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3 years ago

Which planet spins on it's side Uranus or neptune?

Physics

2 answers:

BartSMP [9]3 years ago

8 0

The answer is Uranus, as it has a tilted axis.

maxonik [38]3 years ago

6 0

I believe its Uranus because if you see some pictures of it theres something spinning around it.
Hope this helps! :)

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A horizontal wire is hung from the ceiling of a room by two massless strings. The wire has a length of 0.11 m and a mass of 0.01

AnnZ [28]

Answer:

Explanation:

The magnetic force acting horizontally will deflect the wire by angle φ from the vertical

Let T be the tension

T cosφ = mg

Tsinφ = Magnetic force

Tsinφ = BiL , where B is magnetic field , i is current and L is length of wire

Dividing

Tanφ = BiL / mg

= .055 x 29 x .11 / .010 x 9.8

= 1.79

φ = 61° .

Tension T = mg / cosφ

= .01 x 9.8 / cos61

= .2 N .

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3 years ago

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mestny [16]

Check-call-Care if you are training to be a life guard.

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3 years ago

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How do you think electricity was discovered

sasho [114]

Italian physicist Alessandro Volta discovered that particular chemical reactions could produce electricity, and in 1800 he constructed the voltaic pile (an early electric battery) that produced a steady electric current, and so he was the first person to create a steady flow of electrical charge.

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3 years ago

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An electron that has a velocity with x component 2.6 × 106 m/s and y component 3.2 × 106 m/s moves through a uniform magnetic fi

elena55 [62]

Answer:

a) $\vec F_{B} = 8.766\times 10^{-14}\,T\,k$ , b) $\vec F_{B} = -8.766\times 10^{-14}\,T\,k$

Explanation:

a) The magnetic force experimented by a particle has the following vectorial form:

$\vec F_{B} = q\cdot \vec v \times \vec B$

The charge of the electron is equal to $-1.602\times 10^{-19}\,C$ . Then, cross product can be solved by using determinants:

$\vec F_{B} = \begin{vmatrix}i&j&k\\-4.165\times 10^{-13}\, C\cdot \frac{m}{s} &-5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}$

The magnetic force is:

$\vec F_{B} = 8.766\times 10^{-14}\,T\,k$

b) The charge of the proton is equal to $1.602\times 10^{-19}\,C$ . Then, cross product has the following determinant:

$\vec F_{B} = \begin{vmatrix}i&j&k\\4.165\times 10^{-13}\, C\cdot \frac{m}{s} &5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}$

The magnetic force is:

$\vec F_{B} = -8.766\times 10^{-14}\,T\,k$

8 0

3 years ago

A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$