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marishachu [46]
3 years ago
13

when a wave crosses a surface at an angle from one medium into another, why does it change directions aS IT MOVES ACROSS THE BOU

NDARY INTO THE NEW MEDIUM
Physics
1 answer:
puteri [66]3 years ago
8 0
 <span>This pivot is called refraction. It happens because different materials have different densities. The more dense the material the more atoms the light collides with and the slower it travels; the less dense, the fewer the collisions and hence a faster velocity. This pivoting, or refraction, is caused by the light either slowing down or speeding up.</span>
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A rotating viscometer consists of two concentric cylinders –an inner cylinder of radius Rirotating at angular velocity (rotation
RSB [31]

Answer:

b)  the result we got can be termed approximation because we are neglecting the shear stress acting on the two ends of the cylinder. Here we have considered only the share stress acting on the curved surface area only.

Explanation:

check attachment for solution to A

4 0
3 years ago
QUESTION 36
mel-nik [20]

Answer:

Explanation:

m = ρV = 1.03( 1000 kg/m³)(π(2² m²)(3.0 m)) = 12360π kg

m ≈ 38,830 kg

5 0
3 years ago
5/137 Under the action of its stern and starboard bow thrusters, the cruise ship has the velocity vB = 1 m/s of its mass center
quester [9]

The image is missing, so i have attached it;

Answer:

A) V_rel = [-(2.711)i - (0.2588)j] m/s

B) a_rel = (0.8637i + 0.0642j) m/s²

Explanation:

We are given;

the cruise ship velocity; V_b = 1 m/s

Angular velocity; ω = 1 deg/s = 1° × π/180 rad = 0.01745 rad/s

Angular acceleration;α = -0.5 deg/s² = 0.5 x π/180 rad = -0.008727 rad/s²

Now, let's write Velocity (V_a) at A in terms of the velocity at B(V_b) with r_ba being the position vector from B to A and relative velocity (V_rel)

Thus,

V_a = V_b + (ω•r_ba) + V_rel

Now, V_a = 0. Thus;

0 = V_b + (ω•r_ba) + V_rel

V_rel = -V_b - (ω•r_ba)

From the image and plugging in relevant values, we have;

V_rel = -1[(cos15)i + (sin15)j] - (0.01745k * -100j)

V_rel = - (cos15)i - (sin15)j - 1.745i

Note that; k x j = - i

V_rel = [-(2.711)i - (0.2588)j] m/s

B) Let's write the acceleration at A with respect to B in terms of a_b.

Thus,

a_a = a_b + (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel

a_a and a_b = 0.

Thus;

0 = (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel

a_rel = - (α*r_ba) - (ω(ω•r_ba)) - (2ω*v_rel)

Plugging in the relevant values with their respective position vectors, we have;

a_rel = - (-0.008727k * -100j) - (0.01745k(0.01745k * -100j)) - (2*0.01745k * [-(2.711)i - (0.2588)j])

a_rel = 0.8727i - (0.01745² x 100)j + 0.0946j - 0.009i

Note that; k x j = - i and k x i = j

Thus,simplifying further ;

a_rel = 0.8637i + 0.0642j m/s²

4 0
3 years ago
A force, F1, of magnitude 2.0 N and directed due east is exerted on an object. A second force exerted on the object is F2 = 2.0
IgorLugansk [536]

Answer:

Magnitude of the force is

F_3 = 2.83

direction of the force is given as

\theta = 45 degree West of South

Explanation:

As we know that force is a vector quantity and in order to find the resultant of two or more forces we need to add them vectorialy

So here we have

\vec F_1 + \vec F_2 + \vec F_3 = 0

here we know that first force is of magnitude 2 N towards east

\vec F_1 = 2 \hat i N

second force is also of 2.0 N due North

\vec F_2 = 2 \hat j

now from above equation

2\hat i + 2\hat j + \vec F_3 = 0

\vec F_3 = -2\hat i - 2\hat j

so magnitude of the force is given as

F_3 = \sqrt{2^2 + 2^2}

F_3 = 2.83

direction of the force is given as

\theta = tan^{-1}\frac{F_y}{F_x}

\theta = tan^{-1}\frac{-2}{-2}

\theta = 45 degree West of South

3 0
3 years ago
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Calculate the potential energy of a five KG object sitting at the top of a 2 m ramp
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5 x 10 x 200 = 10000
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