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Otrada [13]
3 years ago
5

Which of the following is not a common property of waves?

Physics
1 answer:
Inessa05 [86]3 years ago
7 0
Mass, as u can't weigh the waves!!
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The atoms ,molecules,or compound present at the start of a chemical reaction that parcitipate in the reaction . which part of a
Tomtit [17]

Answer:

The atoms ,molecules, or compound present at the start of a chemical reaction that parcitipate in the reaction are <u><em>the reactants.</em></u>

Explanation:

The chemical reaction is the way in which one substance reacts against another. So, a chemical reaction consists of the transformation of some substances into others, that is, the process of arranging atoms and bonds when chemical substances come into contact.

In a chemical reaction, the initial substances are called reactants, while the new substances obtained are called products.

So, <u><em>the atoms ,molecules, or compound present at the start of a chemical reaction that parcitipate in the reaction are the reactants.</em></u>

8 0
4 years ago
NEED HELP ASAP PLEASE
Zarrin [17]

Answer:

D. 3.0 m/s

Explanation:

because I did this in my class

8 0
3 years ago
An automobile traveling 50km/hr. It decelerates at 5 meter per second square. How long will it take to stop the car? How far wil
const2013 [10]
We know, a = v/t
Here, a = 5 m/s²
v = 50 km/h= 13.88 m/s

Substitute their values into the expression:
5 = 13.88 / t
t = 13.88/5
t = 2.78 sec

Now, we know, v = d/t
13.88 = d/2.78
d = 13.88 * 2.78
d = 38.53 meter

In short, Your Answers would be:
i) It will take 2.78 sec
ii) It will travel for 38.53 m after a brakes applied.

Hope this helps!
3 0
3 years ago
A spring with a pointer attached is hanging next to a scale marked in millimeters. Three different packages are hung from the sp
SOVA2 [1]

solution;

the expression for force applied on the spring due to the load is\\f=k\Delta x\\here,\Delta x is the extension in the spring due to appling force\\given three case as following\\110N=k(40-x_{o})----------1\\240N=k(60-x_{o})----------2\\w=k(30-x_{o})-------------3\\To calculate the accrual length of the spring,solve th equation 1 and 2\\\frac{110N}{240N}=\frac{k(40-x_{o})}{k(60-x_{o})}\\0.458=\frac{k(40-x_{o})}{k(60-x_{o})}\\0.458(60mm-x_{o})=(40mm-x_{o})\\x_{o}(1-0.458)=(40-60(0.458))mm\\x_{o}\frac{12.52}{0.542}\\=23.1mm\\to calculate the force on the spring in case,\\solve the equation 1 and 2\\\frac{110}{w}=\frac{k(40-x_{o})}{k(60-x_{o})}\\\frac{110}{w}=\frac{(40mm-23.1mm)}{30mm-23.1mm}\\w=\frac{110}{2.45}=44.9N

8 0
3 years ago
Light rays from stars bend toward smaller angles as they enter Earth's atmosphere. a. Explain why this happens using Snell's law
Grace [21]

Answer:

Following are the answer to this question:

Explanation:

In option (a):

  • The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle.  
  • Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.

In option (b):

  • Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth.  
  • Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.

8 0
3 years ago
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