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Advocard [28]
3 years ago
15

Give a specific example of a scalar or vector quantity. Explain why thus type of quantity is used to describe this event.

Physics
1 answer:
dlinn [17]3 years ago
3 0

Example of scalar: speed. Example of vector: velocity

Explanation:

In physics, there are two types of quantities:

- Scalar: a scalar quantity is a quantity having only magnitude, so it is just a number followed by a unit. Examples of scalar quantities in physics are:

Mass

Time

Speed

- Vector: a vector quantity is a quantity having both a magnitude and a direction. Examples of vector quantities in physics are:

Force

Acceleration

Velocity

The two types of quantities can be used in the same event, but in a different way. One of the most common example is the difference between speed and velocity.

In fact, let's consider an object moving in a uniform circular motion: it means that it is moving in a circle at a constant speed. The speed of the object measures only how fast the object is moving, but without telling anything about its direction of motion. The velocity, viceversa, also takes into account the direction of motion, and exactly for this reason, the velocity in a uniform circular motion is not constant, because the direction (it is a vector) is constantly changing. So, in a uniform circular motion, the speed is constant but the velocity is not.

Learn more about vectors:

brainly.com/question/2678571  

brainly.com/question/4945130  

#LearnwithBrainly

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To compare and rank each one's acceleration, get the acceleration of each person by using the formula, Acceleration = Velocity/Time

Xander = 4.5mps / 3.5s = 1.29m/s^2
Finley = 3.6mps/4.2s = 0.86m/s^2
Max = 7.3mps/1.2s = 6.08m/s^2

Based on the answers above, the least acceleration was done by Finley, then Xander had higher acceleration than Finley, while most acceleration was done by Max.

3 0
3 years ago
Work of 2 joules is done in stretching a spring from its natural length to 11 cm beyond its natural length. what is the force (i
Sidana [21]
The law applied here is Hooke's Law which describes the force exerted by the spring with a given distance. The equation for this is F = kΔx, where F is the force in Newtons, k is the spring constant in N/m while Δx is the displacement in meters.

If you want to find work done by a spring, this can be solved by using differential equations. However, derived equations are already ready for use. The equation is

W = k[{x₂-x₁)² - (x₁-xn)²],

where 
xn is the natural length
x₁ is the stretched length 
x₂ is also the stretched length when stretched even further than x₁

In this case xn =x₁. So, that means that (x₁-xn) = 0 and (x₂-x₁) = 11 cm or 0.11 m.

Then, substituting the values,

2 J = k (0.11² -0²)
k = 165.29 N/m

Finally, we use the value of k to the Hooke's Law to determine the Force.

F = kΔx = (165.29 N/m)(0.11 m)
F = 18.18 Newtons
5 0
4 years ago
What is stomato and write its functions ​
Nataly_w [17]

Answer:

Stomata- In botany, a stoma (also stomate; plural stomata) is a tiny opening or pore that is used for gas exchange. They are mostly found on the under-surface of plant leaves. Its functions are- The gas exchange that occurs when stomata are open facilitates photosynthesis. Mark Brainliest Pls

Explanation:

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4 years ago
A __________ change involves a change in one or more physical properties, but no change in the fundamental components that make
ch4aika [34]
Physical change :) Hope you get it right! :D Good luck! let me know what else I can help you with. Have a great day! :D 
3 0
3 years ago
A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac
RoseWind [281]

Answer:

a) \Delta U_{g} = 12.945\,J, b) \Delta U_{k} = 12.945\,J, c) k = 2930.059\,\frac{N}{m}

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)

\Delta U_{g} = 12.945\,J

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

\Delta U_{k} = 12.945\,J

c) The spring constant of the gun is:

\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}

k = \frac{2\cdot \Delta U_{k}}{x^{2}}

k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}

k = 2930.059\,\frac{N}{m}

4 0
3 years ago
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