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3 years ago

26. The speed of light in a certain medium is

1 answer:

6 0

The number we need in order to answer the question belongs in the space between the words "is" and "of". You left that blank blank, so there really isn't any question here to answer.

HOWEVER ... the refractive index of a medium can never be less than 1.0 , so we know for sure that <em>choice-a can't be</em> the correct answer.

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A 5.00 kilogram mass is traveling at 100. meters per second. Determine the speed of the mass after an impulse with a magnitude o

faltersainse [42]

m = mass = 5 kg

$v_{i}$ = initial velocity = 100 m/s

$v_{f}$ = final velocity = ?

I = impulse = 30 Ns

Using the impulse-change in momentum equation

I = m( $v_{f}$ - $v_{i}$ )

30 = 5 ( $v_{f}$ - 100)

$v_{f}$ = 106 m/s

5 0

3 years ago

Read 2 more answers

The atmospheric pressure on the top of the Engineering Sciences Building (ESB) is 97.6 kPa, while that in Room G39-ESB (ground f

Stells [14]

Answer:

Δ h = 52.78 m

Explanation:

given,

Atmospheric pressure at the top of building = 97.6 kPa

Atmospheric pressure at the bottom of building = 98.2 kPa

Density of air = 1.16 kg/m³

acceleration due to gravity, g = 9.8 m/s²

height of the building = ?

We know,

Δ P = ρ g Δ h

(98.2-97.6) x 10³ = 1.16 x 9.8 x Δ h

11.368 Δ h = 600

Δ h = 52.78 m

Hence, the height of the building is equal to 52.78 m.

6 0

3 years ago

Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros

kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

$v^2-u^2=2gh$

where

v = 0 is the velocity at the maximum height

u is the initial velocity

$g=-9.8 m/s^2$ is the acceleration of gravity

Solving for u,

$u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s$

The time needed to reach the maximum height can now be found by using the equation

$v=u+gt$

Solving for t,

$t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s$

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

$v'^2-u^2=2gh'$

we find

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s$

So the time to go from h' to h is

$\Delta t = t-t'=0.52-0.32=0.20 s$

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

$\Delta t=2\cdot 0.20 = 0.40 s$