Help please (about projectile motion at an angle)

Physics

1 answer:

PolarNik [594]3 years ago

7 0

Answer:

Since this is old, im just gonna get these points, don't wan't them to go to waste lm.ao

Explanation:

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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.