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mafiozo [28]
3 years ago
11

Which one is it cause I have no idea

Physics
2 answers:
Tasya [4]3 years ago
7 0
The play is stopped when went out of bounds bounds When there is a penalty or injury or when a goal is scored. If the ball goes over the goal line ending line but now out into the goal and was last touched by the opposite team it is put back into play by the defending team (it’s A)
Stella [2.4K]3 years ago
4 0

it depends on the game but in most cases, it will be the opposing team.


ans is a.


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In this unit, the amount of solute is measured in?
Hoochie [10]
Try liters if you haven’t done it yet. I’m so sorry if i’m incorrect.
5 0
3 years ago
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Which of the following types of light microscopy improves the resolution of thick specimens by illuminating one plane of the spe
Vilka [71]

Answer:

confocal microscopy

Explanation:

According to my research on different types of microscopes, I can say that based on the information provided within the question the tool being mentioned in this situation is a confocal microscopy. This is an extremely powerful microscope used to develop extremely sharp images of cells and tissues by viewing one plane of the specimen at a given time.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

7 0
3 years ago
An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,
insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

Where

t_{hf} = Temperature at the inside of the pipe = 300 °C

t_{f} = Temperature at the outside of the pipe = 20 °C

r₁ =internal  radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

k_A = 15 W/m·K

k_B = 0.038 W/m·K

h_{hf} = 75 W/m²·K

h_{cf} = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

Q = \frac{t_A-t_B}{R_T} Where R_T for the air film on the pipe outer surface is given by

R_T= \frac{1}{\alpha A}

where A =area of the outside of the pipe

= \frac{1}{10*2\pi*0.07*1 } = 0.227 K/W

Therefore

131.32 W = \frac{t_A-20}{0.227} which gives

t_A = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

8 0
3 years ago
The big bang produced an imprint of leftover heat called
tensa zangetsu [6.8K]
That's called the "Cosmic Microwave Background".  (CMB)
It was discovered in 1965, and its discoverers were awarded
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Doe anyone get this ​
8_murik_8 [283]

Answer:

we know a = F/ M

Explanation:

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4 0
3 years ago
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