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3 years ago

If the refractive index of a medium is 1.3,

Physics

2 answers:

Nataly [62]3 years ago

3 0

Explanation:

see the above attachment to solve the question and get the answer.

hope this helps you.

Lena [83]3 years ago

3 0

Answer:

We know that refractive index of medium is n=

speedoflightinvacuum

speedoflightinvaccum

c is the velocity of light in vacuum and v is velocity in that medium.

For water :

1.3=

2.25×10

c=2.925×10

m/s

For glass :

c=2.925×10

m/s

glass

1.5=

glass

2.925×10

glass

=1.95×10

m/s

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(1) A net force of 265 N accelerates a bike and rider at 2.30 m/s. What is the mass of the

kati45 [8]

Answer: 115.2kg

Explanation:

Net force = 265 N

Acceleration of bike & rider = 2.30m/s2 (The SI unit of acceleration is m/s2)

Mass of the bike and rider together = ?

Since force is the product of the mass of an object and the acceleration by which it moves, Force = Mass x Acceleration

265N = Mass x 2.30m/s2

Mass = (265N/2.30m/s2)

Mass = 115.2 kg

Thus, the Mass of the bike and rider together is 115.2kg

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4 years ago

Ben Rushin is waiting at a stop light. Turns green, ben accelerated from rest at a rate of 6.00 m/s squared for a time of 4.10 s

Lera25 [3.4K]

D= vt +.5at^2
since he started at rest, v (initial velocity) is 0
so d=.5at^2
d = .5 (6m/s^2) (4.1s)^2
then put that into a calculator.

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3 years ago

A car with a velocity of 6.4 m/s, forward, accelerates to a velocity of 10.6 m/s, forward, in 16 s. What is the card acceleratio

tatuchka [14]

Answer:

acceleration = 0.2625 m/s²

Explanation:

acceleration = ( final velocity - initial velocity ) / time

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using the equation:

acceleration = ( 10.6 - 6.4 ) / 16

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6 0

3 years ago

Read 2 more answers

A 15 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60° angle with horizontal. (a) Find t

scoray [572]

Answer:

a) F₁ = 267.3 N, N₁ = 1300 N, b) μ = 0.324

Explanation:

For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces

let's use subscript 1 for the ladder and 2 for the firefighter

∑ τ = 0

-W₁ x₁ - W₂ x₂ + N₁ y = 0

N₁ = $\frac{W_1 x_1 + W_2 x_2}{y}$ (1)

the center of mass of the ladder is at its geometric center,

d = L / 2 = 15/2 = 7.5 m

cos 60 = x₁ / d₁

x₁ = d₁ cos 60

x₁ = 7.5 cos 60

x₁ = 3.75 m

for the firefighter d₂ = 4 m

cos 60 = x₂ / d₂

x₂ = d₂ cos 60

x₂ = 4 cos 60 = 2 m

for the fulcrum d₃ = 15 m

sin 60 = y / d₃

y = d₃ sin 60

y = 15 sin 60

y = 13 m

we look for the Normal by substituting in equation 1

N₂ = $\frac{500 \ 3.75 \ + 800 \ 2}{13}$

N₂ = 267.3 N

now let's use the translational equilibrium relations

X axis

F₁ - N₂ = 0

F₁ = N₂

F₁ = 267.3 N

Axis y

N₁ - W₁ -W₂ = 0

N₁ = W₁ + W₂

N₁ = 500 + 800

N₁ = 1300 N

b) for this case change the firefighter's distance d₂ = 9 m

x₂ = 9 cos 60

x₂ = 4.5 m

we substitute in 1

N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}

N₂ = 421.15 N

of the translational equilibrium equation on the x-axis

fr = F₁ = N₂

fr = 421.15 N

friction force has the expression

fr = μ N

in this case the reaction of the Earth to the support of the ladder is N1 = 1300N

μ = fr / N₁

μ = 421.15 / 1300

μ = 0.324

8 0

2 years ago

A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision

Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is $a=2,800m/s^{2}$

Explanation:

Known Data

We will asume initial speed has a negative direction, $v_{i}=-30m/s$ , final speed has a positive direction, $v_{f}=26m/s$ , $\Delta t=20ms=0.020s$ and mass $m_{b}$ .