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Ugo [173]
3 years ago
5

Will ice melt faster in salt, sugar or water?

Physics
1 answer:
viva [34]3 years ago
3 0

Answer:

Water because the salt will just make it colder and the sugar won't do anything

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just as rea was situated at a distance in picture and a half dozen years old in his first comment in picture the actor was the first thing he had ever heard about it he reati was the only person who has

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plz help you get the best in your house but muje has been in a relationship since the last

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3 years ago
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A 4.6-kg block of ice originally at 263 K is placed in thermal contact with a 15.7-kg block of silver (cAg = 233 J/kg-K) which i
Viktor [21]

Answer:

temperature at  326.44 K system achieve equilibrium

Explanation:

given data

mass of block of ice = 4.6 kg

temperature = 263 K

thermal contact =  15.7-kg

specific heat of silver  cAg = 233 J/kg-K

initially temperature = 1052 K

to find out

what temperature will the system achieve equilibrium

solution

first we consider final temperature of the system  is T

we know that specific heat of water (C w) = 4186 J(kg K)

and

specific heat of ice ( C i )  = 2030 J/(kg K)

and

latent heat of fusion of ice ( Lf ) = 3.33 × 10^{5} J/kg

and we know that system is insulated

so  heat lost by silver = heat generated by ice    .................1

so we can say

mass of silver × specific heat of silver  × ( initial temp - final temp ) = mass of  ice × specific heat of ice  × ( ice temp ) + mass of  ice × latent heat of fusion of ice + mass of  ice × specific heat of water  × (final temperature )  

put here value we get

mass of silver × specific heat of silver  × ( initial temp - final temp ) = mass of  ice × specific heat of ice  × ( ice temp ) + mass of  ice × latent heat of fusion of ice + mass of  ice × specific heat of water  × (final temperature )

15.7  × 233 × ( 1052 - T ) = 4.6 × 2030 × 10 + 4.6 × 3.33 × 10^{5} + 4.6 × 4286 × ( T - 273 )

solve we get

T =  326.44 K

so temperature at  326.44 K system achieve equilibrium

7 0
3 years ago
Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

3 0
3 years ago
What are the uses of evaporative salts?
Makovka662 [10]
<span>to preserve foods, dye fabric, and DE-ice roads i hopes this helps

</span>
4 0
3 years ago
A 6.00-μf parallel-plate capacitor has charges of 40.0 μc on its plates. how much potential energy is stored in this capacitor?
dedylja [7]

Thew energy stored in a capacitor of capacitance C and voltage between the plates V is

E=\frac{1}{2} CV^2=\frac{1}{2C} Q^2.

Substituting numerical value

E=\frac{1}{2*6*10^{-6}} (40*10^{-6})^2\\ E=133.33\; \mu J

7 0
3 years ago
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