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nikitadnepr [17]
3 years ago
11

Groups of college students pursue the study of chemical compounds with similar reactivity. If they wish to do an in-depth backgr

ound study on periodic trends, they should MOST LIKELY
A) consult only their chemistry professors.
Eliminate
B) rely exclusively on each other’s background knowledge.
C) consult chemistry professors, chemistry textbooks, and scientific journals.
D) consult chemistry professors and further study through scientific-fiction movies.
Physics
1 answer:
mihalych1998 [28]3 years ago
7 0
C, because its the best answer

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At high noon, the sun delivers 1150 w to each square meter of a blacktop road. if the hot asphalt loses energy only by radiation
Pie
Stephen`s Law:
P = (Sigma) · A · e · T^4
P in = P out
e = 1 for blacktop;
1150 W = (Sigma) · T^4
(Sigma) = 5.669 · 10 ^(-8) W/m²K^4
T^4 = 1150 : ( 5.669 · 10^(-8) )
T^4 = 202.875 · 10^8
T =  \sqrt[4]{202.857 * 10 ^{8} }
T = 3.774 · 10² = 377.4 K
Answer: Equilibrium temperature is 377.4 K. 
3 0
3 years ago
Matter is anything that...
Lisa [10]
The answer is b, anything that has mass and takes up space

8 0
3 years ago
Read 2 more answers
A helicopter blade spins at exactly 110 revolutions per minute. Its tip is 4.50 m from the center of rotation. (a) Calculate th
NeX [460]

Answer:

(a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

Explanation:

Given that,

Angular velocity = 110 rev/m

Radius = 4.50 m

(a). We need to calculate the average speed

Using formula of average speed

v=r\omega

v = 4.50\times110\times\dfrac{2\pi}{60}

v=51.83\ m/s

(b). The average velocity over one revolution is zero because the net displacement is zero in one revolution.

Hence, (a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

8 0
3 years ago
Why does the spectroscopic parallax method only work for main sequence stars?.
Brums [2.3K]

Answer:

Only main sequence stars have a well-defined relationship between spectral type and luminosity.

Explanation:

Low-mass stars have much longer lifetimes than high-mass stars.

5 0
2 years ago
A spherical shell is rolling without slipping at constant speed on a level floor. What percentage of the shell's total kinetic e
IgorC [24]

Answer:

41.667 per cent of the total kinetic energy is translational kinetic energy.

Explanation:

As the spherical shell is rolling without slipping at constant speed, the system can be considered as conservative due to the absence of non-conservative forces (i.e. drag, friction) and energy equation can be expressed only by the Principle of Energy Conservation, whose total energy is equal to the sum of rotational and translational kinetic energies. That is to say:

E = K_{t} + K_{r}

Where:

E - Total energy, measured in joules.

K_{r} - Rotational kinetic energy, measured in joules.

K_{t} - Translational kinetic energy, measured in joules.

The spherical shell can be considered as a rigid body, since there is no information of any deformation due to the motion. Then, rotational and translational components of kinetic energy are described by the following equations:

Rotational kinetic energy

K_{r} = \frac{1}{2}\cdot I_{g}\cdot \omega^{2}

Translational kinetic energy

K_{t} = \frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2}

Where:

I_{g} - Moment of inertia of the spherical shell with respect to its center of mass, measured in kg\cdot m^{2}.

\omega - Angular speed of the spherical shell, measured in radians per second.

R - Radius of the spherical shell, measured in meters.

After replacing each component and simplifying algebraically, the total energy of the spherical shell is equal to:

E = \frac{1}{2}\cdot (I_{g} + m\cdot R^{2})\cdot \omega^{2}

In addition, the moment of inertia of a spherical shell is equal to:

I_{g} = \frac{2}{3}\cdot m\cdot R^{2}

Then, total energy is reduced to this expression:

E = \frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2}

The fraction of the total kinetic energy that is translational in percentage is given by the following expression:

\%K_{t} = \frac{K_{t}}{E}\times 100\,\%

\%K_{t} = \frac{\frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2} }{\frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2} } \times 100\,\%

\%K_{t} = \frac{5}{12}\times 100\,\%

\%K_{t} = 41.667\,\%

41.667 per cent of the total kinetic energy is translational kinetic energy.

7 0
3 years ago
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