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3 years ago

Light travels faster in a material with a lower index of refraction.

Physics

2 answers:

DochEvi [55]3 years ago

6 0

Answer:

false

Explanation:

lidiya [134]3 years ago

4 0

Answer:true

Explanation:

The lower the refractive index,the faster the velocity of light

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Technician A says that cabin filters are accessible behind the glove compartment. Technician B says that cabin filters are acces

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Answer:

but it depends on the car

Explanation:

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3 years ago

The photo shows a pair of figure skaters performing a spin maneuver. The

solmaris [256]

Answer: the skater on the left pulls the skater on the right closer to him

Explanation:

A PEX this is for increasing their velocity so read carefully

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3 years ago

A 1200-kg car is travelling east at a rate of 9 m/s. A 1600-kg truck is travelling south at a rate of 13 m/s. The truck accident

weeeeeb [17]

Answer:

Around 3.57m/s

Explanation:

p=mv

Let's denote the momentum, mass, and velocity of the car with the subscript 1, and for the truck use 2. After the collision, the combined momentum can be denoted with the subscript 3.

$p_1=1200\cdot 9=10800 \\\\p_2=1600\cdot 13=20800 \\\\20800-10800=10000 \\\\1200kg+1600kg=2800kg \\\\10000=2800v_3 \\\\v_3\approx 3.57m/s$

Hope this helps!

3 0

3 years ago

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

3 years ago

A car travels at a constant speed up a ramp making an angle of 28 degrees with the horizontal component of velocity is 40 kmh^-1

Hoochie [10]

Answer:

Vx = 35.31 [km/h]

Vy = 18.77 [km/h]

Explanation:

In order to solve this problem, we must decompose the velocity component by means of the angle of 28° using the cosine function of the angle.

$v_{x} = 40*cos(28)\\V_{x} = 35.31 [km/h]$

In order to find the vertical component, we must use the sine function of the angle.

$V_{y}=40*sin(28)\\V_{y} = 18.77 [km/h]$

7 0

3 years ago