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3 years ago

A particle moves in a straight line. Its speed(v)

Physics

1 answer:

Alina [70]3 years ago

3 0

Answer:

$6v_0$

Explanation:

The speed of the particle increases linearly with time, so we can write it as

$v(t) = kt+v_0$

where k is the a certain constant of proportionality and $v_0$ is the initial speed.

We also know that at t = 4 s, the speed is $v=2v_0$ , so we can use this information to find k:

$v(4) = 4k+v_0 = 2v_0 \rightarrow k = \frac{v_0}{4}$

So the complete expression of v(t) is

$v(t) = \frac{v_0}{4}t+v_0 = v_0 (\frac{t}{4}+1)$

Now we integrate the quantity $v(t)$ , and we find:

$\int {v(t)} \, dt =\int (\frac{v_0}{4}t+v_0) \, dt =\frac{v_0}{8}t^2 + v_0 t$

Substituting the limit of integration t = 4, we find:

$\frac{v_0}{8}4^2 + 4v_0 =2v_0+4v_0 = 6v_0$

And substituting t = 0, we find 0. So, the result of the integration from 0 to 4 is

$6v_0$

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2 years ago

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7 0

3 years ago

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

matrenka [14]

(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
 $F=qE$
As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

(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
 $F=qvB \sin \theta$
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.

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

(d) does a magnetic field do so?
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.

(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.

(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.

(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
 $F=qE$
So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.

(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
 $F=qvB \sin \theta$
is different from zero because v is different from zero.

6 0

3 years ago

In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?