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d1i1m1o1n [39]
3 years ago
5

Since acid solutions are conductors of electricity, they can be used in batteries. true false

Physics
2 answers:
Eduardwww [97]3 years ago
8 0
This is true, I believe. 

Hope this helps!
Aleks [24]3 years ago
4 0

Answer:Explained

Explanation:

Since acid solutions are conductors of electricity ,they can be used in batteries as batteries need ions to conduct electricity and acids dissociates to gives ions which is necessary for electricity conduction.But not all acids can be used in batteries as some acids are not economical or highly volatile to be put in batteries therefore only certain type of acids is used in batteries.ex. sulfuric acid is used in batteries  .

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Using the scientific definition of work, does moving an object a greater amount of distance always require a greater amount of w
tester [92]
The answer is no. If you are dealing with a conservative force and the object begins and ends at the same potential then the work is zero, regardless of the distance travelled. This can be shown using the work-energy theorem which states that the work done by a force is equal to the change in kinetic energy of the object.
W=KEf−KEi
An example of this would be a mass moving on a frictionless curved track under the force of gravity.
The work done by the force of gravity in moving the objects in both case A and B is the same (=0, since the object begins and ends with zero velocity) but the object travels a much greater distance in case B, even though the force is constant in both cases.

3 0
3 years ago
4. While cleaning your bedroom, you move your mattress to vacuum underneath your bed. You use a force of 48 N to move the mattre
Lerok [7]

Answer:

72 J

Explanation:

Use the Work formula

W= F x d

Given:

F - 48 N

d - 1.5 m

Solution:

W= F x d

W=  48 N x 1.5 m

W= 72 J

8 0
3 years ago
The drawing shows two strings that have the same length and linear density. The left end of each string is attached to a wall, w
Readme [11.4K]

Answer:

WB = 12 N

Explanation:

Solution:

- The relationship between wavelength (λ) and velocity (v) is given as:

                                       v  ∝ λ

- Both wavelength and and velocity are proportional. For the case B, we see that wavelength λ is halved. From using the above relation we can also say that velocity (v) is also halved.

- The relationship between velocity (v) and tension force (T) in the string is given by:

                                      v^2 ∝ T  

- We saw that if wavelength (λ) is halved, velocity (v) is halved then from the above relationship we see that Tension force T is reduced 4 times.

- Since the tension force T in part A is:

                                      TA = m*g = WA = 48

- The Tension force in part B, TB is reduced 4 times as follows:

                                      TB = 0.25*m*g = WB = 12 N

8 0
3 years ago
6. a. find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a w
nasty-shy [4]

Complete Question

(A) Find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a width of 0.125 nm .

(B) The electron makes a transition from the n=1 to n= 4 level by absorbing a photon. Calculate the wavelength of this photon.

Answer:

A

  \Delta  E =  337 \  eV

B

  \lambda  = 3.439 *10^{-9} \  m

Explanation:

Considering question a

From the question we are told that

 The width of the box is  w = 0.125 \  nm  =  0.125 *10^{-9} \  m

Generally the energy level of a particle confined to a box is mathematically represented as

         E_n  =   \frac{n^2 h^2}{8 m L^2 }

Generally the excitation energy is mathematically represented as

         \Delta  E =  \frac{h^2 }{ 8 m L^2 }  [n_2^2 - n_1 ^2  ]

From the question  n_2 =  3\ (Third \  excited \ level ) \ \ and  \ \ n_1 = 1  

   Here h  is the Planck's constant with a value  of  h =  6.62607015 * 10^{-34} J \cdot s

         m is the mass of  electron with value  m  =  9.11 *10^{-31} \  kg

So

        \Delta  E =  \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 }  [3^2 - 1^2  ]

=>     \Delta  E = 539 *10^{-19} \  J

=>     \Delta  E = \frac{539 *10^{-19}}{1.60 *10^{-19}}  \  J

=>     \Delta  E =  337 \  eV

Considering question b

Generally the energy level of a particle confined to a box is mathematically represented as

         E_n  =   \frac{n^2 h^2}{8 m L^2 }

Generally the excitation energy is mathematically represented as

         \Delta  E =  \frac{h^2 }{ 8 m L^2 }  [n_2^2 - n_1 ^2  ]

From the question  n_2 =  4 \ \ and  \ \ n_1 = 1

 Here h  is the Planck's constant with a value  of  h =  6.62607015 * 10^{-34} J \cdot s

         m is the mass of  electron with value  m  =  9.11 *10^{-31} \  kg

So

        \Delta  E =  \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 }  [4^2 - 1^2  ]

=>      \Delta  E = 578 *10^{-19} \  J

=>      \Delta  E =  \frac{ 578 *10^{-19}}{1.60 *10^{-19 }}  

=>      \Delta  E =  361.45 \ eV  

Gnerally the wavelength is mathematically represented as

          \lambda  =  \frac{hc}{\Delta E }

=>       \lambda  =  \frac{ 6.626 *10^{-34} * (3.0 *10^{8})}{578 *10^{-19}  }

=>       \lambda  = 3.439 *10^{-9} \  m

5 0
3 years ago
Would sound travel faster at the north pole or in the sahara desert? Explain ur answer
gtnhenbr [62]

Sound needs a medium to propagate through and this medium which is moves through is the air. The colder the temperature is, the less excited (meaning kinetic energy) the air particles will be meaning that the sound wave will travel slower. In the sahara desert, since the temperatures are warmer, and more excited, the sound will propagate faster through the air.

*Note that sound waves rely on vibration to move and if the air is hotter the molecules in the air are vibrate more than compared to lower temperatures.  

6 0
3 years ago
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