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Darya [45]
3 years ago
6

A spring has an unstretched length of

Physics
1 answer:
koban [17]3 years ago
8 0

Answer:

24 N

Explanation:

Given,

Unstretched length or original length , x1 = 12cm

Stiffness constant, k = 8 N/cm

Load required to take spring to a length , x2 of 15cm

Recall the relation :

F = Ke

Where, e = extension

e = x2 - x1

e = (15 - 12) = 3

F = ke

F = 8 N/cm * 3cm

F = 24 N/cm*cm

F = 24 N

Hence, required load = 24 N

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How much time does it take light from a flash camera to reach a subject 6.0 meters across a room in scientific notation?
wariber [46]

The time it takes light from a flash camera to reach a subject 6.0 meters across a room in scientific notation is 2.0 *10^-8 s.

<u>Explanation:</u>

<u>Given</u>

t=?

d=6m

v=3*10^8 m/s

we  have,  v=d/t

here t=d/v

t=6m/3*10^8 m/s

v=2*10^-8 m/s

The time it takes light from a flash camera to reach a subject 6.0 meters across a room in scientific notation is 2.0 *10^-8 s.

<u></u>

6 0
3 years ago
The diver has least gravitational potentail engery at position
ankoles [38]
I'm not sure if a figure or some choices go along with this, but the closer to the sea floor the diver is, the lower the potential energy
7 0
3 years ago
State Pascal's principle of transmission of pressure​
bulgar [2K]

Answer:

Pascal's law (also Pascal's principle or the principle of transmission of fluid-pressure) is a principle in fluid mechanics given by Blaise Pascal that states that a pressure change at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere.

4 0
3 years ago
A system initially has an internal energy U of 504 J. It undergoes a process during which it releases 111 J of heat energy to th
Likurg_2 [28]

Answer:

615 J

Explanation:

internal energy (U) = 504 J

heat lost (q) = 111 J = - 111 J (negative sign is because heat is lost)

work done = 222 J

what is the final energy in the system

total energy = final energy - initial energy

final energy = total energy + initial energy

where

initial energy = 504 J

total energy = 222 - 111 = 111 J

final energy = 504 + 111 = 615 J

6 0
3 years ago
A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.
noname [10]

Answer:

The value is  \rho_s  =  4.026 *10^{-6} \  C/m^2

Explanation:

From the question we are told that

   The radius of the inner conductor  is  r_1 = 1 \ mm =  0.001 \ m

    The radius of the outer conductor is  r_2 = 3 \ mm = 0.003 \  m

    The potential at the outer conductor is  V = 1.5 kV  =  1.5 *10^{3} \  V

Generally the capacitance per length of the capacitor like set up of the two conductors is

      C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}

Here \epsilon_o is the permitivity of free space with value  \epsilon_o =  8.85*10^{-12} C/(V \cdot m)

=>   C= \frac{2 *  3.142  * 8.85*10^{-12}  }{ ln [\frac{0.003}{0.001} ]}

=>   C= 50.6 *10^{-12} \  F/m

Generally given that the potential  of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line  charge density of the outer  conductor is mathematically represented as

      \rho_l  =  C *  V

=>   \rho_d  =  50.6*10^{-12} *  1.5*10^{3}

=>   \rho_d  =  7.59*10^{-8} \  C/m

Generally the surface charge density is mathematically represented as

        \rho_s  =  \frac{\rho_l }{2 \pi * r_2 }    here 2 \pi r = (circumference \ of \ outer \  conductor  )

=>    \rho_s  =  \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }

=>    \rho_s  =  4.026 *10^{-6} \  C/m^2

3 0
2 years ago
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