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kodGreya [7K]
3 years ago
14

A ray of light approaches a jar of honey at an angle of 30.0. If the angle of refraction is 19.5, what is the refractive index o

f honey?
Physics
1 answer:
Stels [109]3 years ago
7 0

Answer:

n = 1.49

Explanation:

Given that,

Angle of incidence = 30 degrees

Angle of refraction = 19.5 degrees

We need to find the refractive index of the honey. Using Snell's law, we can find it as follows :

n=\dfrac{\text{sin i}}{\text{sin r}}\\\\=\dfrac{\text{sin 30}}{\text{sin 19.5}}\\\\n=1.49

So, the refractive index of honey is 1.49.

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How is energy transferred when<br>hitting a nail?​
Juli2301 [7.4K]

Answer:

kinetic energy

kinetic energy

7 0
3 years ago
A projectile is launched from the ground with an initial velocity of 12ms at an angle of 30° above the horizontal. The projectil
netineya [11]

vi^{2}sin2thita/g =12^{2}sin2[30]/9.8=12.7Answer:

Explanation:

range is given as

6 0
3 years ago
Read 2 more answers
A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.
babunello [35]
A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.


Read more on Brainly.com - brainly.com/question/18743384#readmore
8 0
3 years ago
A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?
Tcecarenko [31]
The weight of the meterstick is:
W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance 
d_1 = 0.50 m - 0.40 m=0.10 m
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
(mg) d_2 = 0.20 Nm
from which we find the value of d2:
d_2 =  \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
4 0
3 years ago
It is known that the kinetics of recrystallization for some alloy obeys the Avrami equation, andthat the value of n in the expon
trapecia [35]

Answer:8.76\times 10^{-3} min^{-1}

Explanation:

Given

n=5

0.3 fraction recrystallize after 100 min

According to Avrami equation

y=1-e^{-kt^n}

where y=fraction Transformed

k=constant

t=time

0.3=1-e^{-k(100)^5}

e^{-k(100)^5} =0.7

Taking log both sides

-k\cdot (10^{10}=\ln 0.7

k=3.566\times 10^{-11}

At this Point we want to compute t_{0.5}\ i.e.\ y=0.5

0.5=1-e^{-kt^n}

0.5=e^{-kt^n}

0.5=e^{-3.566\times 10^{-11}\cdot (t)^5}

taking log both sides

\ln 0.5=-3.566\times 10^{-11}\cdot (t)^5

t^5=1.943\times 10^{10}

t=114.2 min

Rate of Re crystallization at this temperature

t^{-1}=8.76\times 10^{-3} min^{-1}

3 0
3 years ago
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