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ExtremeBDS [4]
3 years ago
7

A medium is a substance or material in which a wave can travel through. True or False

Physics
2 answers:
pogonyaev3 years ago
8 0

The answer is:

True

Hope this Helps

Please mark my nswer as brainliest?

Andru [333]3 years ago
7 0
The answer is true , for your questions hope this helps
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This tasty snack starts as a kernel and is “popped” then
Veronika [31]
Popcorn can be popped by either of the three forms of heat transfer:

Conduction in a pan with hot oil on a stove element.

Convection by an air popper and warm air rising over a heating element... no direct contact with heat source.

Radiation is what occurs in a microwave. Invisible radiant heat activates water molecules in the popcorn.

All the above heat the kernel over 100 Celsius. Water vaporizes/boils (latent heat) and erupts through the kernel.

Mmmm popcorn watch out for the lipids (fat in the oil and butter)
6 0
3 years ago
Which equations could be used as is, or rearranged to calculate for frequency of a wave? Check all that apply.
amm1812
-- Equations  #2  and  #6  are both the same equation,
and are both correct.

-- If you divide each side by  'wavelength', you get Equation #4,
which is also correct.

-- If you divide each side by  'frequency', you get Equation #3,
which is also correct. 
With some work, you can rearrange this one and use it to calculate
frequency.

Summary:

-- Equations #2, #3, #4, and #6 are all correct statements,
and can be used to find frequency.

-- Equations #1 and #5 are incorrect statements.
7 0
3 years ago
Read 2 more answers
A point charge Q is placed at the center of a conducting spherical shell (inner radius a, outer radius b). What is the electric
sp2606 [1]

Answer:

a)E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

b)E=0

c)E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

Explanation:

Given that

A point charge Q is placed at the center of a conducting spherical shell .Due to this - Q charge will induce on the inner sphere surface and +Q will induce on the outer sphere surface .

a)    r < a

At a radius r ,from gauss theorem

E.ds=\dfrac{q_i}{\varepsilon _o}

E\times 4\pi r^2=\dfrac{Q}{\varepsilon _o}

E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

b)   a < r < b

E.ds=\dfrac{q_i}{\varepsilon _o}

The total induce in this surface = - Q+ Q =0

E.ds=\dfrac{0}{\varepsilon _o}

E = 0

c)   r > b

E.ds=\dfrac{q_i}{\varepsilon _o}

E\times 4\pi r^2=\dfrac{Q}{\varepsilon _o}

E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

4 0
3 years ago
A voltaic cell is constructed with two Zn2+-Zn electrodes, where the half-reaction is Zn2+ + 2e− → Zn (s) E° = -0.763 V The conc
wariber [46]

Answer : The cell emf for this cell is 0.077 V

Solution :

The balanced cell reaction will be,  

Oxidation half reaction (anode):  Zn(s)\rightarrow Zn^{2+}+2e^-

Reduction half reaction (cathode):  Zn^{2+}+2e^-\rightarrow Zn(s)

In this case, the cathode and anode both are same. So, E^o_{cell} is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}{diluted}}{[Zn^{2+}{concentrated}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

[Zn^{2+}{diluted}] = 0.0111 M

[Zn^{2+}{concentrated}] = 4.50 M

Now put all the given values in the above equation, we get:

E_{cell}=0-\frac{0.0592}{2}\log \frac{0.0111M}{4.50M}

E_{cell}=0.077V

Therefore, the cell emf for this cell is 0.077 V

7 0
3 years ago
Steve is recording the temperature of an object on Celsius and Fahrenheit scales. Relationship between the two scales is given b
Vlad [161]

Answer:

False

Explanation:

7 0
3 years ago
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