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konstantin123 [22]
3 years ago
6

A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. the wire ful

ly lies in a magnetic field given by (0.3y)i + (0.4y)j Tesla. The magnetic force on the wire is *
Physics
1 answer:
Bess [88]3 years ago
7 0

Answer:

The force is "19 µN".

Explanation:

The lane's j-component is meaningless, as the current is flowing in the -j line.  

Therefore the power is now in the direction of + z (out of the page if x and y are in the page plane) and has the magnitude.

\ formula: \\\\\ Forec (F) = mA \\\\ \ F \ = 2.0mA \times \int {0.3} \ y \ dy  \rightarrow  \ from\ 0 \ to \ 0.25 \\\\\ F \ = \ 2.0mA \times 0.15 * 0.25^{2} m\cdot T \\\\ F = 19 \µN

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Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could
Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}

Where:

d=17.91 mm =17.91(10)^{-3}  m is the diameter of a dime

D is the diameter of the Sun

x_{sun-pinhole}=150,000,000 km=1.5(10)^{11}  m is the distance between the Sun and the pinhole

x_{sunball-pinhole}=100 d=1.791 m is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding D:

D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}

D=\frac{17.91(10)^{-3}  m}{1.791 m} 1.5(10)^{11}  m

D=1.5(10)^{9}  m This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

1 AU=149,597,870,700 m

So, we have to divide this distance between D in order to find how many suns could it fit in this distance:

\frac{149,597,870,700 m}{1.5(10)^{9}  m}=99.73 suns \approx 100 suns

8 0
3 years ago
two clear, colorless liquids are poured together. A bright yellow solid forms. What physical properties have changed?
jek_recluse [69]
Change in state(from liquid to solid) and change in colour I believe.
3 0
3 years ago
If a proton and an electron are released when they are 7.00×10−10 m apart (typical atomic distances), find the initial accelerat
Ugo [173]

Answer:

The acceleration of the proton is 2.823 x 10¹⁷ m/s²

The acceleration of the electron is 5.175 x 10²⁰ m/s²

Explanation:

Given;

distance between the electron and proton, r = 7 x 10⁻¹⁰ m

mass of proton, m_p = 1.67 x 10⁻²⁷ kg

mass of electron, m_e = 9.11 x 10⁻³¹ kg

The attractive force between the two charges is given by Coulomb's law;

F = \frac{k(q_p)(q_e)}{r^2}

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/c²

F = \frac{k(q_p)(q_e)}{r^2} \\\\F = \frac{(9*10^9)(1.602*10^{-19})(1.602*10^{-19})}{(7*10^{-10})^2} \\\\F = 4.714 *10^{-10} \ N

Acceleration of proton is given by;

F = ma

F = m_pa_p\\\\a_p = \frac{F}{m_p}\\\\a_p = \frac{4.714*10^{-10}}{1.67*10^{-27}}\\\\a_p = 2.823 *10^{17} \ m/s^2

Acceleration of the electron is given by;

F = m_ea_e\\\\a_e = \frac{F}{m_e}\\\\a_e = \frac{4.714*10^{-10}}{9.11*10^{-31}}\\\\a_e = 5.175 *10^{20} \ m/s^2

7 0
3 years ago
Calculate the potential energy of 5.0g (0.005 kg) paper airplane 5.0 meters above the ground
Readme [11.4K]

PE = (mass) (gravity) (height)

PE = (0.005 kg) (9.8 m/s²) (5 m)

<em>PE = 0.245 Joule</em>

4 0
3 years ago
True or False- For resistors in series, the larger the resistance is, the larger the voltage drop that is required.​
WINSTONCH [101]

Answer:

I think its true

8 0
2 years ago
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