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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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‼️‼️ Please help, urgent ‼️‼️ (check photo)
Alex787 [66]

Answer: The force constant k is 10600 kg/s^2

Step by step:

Use the law of energy conservation. When the elevator hits the spring, it has a certain kinetic and a potential energy. When the elevator reaches the point of still stand the kinetic and potential energies have been transformed to work performed by the elevator in the form of friction (brake clamp) and loading the spring.

Let us define the vertical height axis as having two points: h=2m at the point of elevator hitting the spring, and h=0m at the point of stopping.

The total energy at the point h=2m is:

E_{tot}=E_{kin}+E_{pot}\\E_{tot}= \frac{1}{2}mv^2+mg\Delta h = \frac{1}{2}2000 kg 4^2\frac{m^2}{s^2}+2000kg\, 9.8\frac{m}{s^2}2m=55200\,kg\frac{m^2}{s^2}

The total energy at the point h=0m is:

E_{tot}=E_{kin}+E_{pot}+Work=0+0+ Work\\E_{tot} =F_{friction}\Delta h+\frac{1}{2}k (\Delta h)^2=17000N\cdot 2m+\frac{1}{2}k\cdot 2^2 m^2

The two Energy values are to be equal (by law of energy conservation), which allows us to determine the only unknown, namely the force constant k:

17000N\cdot 2m+\frac{1}{2}k\cdot 2^2 m^2 = 55200 \,kg\frac{m^2}{s^2}\\k = \frac{55200-34000}{2}\,\frac{kg}{s^2}=10600\frac{kg}{s^2}

5 0
3 years ago
An object is moving across a surface, but it is not gain or lose speed. Which best describes the objects force?
raketka [301]

Answer:

Gravity. An object is moving across a surface, but it does not gain or lose speed.

Explanation:

The basic idea. Physicists see gravity as one of the four fundamental forces that govern the universe, alongside electromagnetism and the strong and weak nuclear forces.

Hope it helps! Brainliest?

3 0
2 years ago
Read 2 more answers
Describe a situation where you can be traveling at a low speed but have an extremely high velocity
Reil [10]

Answer:

Cruising at 35,000 feet in an airliner, straight toward the east,

at 500 miles per hour

Explanation:

3 0
2 years ago
Read 2 more answers
Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t
sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

A_1 v_1 = A_2 v_2 (1)

where

A_1 is the cross-sectional area of the 1st section of the pipe

A_2 is the cross-sectional area of the 2nd section of the pipe

v_1 is the velocity of the 1st section of the pipe

v_2 is the velocity of the 2nd section of the pipe

In this problem we have:

v_1=0.15 m/s is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

d_2=0.74d_1

The cross-sectional area is proportional to the square of the diameter, so:

A_2=(0.74)^2 A_1=0.548 A_1

And solving eq.(1) for v2, we find the final velocity:

v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s

Now we can use Bernoulli's equation to find the pressure drop:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2

where

\rho=1025 kg/m^3 is the blood density

p_1,p_2 are the initial and final pressure

So the pressure drop is:

p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa

8 0
3 years ago
What causes the rider on a roller coaster to experience the + or - g’s if gravity always remains the same on earth
MA_775_DIABLO [31]

Answer:

Yes both = and - g can be felt by a rider in a roller coaster.

Explanation:

It is crucial to understand how we feel gravity in this case.

We humans have no sensory organs to directly detect magnitude and direction like some birds and other creatures, but then how do we we feel gravity?

When we stand on our feet we feel our weight due to the normal reaction of floor on our feet trying to keep us stand and our weight trying to crush us down. In an elevator we feel difference in our weight (difference magnitudes of gravity) but actually we are feeling the differences in normal reactions under different accelerations of the elevator.

In the case of roller coaster you will feel +g as you  sit on a chair in it, but will feel -g when you are in upside down position as roller coaster move.

When you are seated you will feel the normal reaction of seat on you giving you the feeling +g and the support of the buckles to stay in the roller coaster when you are upside down will give you the -g feeling.

<u>This is just the physics approach</u>, a biological approach can be given in association with sensors relating to ears.

8 0
3 years ago
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