A magnetic field would be produced by a beam of

1 answer:

7 0

<span>A magnetic field would be produced by a beam of (3) protons.
Although you could easily make a case for any of these four options (the question is pretty ambiguous), there can be only one correct answer, which is the third option, in this case.
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When a falling meteoroid is at a distance above the earth's surface of 2.60 times the earth's radius, what is its acceleration d

Mice21 [21]

The gravitational acceleration at any distance r is given by

$g= \frac{GM}{r^2}$

where G is the gravitational constant, M the Earth's mass and r is the distance measured from the center of the Earth.

The Earth's radius is $r_e=6.37 \cdot 10^6 m$ , so the meteoroid is located at a distance of:

$r=r_e+2.60 r_e =3.60 r_e = 2.29 \cdot 10^7 m$

And by substituting this value into the previous formula, we can find the value of g at that altitude:

$g= \frac{GM}{r^2} = \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(2.29 \cdot 10^7 m)^2} =0.75 m/s^2$

5 0

3 years ago

Read 2 more answers

A tube with a cap on one end, but open at the other end, has a fundamental frequency of 130.8 Hz. The speed of sound is 343 m/s

sergey [27]

Answer:

Y = V / f where Y equals wavelength

4 Y1 = V / f1 for a closed pipe the wavelength is 1/4 the length of the pipe

2 Y2 = V / f2 for the open pipe the wavelength is 1/2 the length of the pipe

Y1 / Y2 = 2 = f2 / f1 dividing equations

f2 = 2 f1

the new fundamental frequency is 2 * 130.8 = 261.6

(The new wavelength is 1/2 the original wavelength so the frequency must double to produce the same speed.

8 0

3 years ago

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$