What activity in orienteering can improve a player's math skills?

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:

a)

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

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

b)

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

3 years ago

An oscillator makes 322.1 oscillations in 4.255 minutes. What is the period of the oscillator?

sergij07 [2.7K]

Ummm I’m not sure let me do the work

4 0

3 years ago

Which is an example of radiation?

Y_Kistochka [10]

A common example of radiation would have to be sunlight

6 0

3 years ago

Read 2 more answers

Ryan is a nine-year-old boy whose father wants him to be a soccer player. Ryan does not have an aptitude for soccer, and keeps f

mihalych1998 [28]

A bad life when Ryan is older and will criticize his kids

8 0

3 years ago

Read 2 more answers

The hour and minute hands of a tower clock like Big Ben in London are 2.79 m and 4.44 m long and have masses of 58.2 kg and 90 k

LuckyWell [14K]

Answer: 895.85 x 10^-6 J or 8.96 x 10^-4 J

Explanation:

Angular kinetic energy E in Joules

E = ½Iw^2

W is angular velocity in radians/sec

1 radian/sec = 9.55 rev/min

I is moment of inertia in kgm^2

I = cMR^2

M is mass (kg), R is radius (meters)

c = 1/3 for a rod around its end, R = length

For minute hand

I = (1/3)(90)(4.44)^2 = 0.33 x 90 x 19.7136 = 585.49

w= 1 rev/hour = 1 rev/3600sec = 2pi/ 3600 = pi/1800 rad/s

KE = 1/2(1/3)(90)(4.44)^2(pi/1800)^2 = 0.5 x 0.33 x 90 x 19.7136 x 3.05 x 10^-6

KE = 0.00089 J

For hour hand

I = (1/3)(58.2)(2.79)^2 = 0.33 x 58.2 x 2.79^2 = 149.5

w = 1 rev/12hour = 1 rev/(12x3600sec) = 2pi/ 12x3600 = pi/21600 rad/s

KE = 1/2(1/3)(90)(4.44)^2(pi/21600)^2 = 0.5 x 0.33 x 90 x 19.7136 x 2.12 x 10^-8

KE = 5.85 x 10^-6 J

Therefore total kinetic energy = 895.85 x 10^-6 J

4 0

4 years ago