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3 years ago

Which statement is true for electric field lines?

Physics

2 answers:

KiRa [710]3 years ago

8 0

The best answer is D. field lines should always be crossing each other.

yawa3891 [41]3 years ago

3 0

The correct statement is

Field lines should always be perpendicular to the equipotential lines.

we know that, along equipotential lines, work done in moving a charge is zero. Since work done is zero, the force on the charge must be perpendicular to the displacement. the electric force on a charge is parallel to the direction of electric field. hence the electric field must be perpendicular to the displacement or equipotential lines.

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A nuclear fission reactor produces electricity. What is the role of the fuel, the control rods, water, and generator in this pro

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The appropriate response is the third one. A generator is utilized to enact the control poles which contain radioactive isotopes. Once initiated, these isotopes start an atomic splitting chain response. Water in a cooling tank monitors the rate of response as electrons radiated from the response are encouraged through wires to homes and organizations.

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3 years ago

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Some cities now implement signal lights designed to specifically apply to _____ rather than motorized vehicles.

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2 years ago

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3 years ago

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8.) If a car moving at 50km/h skids 15m with locked brakes, how far does the same car moving at 100km/h

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(8) A car starting with a speed v skids to a stop over a distance d, which means the brakes apply an acceleration a such that

0² - v² = 2 a d → a = - v² / (2d)

Then the car comes to rest over a distance of

d = - v² / (2a)

Doubling the starting speed gives

- (2v)² / (2a) = - 4v² / (2a) = 4d

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration a such that

0² - (13.9 m/s)² = 2 a (15 m) → a ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance d such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) d → d ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass m such that

60 J = m g h

where g = 10 m/s² and h is the height it is lifted, 1.2 m. Solving for m gives

m = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

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3 years ago

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$