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viva [34]
3 years ago
5

What was your electric potential relative to a metal pipe if a spark jumped 1.1 cm through dry air from your finger to the pipe?

Physics
2 answers:
aleksley [76]3 years ago
7 0

Answer:

Electric potential = V = 33 KV

Explanation:

The breakdown is an electric stress phenomenon. The value of electric field at which breakdown of dry air occur is 3 × 10^6 V/m. So,

E = 3 × 10^6 V/m

d = 1.1 cm = 1.1 × 10^-2 m  

We know that:

E = V/d  

V = Ed  

V = (3 × 10^6)( 1.1 × 10^-2)

V = 33000 V  

V = 33 KV

romanna [79]3 years ago
6 0
The electric field is given by volts/distance: E= \frac{V}{d}.  The breakdown voltage of dry air is about 3x10^6V/m.  So solving for V we get
V=Ed
or V=(3e6V/m)(0.011m)=33,000V
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how much force would be required to produce 88 j of work when pushing a box 1.1meters at an angle of 10 degrees?
ycow [4]

Answer:81.235N

Explanation:

Work=88J

theta=10°

distance=1.1 meters

work=force x cos(theta) x distance

88=force x cos10 x 1.1 cos10=0.9848

88=force x 0.9848 x 1.1

88=force x 1.08328

Divide both sides by 1.08328

88/1.08328=(force x 1.08328)/1.08328

81.235=force

Force=81.235

5 0
3 years ago
The net horizontal force on a car is 981 N. The car has a mass of 1550 kg and the force is applied when the car has a speed of 2
viktelen [127]

Answer:

Distance, d = 778.05 m                          

Explanation:

Given that,

Force acting on the car, F = 981 N

Mass of the car, m = 1550 kg

Initial speed of the car, v = 25 mi/h = 11.17 m/s

We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

F=ma\\\\a=\dfrac{F}{m}\\\\a=\dfrac{981}{1550}\\\\a=0.632\ m/s^2

Let d is the distance covered by car. Using second equation of motion as :

d=ut+\dfrac{1}{2}at^2\\\\d=11.17\times 35+\dfrac{1}{2}\times 0.632\times (35)^2\\\\d=778.05\ m

So, the car will cover a distance of 778.05 meters.

5 0
3 years ago
How do crystalline solids differ from amorphous solids?
LUCKY_DIMON [66]
Crystalline crystals have sharp, well-defined melting points. Amorphous Solids don't have melting points.
6 0
3 years ago
A truck accelerating at 0.0083 meters/secondÆ covers a distance of 5.8 × 10Ê meters. If the truck's mass is 7,000 kilograms, wha
tigry1 [53]

work done = force * distance moved (in direction of the force)

force= mass* acceleration 

force=58.1N

58.1*(5.8*10^4)
=3,369,800 J
3 0
3 years ago
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zimovet [89]
The velocity of the mass after 9 second is 88 m/s
7 0
3 years ago
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