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Leto [7]
3 years ago
7

Assemblage is ________________ a. a technique for casting where wax is carved then heated after a mold is made, then heated so t

hat the wax melts, leaving an impression. b. an additive process where artists construct work by putting together objects and attaches them in some way. c. a substitution process in which the artist models an image in a soft medium such as wax, and then replaces it with something more durable such as bronze. d. the use of tools to chip away at a base in order to ...
Physics
2 answers:
labwork [276]3 years ago
7 0
Assemblage is an additive process where artists construct work by putting together objects and attaches them in some way. The correct option among all the options that are given in the question is the second option or option "b". This kind of artistry actually gives a three dimensional impression to the work that is done by the artist.
My name is Ann [436]3 years ago
7 0

Answer:

b

Explanation:

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Every year _______ species go extinct.<br> a. 3<br> b. 72<br> c. 1000<br> d. 20000
Tems11 [23]
Well, the thing is: we don't really know, as we don't even know how many species there are on earth.

If we take a look at the estimates of <span>World Wide Fund for Nature, an organization that works toward combating species extinction, their estimates vary from 200 to 100 000 - but a probable number is 20 000 (d). </span>
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7 0
3 years ago
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If Mrs. Reichelt throws a chromebook, because it won't login correctly, with a force of 8N, and the chromebook accelerates at 5m
suter [353]

Answer:

1.6 kg

Step-by-step Solution:

Since Force = mass × acceleration we have:

F = 8N

a= 5 m/s^2

m = ?

By plugging the values above into F=ma we obtain:

F=ma\\\\8=m(5)\\\\\frac{8}{5}=\frac{m(5)}{5}\\\\m=\frac{8}{5}=1.6

Therefore, the Chromebook has a mass of 1.6 kilograms.



7 0
3 years ago
A cosmic ray (an electron or nucleus moving ar speeds close to the speed of light) travels across the Milky Way at a speed of 0.
Fiesta28 [93]

Answer:

Cosmic ray's frame of reference: 99,875 years

Stationary frame of reference: 501,891 years

Explanation:

First of all, we convert the distance from parsec into metres:

d=30,000 pc =9.26\cdot 10^{20} m

The speed of the cosmic ray is

v=0.98 c

where

c=3.0 \cdot 10^8 m/s is the speed of light. Substituting,

v=(0.98)(3.0\cdot 10^8)=2.94\cdot 10^8 m/s

And so, the time taken to complete the journey in the cosmic's ray frame of reference (called proper time) is:

T_0 = \frac{d}{v}=\frac{9.26\cdot 10^{20}}{2.94\cdot 10^8}=3.15\cdot 10^{12} s

Converting into years,

T_0 = \frac{3.15\cdot 10^{12}}{(365\cdot 24\cdot 60 \cdot 60}=99,875 years

Instead, the time elapsed in the stationary frame of reference is given by Lorentz transformation:

T=\frac{T_0}{\sqrt{1-(\frac{v}{c^2})^2}}

And substituting v = 0.98c, we find:

T=\frac{99,875}{\sqrt{1-(\frac{0.98c}{c})^2}}=501,891 years

3 0
3 years ago
Which classification scheme is based on the brightness of a star as it appears to you in the night sky?
anastassius [24]
The apparent magnitude scale is a classification scheme which is based on the brightness of stars. The range of brightness values is from 1 to 6.
The stars which are the most brightest are ranked as number 1 and also called first magnitude stars, stars which are little dimmer than number 1 are ranked as number 2 and also called second magnitude stars. Similarly the most faintest stars are ranked number 6 and also called as the sixth magnitude stars.
3 0
3 years ago
In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.1).Find the speed for a star in which this
soldier1979 [14.2K]

Answer:

1). v = - 2960526m/s

2). Toward us

3). v = - 493421m/s

4). Toward us

5). v = 1480263m/s

6).  Away from us

7). v = 3207236m/s

8). Away from us

Explanation:

Spectral lines will be shifted to the blue part of the spectrum 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).

The wavelength at rest is 121.6 nm (\lambda_{0} = 121.6nm)

Redshift: \lambda_{measured} > \lambda_{0}

Blueshift: \lambda_{measured} < \lambda_{0}

Then, for this particular case it is gotten:

Star 1: \lambda_{measured} = 120.4nm

Star 2: \lambda_{measured} = 121.4nm

Star 3: \lambda_{measured} = 122.2nm

Star 4: \lambda_{measured} = 122.9nm

Star 1:

Blueshift: 120.4nm < 121.6nm

Toward us

Star 2:

Blueshift: 121.4nm < 121.6nm

Toward us

Star 3:

Redshift: 122.2nm > 121.6nm

Away from us

Star 4:

Redshift: 122.9nm > 121.6nm

Away from us

Due to that shift the velocity of the star can be determine by means of Doppler velocity.

v = c\frac{\Delta \lambda}{\lambda_{0}}  (1)

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.

v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}}) (2)

<em>Case for star 1 \lambda_{measured} = 120.4 nm:</em>

<em></em>

v = (3x10^{8}m/s)(\frac{120.4nm-121.6nm}{121.6nm})

v = - 2960526m/s

Notice that the negative velocity means that is approaching to the observer.

<em>Case for star 2 \lambda_{measured} = 121.4 nm:</em>

v = (3x10^{8}m/s)(\frac{121.4nm-121.6nm}{121.6nm})

v = - 493421m/s

<em>Case for star 3 \lambda_{measured} = 122.2 nm:</em>

v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})

v = 1480263m/s

<em>Case for star 4 \lambda_{measured} = 122.9 nm:</em>

v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})

v = 3207236m/s

4 0
3 years ago
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