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serg [7]
3 years ago
9

Which three characteristics make a glowing piece of charcoal a blackbody radiator? A. The frequencies of EMR it emits depend on

its temperature. B. It emits only one frequency of EMR. C. It absorbs most of the EMR it receives. D. It emits a range of wavelengths of EMR.
Physics
1 answer:
Eva8 [605]3 years ago
4 0

Answer:

A. The frequencies of EMR it emits depend on its temperature

B. It emits only one frequency of EMR

C. It absorbs most of the EMR it receives

Explanation:

  • A blackbody is an object that absorbs most of the electromagnetic spectrum of energy that falls on it.
  • According to law to reradiates most of the available energy back on top the outer space at an efficiency of 100% and thus radiation may be in the visible range of temperature than are in 1000K.
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The lower atmosphere is mostly warmed by radiated heat from Earth's surface. However, water heats up and cools down more slowly
JulsSmile [24]
The answer is B. On a sunny day, the air over a lake will be cooler than the air over the bordering land.
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What is a tool that measures the size of a force
MakcuM [25]
<span>this is a dynamometer</span>
3 0
3 years ago
A mercury thermometer has a glass bulb of interior volume 0.100 cm3 at 10°c. the glass capillary 10) tube above the bulb has an
Nadya [2.5K]
Initial volume of mercury is
V = 0.1 cm³

The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.

Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is 
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³

The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
   = (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
   = 4.5 cm

Answer: 4.5 cm
7 0
3 years ago
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You observe that a mass suspended by a spring takes 0.25 s to make a full oscillation. What is the frequency of this oscillation
Katarina [22]

Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, \omega = 25.13 rad/s

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

f = \frac{1}{T}

f = \frac{1}{0.25}

Frequency of oscillation, f = 4 Hz

The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

Therefore, time period, T = 0.25 s

Angular frequency of oscillation is given by:

\omega = 2\pi \times f

\omega = 2\pi \times 4

\omega = 25.13 rad/s

5 0
3 years ago
A person has legs of length L = 1.10 m. (a) If the maximum angle between the legs during walking is ϕ = 50o , what is the person
vampirchik [111]

Answer:

a). Maximum Length L=0.929m

b). T=0.83 Hz or 1.2s

c). Longer, the effortless waling T=2.1 Hz or t=0.475s

d). t=1.2s V=0.774 \frac{m}{s}

t=0.475s V=1.95 \frac{m}{s}

Explanation:

Length legs=L=1.1m

angle=50

the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image

Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length

a).

sin(\alpha) =\frac{op}{h}

op=h*sin(\frac{\alpha }{2})\\ op=1.1m*sin(\frac{50}{2})\\op=0.464m

Length of the step

L=0.464m*2

L=0.928m

b).

period=T

T=\frac{1}{time}=\frac{1}{t}\\ T=\frac{1}{1.2s}\\T=0.83 s^{-1}\\ T=0.83Hz

c).

T1=2\pi *\sqrt{\frac{L}{g}} \\T1=2\pi *\sqrt{\frac{1.1m}{9.8\frac{m}{s^{2}}}}\\ T1=2.1 Hz

The period is the inverse of the time of the motion so, the T1 is faster that the T because

t=\frac{1}{T1}=t= \frac{1}{2.1}=0.47s

d).

The speed is the relation between the distance with time so:

Vt=\frac{0.928m}{1.2s} \\Vt=0.773 \frac{m}{s} \\Vt=\frac{0.928m}{0.475s} \\Vt=1.953 \frac{m}{s}

3 0
3 years ago
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