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3 years ago

If a skydiver weighs 800 N, what would air resistance be when he is at terminal velocity?

2 answers:

7 0

Let his weight be mg.

For a falling body in a fluid.

W - V - U = ma. Where
   W = Weight = 800N.
      m = mass
       V = Viscous Force = Air Resistance.
   U = upthrust = Negligible for air.
   a = acceleration
   g = acceleration due to gravity.
Note also at terminal velocity a = zero.

W - V - 0 = m(0)
800 - V = 0
800 = V.

Therefore the viscous force which is equivalent to air resistance is also 800N.
Cheers.

serg [7]3 years ago

5 0

We call "terminal velocity" the constant speed of a falling body
when it is no longer accelerating.

We know that if a body is not accelerating, then the net force
on it is zero.

From the question, we know that the downward force of gravity
on the skydiver is 800 N.

If the 800 N downward plus the air resistance upward add up
to zero, then the air resistance upward must also be 800 N.

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A 12,500 N alien UFO is hovering about the surface of Earth. At time , its position can be given as () = ((0.24 m/s^3)^3 + 25 m)

stepladder [879]

a) $F=(3675i-4543k)N$

b) 5843 N

Explanation:

The position of the UFO at time t is given by the vector:

$r(t)=(0.24t^3+25)i+(4.2t)j+(-0.43t^3+0.8t^2)k$

Therefore it has 3 components:

$r_x=0.24t^3+25\\r_y=4.2t\\r_z=-0.43t^3+0.8t^2$

We start by finding the velocity of the UFO, which is given by the derivative of the position:

$v_x=r'_x=\frac{d}{dt}(0.24t^3+25)=3\cdot 0.24t^2=0.72t^2\\v_y=r'_y=\frac{d}{dt}(4.2t)=4.2\\v_x=r'_z=\frac{d}{dt}(-0.43t^3+0.8t^2)=-1.29t^2+1.6t$

And then, by differentiating again, we find the acceleration:

$a_x=v'_x=\frac{d}{dt}(0.72t^2)=1.44t\\a_y=v'_y=\frac{d}{dt}(4.2)=0\\a_z=v'_z=\frac{d}{dt}(-1.29t^2+1.6t)=-2.58t+1.6$

The weight of the UFO is W = 12,500 N, so its mass is:

$m=\frac{W}{g}=\frac{12500}{9.8}=1276 kg$

Therefore, the components of the force on the UFO are given by Newton's second law:

$F=ma$

So, Substituting t = 2 s, we find:

$F_x=ma_x=(1276)(1.44t)=(1276)(1.44)(2)=3675 N\\F_y=ma_y=0\\F_z=ma_z=(1276)(-2.58t+1.6)=(1276)(-2.58(2)+1.6)=-4543 N$

So the net force on the UFO at t = 2 s is

$F=(3675i-4543k)N$

The magnitude of a 3-dimensional vector is given by

$|v|=\sqrt{v_x^2+v_y^2+v_z^2}$

where

$v_x,v_y,v_z$ are the three components of the vector

In this problem, the three components of the net force are:

$F_x=3675 N\\F_y=0\\F_z=-4543 N$

Therefore, substituting into the equation, we find the magnitude of the net force:

$|F|=\sqrt{3675^2+0^2+(-4543)^2}=5843 N$

7 0

2 years ago

The mass of an object is 275.32g and its density is 7.562g/cm 3 . Calculate its volume by keeping significant figures in view.

Genrish500 [490]

Answer:

36.408cm3

Explanation:

Since we acknowledge that density is d= m/v, once we switch it up to maintain v as the number to be found it will change to v=m/d. Therefore, 275.32/7.562 is 36.408 and the unit is cm cube!

Hope that helped!!

8 0

3 years ago

A race car traveling at 44m/s slows at a constant rate to a velocity of 22m/s over 11 seconds , how far does it move during this

krok68 [10]

Answer:

add 44m/s and 22m/s then multiply it by 11

Explanation:

8 0

2 years ago

Find the intensity of the electromagnetic wave described in each case. (a) an electromagnetic wave with a wavelength of 630 nm a

qaws [65]

Find the intensity of the electromagnetic wave described in each case.

(a) an electromagnetic wave with a wavelength of 645 nm and a peak electric field magnitude of 8.5 V/m.

(b) an electromagnetic wave with an angular frequency of 6.3 ✕ 1018 rad/s and a peak magnetic field magnitude of 10−10 T.

3 0

3 years ago

X-rays are used instead of visible light because X-rays have shorter ______________ than visible light, allowing them to produce

Novay_Z [31]

X-rays have shorter wavelength than visible light. But that's hardly the reason that they're used for medical imaging. xrays have much higher frequencies then visible light which means they have much greater penetrating ability. with xrays you can see inside the body. you can't do that with a visible flashlight no matter how bright and powerful it is.

4 0

3 years ago

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