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2 years ago

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:

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

What is kinematics ?explain ~​

What is kinematics ?explain ~