1. Order the materials from smallest refractive index to largest refractive index.

Which of the following planets helped astronomers locate another planet?

worty [1.4K]

Answer: Pluto,Mercury

Explanation:

6 0

2 years ago

An end of a light wire rod is bent into a hoop of radius r. The straight part of the rod has length l; a ball of mass M is attac

Roman55 [17]

Answer:

$arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})$

Explanation:

By the Law of Sines,

$sin \theta = \frac{sin \phi R}{ l + R}$

From Newton's Law,

$mg = N\sqrt{\mu^2+1}$

And the last equation again from Newton's Law,

$\mu N = mgsin\phi$

Then if we collect all equations together,

$\mu N = mgsin\phi = N\sqrt{\mu^2+1}sin\phi\\$

$sin\theta = \frac{\mu R}{ (l + R)\sqrt{\mu^2+1}}$

Thus,

$\theta = arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})$

8 0

2 years ago

Describe the energy transformations that take place when a skier starts skiing down a hill but after a time is brought to rest b

Andrews [41]

<span>The skier will transform their gravitational energy into mostly kinetic energy (with a minor amount transformed into heat from the friction of the skis across the snow and air friction). Once the skier hits the snowdrift, their kinetic energy is transferred into the snow which moves when they strike it due to the kinetic energy that is now in the snow. Along with again a minor amount of heat energy transferred as they move through the snowdrift.</span>

6 0

2 years ago

The weight of spaceman Speff at the surface of planet X, solely due to its gravitational pull, is 389 N. If he moves to a distan

miv72 [106K]

Answer:

mass of the planet X = 5.6 × 10²³ kg.

Explanation:

According to Newtons law of universal gravitation,

F = GM₁M₂/r²

Where F = gravitational force, M₁ = mass of the speff, M₂ = mass of the planet X, G = gravitational constant r = distance between the speff and the planet X

making M₂ The subject of the equation above,

M₂ = Fr²/GM₁ .......................... equation 2

Where F = 24.31 N, r = 1.08×18⁴km ⇒( convert to m ) =1.08 × 10⁴ × 1000 m

r = 1.08 × 10⁷ m, G = 6.67 × 10 ⁻¹¹ Nm²/kg², M₁ = 75 kg

Substituting this values in equation 2,

M₂ = 24.13(1.08 × 10⁷ )²/75( 6.67 × 10 ⁻¹¹)

M₂ = 24.13 × 1.17 × 10¹⁴/500.25 × 10⁻¹¹

M₂ = (28.23 × 10¹⁴)/(500.25 × 10⁻¹¹)

M₂ = 0.056 × 10²⁵

M₂ = 5.6 × 10²³ kg.

Therefore mass of the planet X = 5.6 × 10²³ kg.

8 0

2 years ago

A 2.5 m -long wire carries a current of 8.0 A and is immersed within a uniform magnetic field B⃗ . When this wire lies along the

leva [86]

Answer:

Explanation:

Let the magnetic field be B = B₁i + B₂j + B₃k

Force = I ( L x B ) , I is current , L is length and B is magnetic field .

In the first case

force = - 2.3 j N

L = 2.5 i

puting the values in the equation above

- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]

= - 20 B₃ j + 20 B₂ k

comparing LHS and RHS ,

20B₃ = 2.3

B₃ = .115

B₂ = 0

In the second case

L = 2.5 j

Force = I ( L x B )

2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )

= - 20 B₁ k + 20B₃ i

2.3i−5.6k = - 20 B₁ k + 20B₃ i

B₃ = .115

B₁ = .28

So magnetic field B = .28 i + .115 B₃

Part A

x component of B = .28 T

Part B

y component of B = 0

Part C

z component of B = .115 T .

8 0

2 years ago