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1 year ago

What is the difference between air resistance and friction?

1 answer:

5 0

Answer:Friction is a force that opposes the motion of objects; friction can cause objects to slow down. Air resistance is a type of friction. Air resistance causes moving objects to slow down. Different physical properties, such as the shape of an object, affect the air resistance on an object.

Explanation:

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A student is trying to determine the acceleration of a feather as she drops it to the ground. if the student is looking to achie

Anna [14]

The coordinate system should have the origin at the point where the feather is dropped and the downward direction is to be taken as positive.

All falling bodies experience acceleration towards the center of the Earth due to the force of gravitational attraction exerted on the object by the Earth. A feather, when dropped experiences an acceleration in the downward direction. Since the acceleration of the feather is in the downward direction, a feather, when dropped with zero initial velocity, has its velocity vector directed in the direction of its acceleration.

If the downward direction is taken as positive, the falling feather can be said to have a positive velocity and a positive acceleration.

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3 years ago

A machine lifts a 35 kg object doing 6860 J worth of work. How much power is produced by the machine if it lifts the object in 4

timofeeve [1]

Answer:

Explanation:

We need the power equation for this which is

P = Work/time

We have everything we need to solve this (the mass of the object is extra information):

P = 6860/4

P = 1715W

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3 years ago

To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r

sweet [91]

Answer:

a). 53.75 N and 101.92 N

b). 381.44 N and 723.25 N

Explanation:

$V= 77 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 21.38 \frac{m}{s} \\V=106 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 29.44 \frac{m}{s}$

a).

ρ $= 1.2 \frac{kg}{m^{3} }$ , $A_{t}= 0.7 m^{2}$ , $D_{t}= 0.28$

$F_{t1} = \frac{1}{2} * D_{t} * A_{t}* p_{t}* v_{t}^{2}$

$F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 53.75 N$

$F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 101.92 N$

b).

ρ $= 1.2 \frac{kg}{m^{3} }$ , $A_{h}= 2.44 m^{2}$ , $D_{h}= 0.57$

$F_{t1} = \frac{1}{2} * D_{h} * A_{h}* p_{h}* v_{h}^{2}$

$F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 381.44 N$

$F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 723.25 N$

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3 years ago

Suppose an object is launched from a point 320 feet above the earth with an initial velocity of 128 ft/sec upward, and the only

Ne4ueva [31]

Answer:

(a)Therefore the highest altitude attained by the object is =576 ft .

(b)Therefore the object takes 6 sec to fall to the ground.

Explanation:

Initial velocity: Initial velocity is a velocity from which an object starts to move.

u is usually used for notation of initial notation.

Final velocity: Final velocity is a velocity of an object after certain second from starting.

The final velocity is denoted by v.

Acceleration: The difference of final velocity and initial velocity per unit time

The S.I unit of acceleration is m/s².

(a)

Given that u= 128 ft\sec and g = 32 ft/sec².

At highest point the velocity of the object is 0 i.e v=0

Since the displacement is opposite to the gravity.

Therefore acceleration( a)= -g = -32 ft/sec².

To find the time this to happen we use the following formula

$v=u+at$

Here v=0

⇒0=128+(-32) t

⇒32t=128

⇒t = 4 sec

To determine the height we use the following formula

$s=ut+\frac{1}{2} at^2$

$\Rightarrow s= (128\times4)+\frac{1}{2}\times (-32) \times4^2$

⇒s= 256 ft

Therefore the highest altitude attained by the object is =(320+256)ft=576 ft .

(b)

At the highest point the velocity of the object is 0.

so u=0. a=g= 32 ft/sec² [ since the direction of gravity and the displacement are same] s= 576 ft

To determine the time to fall we use the following formula

$s=ut+\frac{1}{2} at^2$

$\Rightarrow 576 = (0\times t)+\frac{1}{2} \times 32 \times t^2$

$\Rightarrow 16\times t^2=576$

$\Rightarrow t^2=\frac{576}{16}$

$\Rightarrow t^2=36$

⇒t=6 sec

Therefore the object takes 6 sec to fall to the ground.

8 0

3 years ago

A wall clock has a second hand 22.0 cmcm long. For related problem-solving tips and strategies, you may want to view a Video Tut

olga55 [171]

Answer:

Explanation:

The tip of the second hand moves on a circular path having radius equal to .22 m . Redial acceleration is given by the expression

ω²R where ω is angular velocity and R is radius of the circular path .

angular velocity of second hand = 2π / T where T is time period of circular motion . For second hand it is 60 s.

ω = 2π / T

= 2π / 60

= .1047

angular acceleration = .1047² x .22

= 2.41 x 10⁻³ rad / s² .

8 0

3 years ago

What is the difference between air resistance and friction? ​

What is the difference between air resistance and friction?