One property that makes electromagnetic waves differ from other types of waves is that they can

My fifth time asking this, but can someone please help me with these four????

givi [52]

I think 3 is 1.5...i kinda hope that helps for one

3 0

3 years ago

Will binoculars work properly if their prisms (assume n = 1.50) are immersed in water (nwater= 1.33)? assume that binoculars wor

Amanda [17]

The binoculars will not work properly.

Please refer to the diagram attached.

Binoculars work on the principle of total internal reflection. The incident ray from air enters the prism through on face normally and it strikes the opposite face at an angle of 45°. The ray undergoes Total internal reflection if its critical angle is less than the angle of incidence,which , in this case is 45°.

The critical angle $i_{c}$ is related to the refractive index of the material of glass μ as shown below.

$\mu =\frac{1}{sini_c}$

For a refractive index equal to 1.5, the critical angle is found as shown below.

$sini_c=\frac{1}{\mu} \\ =\frac{1}{1.5} \\ =0.666$

Take the inverse of the value 0.666 to determine the critical angle.

$sin^{-1} (0.666)=41.8^o$

The critical angle in air is less than the angle of incidence, and hence total internal reflection occurs in air.

When the binoculars are immersed in water, the ray passes into the glass through water. therefore, the refractive index of the prism when immersed in water is given by,

$\mu_g_w=\frac{\mu_g}{\mu_w}$

Therefore the critical angle <em>c</em> in this case is given by,

$sinc=\frac{\mu_w}{\mu_g}$

Substitute 1.33 for $\mu_w$ and 1.5 for $\mu_g$ .

$sinc=\frac{\mu_w}{\mu_g} \\ =\frac{1.33}{1.5} \\ =0.8866$

Take the inverse of the value 0.8866.

$c=sin^{-1} (0.8866)\\ =62^o$

Since the critical angle of the prism when immersed in water, is 62°, which is greater than the angle of incidence of 45° required for viewing the object, the binoculars which are set for Total reflection in air, will not function when immersed in water.

4 0

4 years ago

PLEASE ITS AN Emergency IF ITS RIGHT I WILL GIVE BRAINLIEST

Art [367]

Uh I think it’s point guard

3 0

3 years ago

6. A car accelerates from rest to 5m/s in 20 s. The force from the engine is 2000N. The force of air resistance and friction act

otez555 [7]

Answer:

net force= 2000-1000=1000

a=v-u/t=5/20=1/4

m=?

m=f/a

=1000÷1/4=4000kg

4 0

3 years ago

The distance between the centers of the wheels of a motorcycle is 156 cm. The center of mass of the motorcycle, including the ri

Drupady [299]

Answer:

a = 9.86 m/s²

Explanation:

given,

distance between the centers of wheel = 156 cm

center of mass of motorcycle including rider = 77.5 cm

horizontal acceleration of motor cycle = ?

now,

The moment created by the wheels must equal the moment created by gravity.

take moment about wheel as it touches the ground, here we will take horizontal distance between them.

then, take the moment around the center of mass. Since the force on the ground from the wheels is horizontal, we need the vertical distance.

now equating both the moment

m g d = F h

d is the horizontal distance

h is the vertical distance

m g d = m a h

term of mass get eliminated

g d = a h

so,

$a = \dfrac{g\ d}{h}$

$a = \dfrac{9.8\times 0.78}{0.775}$

a = 9.86 m/s²

4 0

3 years ago