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1 year ago

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A skateboarder rides swiftly up the edge of a bowl-shaped surface and leaps into the air. While in the air, the skateboarder fli

ps upside and tosses the skateboard from hand to hand. The skateboarder then rides safely back down the bowl. During the time that the skateboarder and skateboard are not touching anything, one aspect of their motion that is constant is their total (or combined) [note: neglect any effects due to the air]
a. angular momentum.
b. angular velocity.
c. velocity.
d. momentum.

1 answer:

Katen [24]1 year ago

3 0

Answer:

Option(a) is the correct answer to the given question .

Explanation:

The main objective of the angular momentum is evaluating however much the rotational movement as well as the angular velocity in the entity does have.The angular momentum is measured in terms of $kgm^{2 }\ / s$ .

In the given question the skateboarder rides quickly up the bottom of a bowl-shaped surface and climb into the air.it means it is rotational movement also it is not touching anything so it is angular momentum.
All the other option is incorrect because it is not follows the given scenario

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Two identical horizontal sheets of glass have a thin film of air of thickness t between them. The glass has refractive index 1.4

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Answer:

the wavelength λ of the light when it is traveling in air = 560 nm

the smallest thickness t of the air film = 140 nm

Explanation:

From the question; the path difference is Δx = 2t (since the condition of the phase difference in the maxima and minima gets interchanged)

Now for constructive interference;

Δx= $(m+ \frac{1}{2} \lambda)$

replacing ;

Δx = 2t ; we have:

2t = $(m+ \frac{1}{2} \lambda)$

Given that thickness t = 700 nm

Then

2× 700 = $(m+ \frac{1}{2} \lambda)$ --- equation (1)

For thickness t = 980 nm that is next to constructive interference

2× 980 = $(m+ \frac{1}{2} \lambda)$ ----- equation (2)

Equating the difference of equation (2) and equation (1); we have:'

λ = (2 × 980) - ( 2× 700 )

λ = 1960 - 1400

λ = 560 nm

Thus; the wavelength λ of the light when it is traveling in air = 560 nm

b)

For the smallest thickness $t_{min} ; \ \ \ m =0$

∴ $2t_{min} =\frac{\lambda}{2}$

$t_{min} =\frac{\lambda}{4}$

$t_{min} =\frac{560}{4}$

$t_{min} =140 \ \ nm$

Thus, the smallest thickness t of the air film = 140 nm

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2 years ago

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Inversely proportional to its frequency. If electromagnetic radiation A has a lower frequency than electromagnetic B, then compared to B, the wavelength of A is...? - equal - shorter - longer - exactly half the length of

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1 year ago

A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init

Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

$h=\frac{gt^2}{2}$

For second stone

$h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$ ---2

Equating 1 &2 we get

$\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$

$\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$26.46t-35.721-52.92t+142.884=0$

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2 years ago

If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g

yulyashka [42]

Answer:

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2 years ago

The weight of an object is the same on two different planets. The mass of planet A is only sixty percent that of planet B. Find

natka813 [3]

Answer:

0.775

Explanation:

The weight of an object on a planet is equal to the gravitational force exerted by the planet on the object:

$F=G\frac{Mm}{R^2}$

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the object

R is the radius of the planet

For planet A, the weight of the object is

$F_A=G\frac{M_Am}{R_A^2}$

For planet B,

$F_B=G\frac{M_Bm}{R_B^2}$

We also know that the weight of the object on the two planets is the same, so

$F_A = F_B$

So we can write

$G\frac{M_Am}{R_A^2} = G\frac{M_Bm}{R_B^2}$

We also know that the mass of planet A is only sixty percent that of planet B, so

$M_A = 0.60 M_B$

Substituting,

$G\frac{0.60 M_Bm}{R_A^2} = G\frac{M_Bm}{R_B^2}$

Now we can elimanate G, MB and m from the equation, and we get

$\frac{0.60}{R_A^2}=\frac{1}{R_B^2}$

So the ratio between the radii of the two planets is

$\frac{R_A}{R_B}=\sqrt{0.60}=0.775$

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1 year ago