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Anastaziya [24]
2 years ago
8

5. Three charges with q = 7.5 x 10-6 C are located as shown, where

Physics
1 answer:
prisoha [69]2 years ago
5 0

F12 = k • q1 • q2 / r12²

F12 = (9 × 10⁹) • (7.5 × 10⁻⁶)² / (0.25)²

F12 = 8.1 N

F13 = k • q1 • q2 / r12²

F12 = (9 × 10⁹) • (7.5 × 10⁻⁶)² / (0.25² + 0.25²)

F12 = 4.05 N

F² = F12² + F13² + 2(F12)(F13) cos θ

F² = (8.1)² + (4.05)² + 2(8.1)(4.05) cos 45°

F = 11.33 N

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FrozenT [24]

the answer is (A) the movement of the magnet relative to the coil

4 0
3 years ago
Read 2 more answers
An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p
igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

Explanation:

Given:

Distance from the wire to the field point r = 2.83 \times 10^{-2} m

Speed of electron v = 35.5 \%c

Current I = 17.7 A

For finding the acceleration,

First find the magnetic field due to wire,

  B = \frac{\mu _{o}I }{2\pi r }

Where \mu_{o} = 4\pi   \times 10^{-7}

  B = \frac{4\pi \times 10^{-7}  \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }

  B = 12.50 \times 10^{-5} T

The magnetic force exerted on the electron passing through straight wire,

  F = qvB  

  F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}

  F = 21.3 \times 10^{-16} N

From the newton's second law

  F = ma

Where m = mass of electron = 9.1 \times 10^{-31} kg

So acceleration is given by,

   a = \frac{F}{m}

   a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

   a = 2.34 \times 10^{15} \frac{m}{s^{2} }

Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

7 0
3 years ago
An ideal refrigerator does 130. 0 j of work to remove 780. 0 j of heat from its cold compartment during each cycle. what is the
zheka24 [161]

The refrigerator's coefficient of performance is 6.

The heat extracted from the cold reservoir Q cold (i.e., inside a refrigerator) divided by the work W required to remove the heat is known as the coefficient of performance, or COP, of a refrigerator (i.e., the work done by the compressor). The required inside temperature and the outside temperature have a significant impact on the COP.

As the inside temperature of the refrigerator decreases, its coefficient of performance decreases. The coefficient of performance (COP) of refrigeration is always more than 1.

The heat produced in the cold compartment, H = 780.0 J

Work done in ideal refrigerator, W = 130.0 J

Refrigerator's coefficient of performance = H/W

                                                                     = 780/130

                                                                     = 6

Therefore, the refrigerator's coefficient of performance is 6.

Energy conservation requires the exhaust heat to be = 780 + 130

                                                                                          = 910 J

Learn more about  coefficient here:

brainly.com/question/18915846

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5 0
2 years ago
A crime suspect fled the scene when police officers entered the building. How far could he run in 10 minutes if he can run 5 mil
emmainna [20.7K]
(5 mi/hr) x (1hr/60min) x (10min) = 5 x 10 / 60 = <em>5/6 mile</em>

(5/6 mile) x (1,760 yd/mile) = <em>1,466 and 2/3 yards</em>
3 0
3 years ago
Consider two circular metal wire loops each carrying the same current I as shown below. In what region(s) could the net magnetic
vova2212 [387]

Answer:

h

Explanation:

6 0
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