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Illusion [34]
2 years ago
11

What can cause electricity to jump across a gap

Physics
1 answer:
Thepotemich [5.8K]2 years ago
3 0

Answer:

It often happens in a circuit which was conducting,and a switch is opened. Any inductance in the circuit will help the voltage increase due to back emf,and this can cause breakdown of the air insulation and an arc can form.

Explanation:

You might be interested in
A lacrosse ball that is thrown straight upwards reaches a maximum height of 4.5 m. At what initial velocity was it thrown? (note
shtirl [24]

Answer:

The initial velocity was 9.39 m/s

Explanation:

<em>Lets explain how to solve the problem</em>

The ball is thrown straight upward with initial velocity u

The ball reaches a maximum height of 4.5 m

At the maximum height velocity v = 0

The acceleration of gravity is -9.8 m/s²

We need to find the initial velocity

The best rule to find the initial velocity is <em>v² = u² + 2ah</em>, where v is

the final velocity, u is the initial velocity, a is the acceleration of

gravity and h is the height

⇒ v = 0 , h = 4.5 m , a = -9.8 m/s²

⇒ 0 = u² + 2(-9.8)(4.5)

⇒ 0 = u² - 88.2

Add 88.2 to both sides

⇒ 88.2 = u²

Take square root for both sides

⇒ u = 9.39 m/s

<em>The initial velocity was 9.39 m/s</em>

5 0
3 years ago
A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.
Pepsi [2]

Answer:

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately \displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}.

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately \displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s.

c) The orbital radius required would be approximately \rm 2.04 \times 10^7\; m.

d) The escape velocity from the surface of that planet would be approximately \rm 5.01\times 10^3\; m \cdot s^{-1}.

Explanation:

<h3>a)</h3>

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

\displaystyle \frac{G \cdot M \cdot m}{r^{2}},

where

  • G is the constant of universal gravitation.
  • M is the mass of the first mass. (In this case, let M be the mass of the planet.)
  • m is the mass of the second mass. (In this case, let m be the mass of the satellite.)  
  • r is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

\displaystyle \frac{m \cdot v^{2}}{r},

where

  • m is the mass of the object (in this case, that's the mass of the satellite.)
  • v is the orbital speed of the satellite.
  • r is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for v:

\displaystyle \frac{G \cdot M \cdot m}{r^{2}} = \frac{m \cdot v^{2}}{r}.

\displaystyle v^2 = \frac{G \cdot M}{r}.

\displaystyle v = \sqrt{\frac{G \cdot M}{r}}.

Take G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2},  Simplify the expression v:

\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}.

<h3>b)</h3>

Since the orbit is a circle of radius R, the distance traveled in one period would be equal to the circumference of that circle, 2 \pi R.

Divide distance with speed to find the time required.

\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx  9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}.

<h3>c)</h3>

Convert 24.6\; \rm \text{hours} to seconds:

24.6 \times 3600 = 88560\; \rm s

Solve the equation for R:

9.62 \times 10^{-7}\, R^{3/2}= 88560.

R \approx 2.04 \times 10^7\; \rm m.

<h3>d)</h3>

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r} (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

\displaystyle \text{KE} = \frac{1}{2} \, m \cdot v^{2}.

Solve for v. The value of m shouldn't matter, for it would be eliminated from both sides of the equation.

\displaystyle -\frac{G \cdot M \cdot m}{r} + \frac{1}{2} \, m \cdot v^{2}= 0.

\displaystyle v = \sqrt{\frac{2\, G \cdot M}{R}} \approx 5.01\times 10^{3}\; \rm m\cdot s^{-1}.

5 0
3 years ago
A pendulum clock gives correct time at 20°c how many sec/day will it gain or loose when the temperature fall to 5°c? cofficient
rewona [7]

Answer:

129.6 seconds

Explanation:

Given that :

α = 0.0002°c-1​

θ1 = 20°C

θ2 = 5°C

Time t = one day ; Converting to seconds ; number of seconds in a day ; (24 * 60 * 60) = 86400 seconds

Let dT= change in time

Using the relation :

dT = 0.5* α * dθ * t

dθ = (20 - 5) = 15°C

dT = 0.5 * 0.0002 * 15 * 86400

dT = 129.6 seconds

5 0
2 years ago
A machine gear consists of 0.10 kg of iron and 0.16 kg of copper.
Natali5045456 [20]

Answer:

option (c)

Explanation:

mass of iron = 0.10 kg

mass of copper = 0.16 kg

rise in temperature, ΔT = 35°C

specific heat of iron = 450 J/kg°C

specific heat of copper = 390 J/kg°C

Heat by iron (H1) = mass of iron x specific heat of iron x ΔT

H1 = 0.10 x 450 x 35 = 1575 J

Heat by copper (H2) = mass of copper x specific heat of copper x ΔT

H1 = 0.16 x 390 x 35 = 2184 J

Total heat H = H1 + H2

H = 1575 + 2184 = 3759 J

by rounding off

H = 4000 J

6 0
3 years ago
Read 2 more answers
A diagram of the carbon-oxygen cycle is shown below.
vovikov84 [41]
Plants through photosynthesis.
3 0
2 years ago
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