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-BARSIC- [3]
2 years ago
5

light traveling through air encounters a second medium which slows the light to 100,000 miles/second. What is the index of the s

econd medium?
Physics
2 answers:
IceJOKER [234]2 years ago
6 0

Answer:

300000 hope it is help full

Explanation:

300000

jenyasd209 [6]2 years ago
4 0
Answer 3000

Explanation:
The formula is n= c/v, “n” means index, “c” means speed of light in vacuum and “v” means speed of light in the material
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An object of mass 800g hangs on a spring. Calculate the force exerted by the object if acceleration due to gravity is 10/s2​
xxTIMURxx [149]

Answer:

8 N

Explanation:

Using the equation F=ma (F: force/ m: mass  in kg/ a: acceleration),

F = (800/1000)(10)

F = 8 N

4 0
3 years ago
Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

The Coyote hits the ground when r_y=0:

15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

4b. Here we evaluate r_x at the time found in (4a).

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

5a. The vertical component of the shell's position vector is

r_y=1.52\,\mathrm m-\dfrac g2t^2

We find the shell hits the ground at

1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

5b. The horizontal component of the bullet's position vector is

r_x=v_0t

where v_0 is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for v_0:

3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

5 0
3 years ago
Walk in a straight line. Now stop. Did you accelerate? Explain
Brums [2.3K]

Answer;

It's about acceleration, right?

4 0
3 years ago
Read 2 more answers
a circular cylinder and isused to maintain a water depth of 4 m. That is, when the water depth exceeds 4 m, thegate opens slight
stiv31 [10]

Answer:

  W / A = 39200 kg / m²

Explanation:

For this problem let's use the equilibrium equation of / newton

           F = W

Where F is the force of the door and W the weight of water

         W = mg

We use the concept of density

        ρ = m / V

        m = ρ V

The volume of the water column is

          V = A h

We replace

         W = ρ A h g

On the other side the cylinder cover has a pressure

          P = F / A

          F = P A

We match the two equations

       P A = ρ A h g

        P = ρ g h

        P = 39200 Pa

The weight of the water column is

       W  = 1000 9.8 4 A

       W / A = 39200 kg / m²

3 0
3 years ago
NEED HELP TODAY
Free_Kalibri [48]

Answer:

I am pretty sure it is B My friend hope you are well

Explanation:

6 0
3 years ago
Read 2 more answers
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