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gtnhenbr [62]
3 years ago
5

The terminal velocity is not dependent on which one of the following properties? the drag coefficient 1 the force of gravity 2 c

ross-sectional area 3 air density 4 the falling time 5 terminal velocity depends on all of the 6 given parameters
Physics
1 answer:
ahrayia [7]3 years ago
6 0
<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.

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Explanation:

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How far (in feet) could a pitcher throw a baseball on flat, level ground if he can throw it at 100mph? (Neglect wind drag and th
ehidna [41]

Answer:

R = 668.19 ft  

Explanation:

given,

speed of the ball thrown by the pitcher = 100 mph

to travel maximum distance θ = 45°

distance traveled by the ball = ?

using formula

1 mph = 0.44704 m/s

100 mph = 44.704 m/s

R = \dfrac{u^2sin2\theta}{g}

R = \dfrac{44.704^2sin2\times 45}{9.81}

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R = 203.71 × 3.28

R = 668.19 ft  

hence, ball will go at a distance of 668.19 ft   when pitcher throw it at 100 mph.

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3 years ago
A parking lot is going to be 60 m wide and 240 m long. what dimensions could be used for a scale model of the lot?​
Zanzabum

Answer:

it is 20cm x 80cm

Explanation:

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4 years ago
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A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a
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(a) 4.0 s

The acceleration of the car is given by

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

For this car, we have

v = 0 (the final speed is zero since the car comes to a stop)

u = 20 m/s is the initial velocity

a=-5.0 m/s^2 is the deceleration of the car

Solving the equation for t, we find the time needed to stop the car:

t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s

(b) 40 m

The stopping distance of the car can be calculated by using the equation

v^2 - u^2 = 2ad

where

v = 0 is the final velocity

u = 20 m/s is the initial velocity

a = -5.0 m/s^2 is the acceleration of the car

d is the stopping distance

Solving the equation for d, we find

d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m

(c) -5.0 m/s^2

The deceleration is given by the problem, and its value is -5.0 m/s^2.

(d) 5000 N

The net force applied on the car is given by

F=ma

where

m is the mass of the car

a is the magnitude of the acceleration

For this car, we have

m = 1000 kg is the mass

a=5.0 m/s^2 is the magnitude of the acceleration

Solving the formula, we find

F=(1000 kg)(5.0 m/s^2)=5000 N

(e) 2.0\cdot 10^5 J

The work done by the force applied by the car is

W=Fd

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F is the force applied

d is the total distance covered

Here we have

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d = 40 m (stopping distance)

So, the work done is

W=(5000 N)(40 m)=2.0\cdot 10^5 J

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when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).

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