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erastova [34]
2 years ago
8

What is kinematics ?explain ~​

Physics
2 answers:
hodyreva [135]2 years ago
7 0

Answer:

the branch of mechanics concerned with the motion of objects without reference to the forces which cause the motion.

and/or

Kinematics is the study of motion of a system of bodies without directly considering the forces or potential fields affecting the motion. In other words, kinematics examines how the momentum and energy are shared among interacting bodies.

Vlad [161]2 years ago
3 0

Answer:

Kinematics is the branch of physics in which we study the motion of an object without studying the cause of it's motion.

<h3>hope it helps</h3>

have a good day Sir

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The terminal velocity is not dependent on which one of the following properties? the drag coefficient 1 the force of gravity 2 c
ahrayia [7]
<h2>Answer: the falling time</h2>

Explanation:

When a body or object falls, basically two forces act on it:  

1. The force of air friction, also called<em> </em><u><em>"drag force"</em></u> D:  

D={C}_{d}\frac{\rho V^{2} }{2}A  (1)

Where:  

C_ {d} is the drag coefficient  

\rho is the density  of the fluid (air for example)

V is the velocity  

A is the transversal area of the object

So, this force is proportional to the transversal area of ​​the falling element and to the square of the velocity.  

2. Its <u>weight </u>due to the gravity force W:  

W=m.g

(2)

Where:  

m is the mass of the object

g is the acceleration due gravity  

So, at the moment <u>when the drag force equals the gravity force, the object will have its terminal velocity:</u>

D=W (3)

{C}_{d}\frac{\rho V^{2} }{2}A=m.g  (4)

V=\sqrt{\frac{2m.g}{\rho A{C}_{d}}}  (5) This is the terminal velocity

As we can see, there is no "falling time" in this equation.

Therefore, the terminal velocity is not dependent on the falling time.

6 0
3 years ago
An airplane is flying from Dallas, Texas to Pensacola, Florida. Flying at maximum velocity, it encounters strong winds moving at
jenyasd209 [6]

Answer:

G. It will take twice as long.

Explanation:

Let's call v the original speed of the plane and d the distance between Dallas and Pensacola. The time the plane originally takes to complete the flight is

t=\frac{d}{v}

In this problem, we are told that the plane encounters wind moving at half of its speed: \frac{v}{2}, in the opposite direction. This means that the new speed of the plane is

v'=v-\frac{v}{2}=\frac{v}{2}

And so, the time the plane takes now to complete the flight is

t'=\frac{d}{v/2}=2\frac{d}{v}=2t

So, the plane takes twice the time as before.

4 0
3 years ago
The flow of electrons through a circuit is measured in which of the following units? A. electrical pressure B. amperes C. volts
damaskus [11]

The total quantity of electrons that have flowed through a circuit is a
quantity of charge, measured in Coulombs, or in Ampere-seconds.

The <em><u>rate</u></em> of flow of electrons, or more accurately the rate of flow of
the charge on them, is electrical current.  Its unit is the Ampere. 
1 Ampere is 1 Coulomb of charge per second.


8 0
3 years ago
After hearing about an accident on his normal route, Mr. Gujral checks for alternate routes to get to work. What type of circuit
shtirl [24]

Answer:

C

Explanation:

a parallel circuit because there is more than one path

3 0
3 years ago
A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra
bija089 [108]

Explanation:

Terminal velocity is given by:

v_t=\sqrt{\frac{2mg}{\rho C_dA}}

Here, m is the mass of the falling object, g is the gravitational acceleration,  C_d is the drag coefficient, \rho is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is 1.28\frac{kg}{m^3}.

v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}

7 0
3 years ago
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