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777dan777 [17]
3 years ago
9

For refraction to occur in a wave, the wave must

Physics
2 answers:
UNO [17]3 years ago
6 0
Enter a new medium at an angle.
fgiga [73]3 years ago
3 0

The answer is:

C - enter a new medium at an angle

Hope this helps :)

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Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p
Marta_Voda [28]

Answer:

Acceleration, a=9.91\times 10^{15}\ m/s^2

Explanation:

It is given that,

Separation between the protons, r=3.73\ nm=3.73\times 10^{-9}\ m

Charge on protons, q=1.6\times 10^{-19}\ C

Mass of protons, m=1.67\times 10^{-27}\ kg

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

ma=\dfrac{kq^2}{r^2}

a=\dfrac{kq^2}{mr^2}      

a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}      

a=9.91\times 10^{15}\ m/s^2

So, the acceleration of two isolated protons is 9.91\times 10^{15}\ m/s^2. Hence, this is the required solution.

3 0
3 years ago
Help!!Help!!Help!!Help!!
DerKrebs [107]

Answer:

y=8

Explanation:

every time you multiply x by 3 you divide y by 3.

x=2, multiply it by 3: x=6

y=24, divide it by 3: y=8

8 0
2 years ago
The magnitude of the momentum of an object is the product of its mass m and speed v. If m = 3 kg and v = 1.5 m/s, what is the ma
Oliga [24]

Answer:

Momentum, p = 5 kg-m/s

Explanation:

The magnitude of the momentum of an object is the product of its mass m and speed v i.e.

p = m v

Mass, m = 3 kg

Velocity, v = 1.5 m/s

So, momentum of this object is given by :

p=3\ kg\times 1.5\ m/s

p = 4.5 kg-m/s

or

p = 5 kg-m/s

So, the magnitude of momentum is 5 kg-m/s. Hence, this is the required solution.

5 0
3 years ago
is dimensionally correct relation necessarily to be a correct physical relation? explain with example.​
Andreas93 [3]

Answer: hope it helps you...❤❤❤❤

Explanation: If your values have dimensions like time, length, temperature, etc, then if the dimensions are not the same then the values are not the same. So a “dimensionally wrong equation” is always false and cannot represent a correct physical relation.

No, not necessarily.

For instance, Newton’s 2nd law is  F=p˙ , or the sum of the applied forces on a body is equal to its time rate of change of its momentum. This is dimensionally correct, and a correct physical relation. It’s fine.

But take a look at this (incorrect) equation for the force of gravity:

F=−G(m+M)Mm√|r|3r  

It has all the nice properties you’d expect: It’s dimensionally correct (assuming the standard traditional value for  G ), it’s attractive, it’s symmetric in the masses, it’s inverse-square, etc. But it doesn’t correspond to a real, physical force.

It’s a counter-example to the claim that a dimensionally correct equation is necessarily a correct physical relation.

A simpler counter example is  1=2 . It is stating the equality of two dimensionless numbers. It is trivially dimensionally correct. But it is false.

4 0
3 years ago
ONE EASY QUESTION
Brrunno [24]

Answer:

its tension force which acts in a string

Explanation:

need a thanks and thats it

8 0
3 years ago
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