Answer:
Δt =
Explanation:
Hi there!
Using the equation of speed for the whole trip, we can obtain the time each one needed to cover the distance D.
The speed (v) is calculated by dividing the traveled distance (d) over the time needed to cover that distance (t):
v = d/t
Rick traveled half of the distance at Vr and the other half at Vw. Then, when v = Vr, the distance traveled was D/2 and the time is unknown, Δt1:
Vr = D/ (2 · Δt1)
For the other half of the trip the expression of velocity will be:
Vw = D/(2 · Δt2)
The total time traveled is the sum of both Δt:
Δt(total) = Δt1 + Δt2
Then, solving the first equation for Δt1:
Vr = D/ (2 · Δt1)
Δt1 = D/(2 · Vr)
In the same way for the second equation:
Δt2 = D/(2 · Vw)
Δt + Δt2 = D/(2 · Vr) + D/(2 · Vw)
Δt(total) = D/2 · (1/Vr + 1/Vw)
The time needed by Rick to complete the trip was:
Δt(total) = D/2 · (1/Vr + 1/Vw)
Now let´s calculate the time it took Tim to do the trip:
Tim walks half of the time, then his speed could be expressed as follows:
Vw = 2d1/Δt Where d1 is the traveled distance.
Solving for d1:
Vw · Δt/2 = d1
He then ran half of the time:
Vr = 2d2/Δt
Solving for d2:
Vr · Δt/2 = d2
Since d1 + d2 = D, then:
Vw · Δt/2 + Vr · Δt/2 = D
Solving for Δt:
Δt (Vw/2 + Vr/2) = D
Δt = D / (Vw/2 + Vr/2)
Δt = D/ ((Vw + Vr)/2)
Δt = 2D / (Vw + Vr)
The time needed by Tim to complete the trip was:
Δt = 2D / (Vw + Vr)
Let´s find the diference between the time done by Tim and the one done by Rick:
Δt(tim) - Δt(rick)
2D / (Vw + Vr) - (D/2 · (1/Vr + 1/Vw))
= Δt
Let´s check the result. If Vr = Vw:
Δt = 2D/2Vr - D/2Vr - D/2Vr
Δt = D/Vr - D/Vr = 0
This makes sense because if both move with the same velocity all the time both will do the trip in the same time.