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2 years ago

What is the y component of a vector defined as 12.2m at 81.5°?

Physics

1 answer:

sergejj [24]2 years ago

3 0

Answer:

Explanation:

This is a displacement vector since it is defined in terms of distance (meters, to be exact). The way you find the y-component is

$V_y=Vsin\theta$ which says that you multiply the magnitude of the vector (its length) by the sin of the direction (the angle):

$V_y=12.2sin(81.5)$ and get

$V_y=$ 12.1 m

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A 60.0-kg skater begins a spin with an angular speed of 6.0 rad/s. By changing the position of her arms, the skater decreases he

AURORKA [14]

Answer:

The final angular speed of the skater is 12 radians per second.

Explanation:

Let consider the skater as a rotating system, given the absence of external forces, the Principle of Angular Momentum Conservation is applied:

$I_{o}\cdot \omega_{o} = I_{f}\cdot \omega_{f}$

Where:

$I_{o}$ , $I_{f}$ - Initial and final moment of inertia, measured in $kg \cdot m^{2}$ .

$\omega_{o}$ , $\omega_{f}$ - Angular speed, measured in radians per second.

The final angular speed is cleared afterwards:

$\omega_{f} = \frac{I_{o}}{I_{f}} \cdot \omega_{o}$

Given that $I_{f} = \frac{1}{2}\cdot I_{o}$ and $\omega_{o} = 6\,\frac{rad}{s}$ , the final angular speed is:

$\omega_{f} = \frac{I_{o}}{\frac{1}{2}\cdot I_{o} } \cdot \omega_{o}$

$\omega_{f} = 2 \cdot \omega_{o}$

$\omega_{f} = 2 \cdot \left(6\,\frac{rad}{s} \right)$

$\omega_{f} = 12\,\frac{rad}{s}$

The final angular speed of the skater is 12 radians per second.

6 0

2 years ago

We can learn a lot about the properties of a star by studying its spectrum. All of the followingstatements are true except one.

Kisachek [45]

Answer:

B. The total amount of light in the spectrum tells us the star’s radius.

Explanation:

The effective temperature of a star can be determined by means of its spectrum¹ and Wien's displacement law.

Since stars behave in a local way as a blackbody, it will take the wavelength at which is the peak of emission greater in the continuum (see the image below).

Then, the maximum peak of emission ( $\lambda_{max}$ ) will be replaced in the next equation of the Wien's displacement law:

$T = \frac{2.898x10^{-3} m. K}{\lambda max}$ (1)

Where T is the effective temperature of the star.

Bodies that are hot enough emits light as consequence of its temperature. For example, a iron bar in contact with fire will start to change colors as the temperature increase, until it gets to a blue color, which its know as Wien's displacement law. Which establishes that the peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase.

The same scenario described above can be found in the stars, a star whit higher temperature will have a blue color and one with lower temperature, a red color.

Since star does not have the same size, they have different brightness, That is because the photons have a free mean path greater in a bigger radius.

So a star brightness is a consequence of its radius.

Spectral lines will be shifted to the blue part of the spectrum1 if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

By using that shift in the spectral lines, the Doppler velocity can be determined.

$v = c\frac{\Delta \lambda}{\lambda_{0}}$ (2)

Where $\Delta \lambda$ is the wavelength shift, $\lambda_{0}$ is the wavelength at rest, v is the velocity of the source and c is the speed of light.

When a photon is absorbed by an electron in an atom of a particular element in the star photosphere, the electron will be pass to a higher state, when it comes back to the ground state, a photon will be emitted again. If the emitted photon does not go in the same direction of the incident photon an absorption line will be created in the spectrum of the star.

This patterns of spectral lines in the spectrum of the star are compared with the patterns that are got by lamps of that element in a laboratory.

Key term:

¹Spectrum: decomposition of light in its characteristic colors (wavelengths).

3 0

2 years ago

If Pete ( mass=90.0kg) weights himself and finds that he weighs 30.0 pounds, how far away from the surface of the earth is he

shutvik [7]

Answer: 9938.8 km

Explanation:

1 pound-force = 4.48 N

30.0 pounds-force = 134.4 N

The force of gravitation between Earth and object on the surface of is given by:

$F = \frac{GMm}{R^2} = mg$

Where M is the mass of the Earth, m is the mass of the object, R (6371 km) is the radius of the Earth.

At height, h above the surface of the Earth, the weight of the object:

$(mg)'= \frac{GMm}{(R+h)^2}$

we need to find "h"

taking the ratio of two:

$\frac{mg}{(mg)'}=\frac{(R+h)^2}{R^2}\\ \Rightarrow \frac{90kg \times 9.8 m/s^2}{134.4 N}=\frac{(R+h)^2}{R^2}\\ \Rightarrow 6.56 R^2= (R+h)^2 \Rightarrow h= (2.56-1)R\\ \Rightarrow h = 1.56 R = 1.56 \times 6371 km = 9938. 8 km$

Hence, Pete would weigh 30 pounds at 9938.8 km above the surface of the Earth.

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2 years ago

A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington

Roman55 [17]

Let $\theta$ be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

$\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

The current has velocity vector (relative to the Earth)

$\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath$

The swimmer's resultant velocity (her velocity relative to the Earth) is then

$\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}$

$\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath$

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

$\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ$

which is approximately 41º west of north.

6 0

2 years ago

Consider a boat heading due east at 15 miles/hour. The water's current is moving at 7.1 miles/hour at 45º south of east. Drag ve

givi [52]

If a boat is going East at 15mph and there is a water current going southeast at 45° then the boat is being drifted southward. So since the current is going at an angle then it has a x and y component. So Rx refers to the x-component force of the current and Ry refers to the y-component of the current, and |R| refers to the magnitude of these forces.

7 0

2 years ago

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