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Aliun [14]
4 years ago
15

Copy of AL

Physics
1 answer:
tamaranim1 [39]4 years ago
3 0

Answer:

I’m so sorry I tried solving it but I don’t understand it can you explain the question a little bit more ty

Explanation:

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What is malware short for?
yaroslaw [1]
The correct answer is C
8 0
3 years ago
The stoplights on a street are designed to keep traffic moving at 26 mi/h. the average length of a street block between traffic ligh
bonufazy [111]

We use the formula,

v = \frac{d}{t}.

Here, v is velocity and its value given 26 mi/h ( in m/s,  \frac{26 \times 1.609 \times 10^{3} m}{60 \times 60 s} =11.62 \ m/s ) and d is distance and its value is given 80 m.

Substituting these values in above formula we get,

t = \frac{80 m}{11.62m/s } = 6.88 \ s

Thus, the time delay between green lights on successive blocks to keep the traffic moving continuously is 6.88 s


3 0
3 years ago
Need a little help here :(
Goshia [24]

Answer:

The output out be 200

Explanation:

Hope this helps :))

8 0
3 years ago
Read 2 more answers
How much heat energy is required to raise the temperature of 0.368kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of cop
Katarina [22]
The change in temperature here corresponds to a sensible heat. The amount of energy required can be calculated by multiplying the specific heat capacity, the amount of the substance and the corresponding change in temperature.

Heat required = mCΔT
Heat required = 0.368 kg (0.0920 cal/g°C) (60 - 23)°C
Heat required = 1.25 cal
3 0
3 years ago
Read 2 more answers
: The maximum theoretical efficiency of a Carnot engine operating between reservoirs at the steam point and at room temperature
Hunter-Best [27]

Answer:

The value is   \eta  =  0.2145  or  21.45%

Explanation:

From the question we are told that

    The first reservoir is at steam point  T_s =  100^o C =  100 + 273 = 373 \ K  

    The  second reservoir is at room temperature T_r  =  20^o C = 293 \ K

Generally the  maximum theoretical efficiency of a Carnot engine  is mathematically evaluated as

     \eta  =  1- \frac{T_r}{T_s}

=>    1 - \frac{ 293}{373}

=>    \eta  =  0.2145

5 0
3 years ago
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