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Burka [1]
2 years ago
5

Which of the following statements is TRUE for high-visibility clothing? A. High-visibility clothing helps to reduce insect probl

ems when exercising. B. Bright colors work to keep you cooler while exercising. C. High-visibility clothing is important to wear in areas with moving vehicles. D. none of the above
Physics
2 answers:
allochka39001 [22]2 years ago
6 0

Answer:

C. High-visibility clothing is important to wear in areas with moving vehicles.

Explanation:

plz mark as brainliest

Mazyrski [523]2 years ago
5 0
C. High-visibility clothing is important to wear in areas with moving vehicles
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in a class where the number of girls is 36% of the total number,there are 48 boys.how many students are there in the class?​
Pavlova-9 [17]

Answer:

There are 75 people in the class. The number of boys is 48 and the number of girls is 27. The percentage of girls is 36% of 75.

Explanation:

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3 years ago
Does anyone know these four?! I really need help and fast pls
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The Beams And Joints That Hold It .
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You calculate the density of a block of aluminum to be 2.68 g/cm3. You look up the density of a block of aluminum at room temper
Hunter-Best [27]

Answer:

Systematic errors.

Explanation:

The density of the aluminium was calculated by a human and this is not natural but can be due to errors in the calibration of the scale for measuring the weight or taking readings from the measuring cylinder.

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3 0
3 years ago
Write a numerical expression for the emissive intensity (in W/m^2.sr) coming out of a tiny hole in an enclosure of surface tempe
stiks02 [169]

Answer:

6.0 × 10^{11} W/m^{2}

Explanation:

From Wien's displacement formula;

Q = e AT^{4}

Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.

The emissive intensity = \frac{Q}{A} = eT^{4}

Given from the question that: e = 0.6 and T = 1000K, thus;

emissive intensity = 0.6 × (1000)^{4}

                             = 0.6 × 1.0 × 10^{12}

                             = 6.0 × 10^{11} \frac{W}{m^{2} }

Therefore, the emissive intensity coming out of the surface is 6.0 × 10^{11} W/m^{2}.

3 0
2 years ago
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