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3 years ago

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Physics

2 answers:

True [87]3 years ago

8 0

The answer is B because the pollinators give pollen from the plant

Roman55 [17]3 years ago

8 0

Answer:

Explanation:

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Suppose your surface body temperature averaged 90 degrees F. How much radiant energy in W/m^2 would be emitted from your body?

Debora [2.8K]

$493 \; \text{W}\cdot \text{m}^{-2}$ .

<h3>Explanation</h3>

The Stefan-Boltzmann Law gives the energy radiation <em>per unit area</em> of a black body:

$\dfrac{P}{A} = \sigma \cdot T^{4}$

where,

$P$ the total power emitted,
$A$ the surface area of the body,
$\sigma$ the Stefan-Boltzmann Constant, and
$T$ the temperature of the body in degrees Kelvins.

$\sigma = 5.67 \times 10^{-8} \;\text{W}\cdot \text{m}^{-2} \cdot \text{K}^{-4}$ .

$T = 90 \; \textdegree{}\text{F} = (\dfrac{5}{9} \cdot (90-32) + 273.15) \; \text{K} = 305.372 \; \text{K}$ .

$\dfrac{P}{A} = \sigma \cdot T^{4} = 5.67 \times 10^{-8} \times 305.372^{4} = 493\; \text{W}\cdot \text{m}^{-2}$ .

Keep as many significant figures in $T$ as possible. The error will be large when $T$ is raised to the power of four. Also, the real value will be much smaller than $493\; \text{W}\cdot \text{m}^{-2}$ since the emittance of a human body is much smaller than assumed.

5 0

3 years ago

Question 30

stellarik [79]

Answer: $0.69\°$

Explanation:

The angular diameter $\delta$ of a spherical object is given by the following formula:

$\delta=2 sin^{-1}(\frac{d}{2D})$

Where:

$d=16 m$ is the actual diameter

$D=1338 m$ is the distance to the spherical object

Hence:

$\delta=2 sin^{-1}(\frac{16 m}{2(1338 m)})$

$\delta=0.685\° \approx 0.69\°$ This is the angular diameter

3 0

3 years ago

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it i

lianna [129]

Answer:

<em>A) the moment of inertia of the system decreases and the angular speed increases. </em>

Explanation:

The complete question is

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, It is true to say that

A) the moment of inertia of the system decreases and the angular speed increases.

B) the moment of inertia of the system decreases and the angular speed decreases.

C) the moment of inertia of the system decreases and the angular speed remains the same.

D) the moment of inertia of the system increases and the angular speed increases.

E) the moment of inertia of the system increases and the angular speed decreases

In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as

$I_{1} w_{1} = I_{2} w_{2}$ ....1

where $I_{1}$ and $I_{2}$ are the initial and final moment of inertia respectively.

and $w_{1}$ and $w_{2}$ are the initial and final angular speed respectively.

Also, we know that the moment of inertia of a rotating body is given as

$I = mr^{2}$ ....2

where $m$ is the mass of the rotating body,

and $r$ is the radius of the rotating body from its center.

We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.

From equation 1, we see that in order for the angular momentum to be conserved, the decrease from $I_{1}$ to $I_{2}$ will cause the angular speed of the system to increase from $w_{1}$ to $w_{2}$ .

From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.

7 0

3 years ago

Which statement represents how light travels? Light given off from a moving source will travel faster in a vacuum. Light given o

Debora [2.8K]

Answer:

the first one

Explanation:

4 0

3 years ago

Read 2 more answers

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