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const2013 [10]
3 years ago
12

Why do some transportation agencies spread a mixture of sand and salt on icy roads in winter?

Physics
1 answer:
timofeeve [1]3 years ago
4 0

Answer:While workers use salt to melt ice they rely on sand for improved traction

Explanation:Sand increases friction and help prevent vehicle tires from slipping on silk roads making traveling easier, Salt can also help prevent new ice from forming on roads

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Why did earth most likely form into distinct, separate layers? (HELP ASAP PLZ)
g100num [7]

Answer:

C

Explanation:

Rocks with hotter temperatures sank to the bottom of the Earth. The rocks with the hottest temperature became the core and the rocks with the least amount of heat became the crust.

Hope this helps!

8 0
3 years ago
Read 2 more answers
It takes a bus driver 30 minutes to pick up students from four stops. The last stop is at the corner of Green Street and Route 7
deff fn [24]

Answer:

No, the distance from the last stop to the school and the time it takes to travel that distance are required.

7 0
3 years ago
in the circuit diagram below, three identical light bulbs are connected to a battery. what happens to the brightness of bulb 1 w
andrezito [222]

Answer:

The brightness of bulb 1 dies because it is switched off.

6 0
3 years ago
What is the resultant of a and b if a = 3i 3j and b = 3i − 3j?
ss7ja [257]
 <span>a+b= ? 
3i +3j + (3i -3j) = ? 
3i + 3j + 3i -3j =? 
= 6i + 0j</span>
3 0
3 years ago
Read 2 more answers
Heat transfers energy from a hot object to a cold object. Both objects are isolated from their surroundings. The change in entro
aniked [119]

To develop this problem we will start from the definition of entropy as a function of total heat, temperature. This definition is mathematically described as

S = \frac{Q}{T}

Here,

Q = Total Heat

T = Temperature

The total change of entropy from a cold object to a hot object is given by the relationship,

\Delta S = \frac{Q}{T_{cold}}-\frac{Q}{T_{hot}}

From this relationship we can realize that the change in entropy by the second law of thermodynamics will be positive. Therefore the temperature in the hot body will be higher than that of the cold body, this implies that this term will be smaller than the first, and in other words it would imply that the magnitude of the entropy 'of the hot body' will always be less than the entropy 'cold body'

Change in entropy \Delta S_{hot} is smaller than \Delta S_{cold}

Therefore the correct answer is C. Will always have a smaller magnitude than the change in entropy of the cold object

5 0
3 years ago
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