All categories

2 years ago

A student uses the right-hand rule as shown.

Physics

2 answers:

Lera25 [3.4K]2 years ago

6 0

Answer:

Right

Explanation:

Just did this on edgen, in the diagram, the fingers are pointing to the right, which indicates that the magnetic field is acting to the right.

Kipish [7]2 years ago

5 0

Answer:

right is the correct answer to the given question .

Explanation:

In this question figure is missing

The main objective right-hand rule to decide the position of the magnetic force on the positive force acting, either the position of the thumb of a right hand with in position of v, the fingers throughout the position of B1, and a right angles throughout the position of F1 to the hand positions.

So $F1 \ =\ q\ v \ B1\ Sin\alpha$

So from the magnetic right hand rule the direction of the magnetic field in front of a wire is right .
All the others options are incorrect because they do not give the direction of the magnetic field in front of a wire is right .

You might be interested in

A mobile phone is pulled northward by a force of 10 n and at the same time pulled southward by another force of 15 n. the result

makvit [3.9K]

5 Southward would be the correct answer.

6 0

3 years ago

You are on your balcony and notice some bad squirrels digging in your garden directly below. You start tossing your pistachios a

Rashid [163]

Answer:

0.43 s

Explanation:

We have the following parameters:

Initial velocity, u = 7.4 m/s

Acceleration of gravity, g = 9.8 $m/s^2$

Distance, s = 43 in + 10 ft = 1.092 m + 3.048 m = 4.14 m

Time, t = ?

Using the equation of motion $s=ut +\frac{1}{2}gt^2$ , we have

$4.14 = 7.4t + 0.5\times9.8t^2$

$4.9t^2 + 7.4t - 4.14 =0$

Using the quadratic formula $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ where a = 4.9, b = 7.4 and c = - 4.14, and solving for the positive value of t only, we have

$t = 0.43$ s

3 0

3 years ago

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$