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madam [21]
3 years ago
8

My fifth time asking this, but can someone please help me with these four????

Physics
1 answer:
givi [52]3 years ago
3 0
I think 3 is 1.5...i kinda hope that helps for one
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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:

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
Suppose you are in total equilibrium in water (in other words you are floating and not moving in any direction)What could you do
Eva8 [605]

Explanation:

According to newtons first law of motion:

'' a body will continue in its state of rest or uniform motion along a path unless it is acted upon by an external force".

A body in equilibrium that is floating will be stable and not move in any direction. Even if it moves, the motion will be constant wouldn't change.

  • To move the body in any direction, one has to swim.
  • Swimming is the application of an external force to counter the balanced forces at equilibrium on a body.
  • This works when the net external force is greater than that the balanced forces.

learn more:

Newton brainly.com/question/11411375

#learnwithBrainly

5 0
3 years ago
. A 1.50kg mass on a spring has a displacement as a function of time given by the equation: x(t) = (7.40cm)cos[(4.16s-1)t – 2.42
rusak2 [61]

Answer:

Solution:

we have given the equation of motion is x(t)=8sint [where t in seconds and x in centimeter]

Position, velocity and acceleration are all based on the equation of motion.

The equation represents the position.  The first derivative gives the velocity and the 2nd derivative gives the acceleration.

x(t)=8sint

x'(t)=8cost

x"(t)=-8sint

now at time t=2pi/3,

position, x(t)=8sin(2pi/3)=4*squart(3)cm.

velocity, x'(t)=8cos(2pi/3)==4cm/s

acceleration, x"(t)==8sin(2pi/3)=-4cm/s^2

so at present the direction is in y-axis.

5 0
3 years ago
Rock candy is made from crystallized sugar. What is crystallization?
ankoles [38]

Answer: Rock candy is candy made of large sugar crystals. To make rock candy, a supersaturated solution of sugar in water is created and left undisturbed for a few days. The driving force behind crystallization is supersaturation.

7 0
3 years ago
Read 2 more answers
Suppose light from a 632.8 nm helium-neon laser shines through a diffraction grating ruled at 520 lines/mm. How many bright line
Leya [2.2K]

Answer:

1 bright fringe every 33 cm.

Explanation:

The formula to calculate the position of the m-th order brigh line (constructive interference) produced by diffraction of light through a diffraction grating is:

y=\frac{m\lambda D}{d}

where

m is the order of the maximum

\lambda is the wavelength of the light

D is the distance of the screen

d is the separation between two adjacent slit

Here we have:

\lambda=632.8 nm = 632.8\cdot 10^{-9} m is the wavelength of the light

D = 1 m is the distance of the screen (not given in the problem, so we assume it to be 1 meter)

n=520 lines/mm is the number of lines per mm, so the spacing between two lines is

d=\frac{1}{n}=\frac{1}{520}=1.92\cdot 10^{-3} mm = 1.92\cdot 10^{-6} m

Therefore, substituting m = 1, we find:

y=\frac{(632.8\cdot 10^{-9})(1)}{1.92\cdot 10^{-6}}=0.330 m

So, on the distant screen, there is 1 bright fringe every 33 cm.

6 0
3 years ago
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