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Novay_Z [31]
3 years ago
14

What sentence(s) is/are true when we talk about equipotential lines?

Physics
1 answer:
rodikova [14]3 years ago
6 0

Answer:

a. True

Explanation:

Equipotential lines are the imaginary lines in the space where actually the electric potential is same at each and every point.

Work is not required to move along such points of the equipotential line because the movement is always perpendicular to the electric field lines because these lines are always perpendicular to the electric field lines.

The electric potential for a point charge is given mathematically as:

V=\frac{1}{4\pi.\epsilon_0}\times \frac{Q}{r}

where:

Q= magnitude of the point charge

r= radial distance form the charge

\epsilon_0= permittivity of free space

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How long would it take for the turtle to move 650 meters at this speed of 0.05 m/s
Charra [1.4K]

Answer:

t=13000 or t= 1.3x10^4 (ten to the power of four)

Explanation:

d= 650m v=0.05 t=? v= d/t   t= d/v 650m/0.05m/s= 13000 if you need to simplify it into scientific notation, move the decimal up, you need to move it up four times, so t= 1.3x10^4s

7 0
3 years ago
Please select the word from the list that best fits the definition
Elenna [48]

Answer:

Hi... Your answer is impulse

8 0
2 years ago
How much has the ayompspheric co2 content in parts per million (ppm) increased since 1958
jeka94

Answer:

c

Explanation:

4 0
2 years ago
Adolf and Ed are wearing harnesses and are hanging at rest from the ceiling by means of ropes attached to them. Face to face, th
Alex

Explanation:

It is given that,

Mass of Adolf, m_1=120\ kg

Mass of Ed, m_2=70\ kg

Adolf swings upward to a height of 0.52 m above his starting point. Initially both men are at rest. Their momentum will remain conserved.

Firstly, finding the speed of Adolf by using the conservation of energy as :

mgh=\dfrac{1}{2}mv^2

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 0.52}

v = 3.19 m/s

Let v' is the speed of Ed. It can be calculated using the conservation of momentum as :

m_1v+m_2v'=0

v'=-\dfrac{m_1v}{m_2}

v'=-\dfrac{120\times 3.19}{70}

v' = -5.46 m/s

Let H is the height above which Ed rise. It can be calculated using the conservation of energy again as:

H=\dfrac{v^2}{2g}

H=\dfrac{(-5.46)^2}{2\times 9.8}

H = 1.52 meters

So, Ed will rise to a height of 1.52 meters. Hence, this is the required solution.

6 0
3 years ago
A 1.4 kg particle moves along an x axis, being propelled by a variable force directed along that axis. Its position is given by
mihalych1998 [28]

Answer:

The value of c is 27.4 m/s²

Explanation:

Hi there!

Let´s write the position function:

x = 3.0 m + (4.0 m/s) · t + c · t² - (1.6 m/s³) · t³

The velocity of the particle is given by the derivative of the position function with respect to time:

dx/dt = v = 4.0 m/s + 2 · c · t - 4.8 m/s³ · t²

The acceleration of the particle is the derivative of the velocity function with respect to t:

dv/dt = a = 2 · c - 9.6 m/s³ · t

The applied force at t = 3.0 s is calculated as follows:

F = m · a

Where:

F = applied force.

m = mass of the particle.

a = acceleration.

Then:

F = m · a

36 N = 1.4 kg · a

36 N / 1.4 kg = a

a = 26 m/s²

We have derived the equation of the acceleration above:

a = 2 · c - 9.6 m/s³ · t

Then, using a = 26 m/s² and t = 3.0 s, we can solve the equation for c:

26 m/s² = 2 · c - 9.6 m/s³ · 3.0 s

26 m/s² + 9.6 m/s³ · 3.0 s  = 2 · c

54.8 m/s² = 2 · c

54.8 m/s² / 2 = c

c = 27.4 m/s²

The value of c is 27.4 m/s²

4 0
3 years ago
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