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LenaWriter [7]
3 years ago
8

Why cant soy sauce freeze

Physics
2 answers:
juin [17]3 years ago
8 0
The salt content in soy
choli [55]3 years ago
7 0
Soy sauce<span> does not </span>freeze<span> due to its high salt content-google

my words-soy sause cant frese because of its salt...salt melts ice but i have no idea why salt melts ice</span>
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When two capacitors are connected in parallel and then connected to a battery, the total stored energy is 6.9 times greater than
Mumz [18]

Answer:

C1/C2 = 0.213 or C2/C1 = 4.68

Explanation:

Please refer to the attached image for step by step explanation.

5 0
3 years ago
1. The thin top layer of the crust is called the Crust. What are the two types of
kolbaska11 [484]
Oceanic and continental
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The average speed of a greyhound bus from lansing to detroit is 104.5 km/h. on the return trip from detroit to lansing the avera
Nitella [24]
The average speed of the bus from Lansing to Detroit is 
v_1 = 104.5 km/h
while the average speed of the bus from Detroit to Lansing is
v_2 =56.2 km/h

The distance covered by the bus in the two trips is the same (the distance between the two cities), therefore, the average speed of the round trip can be calculated as the mean of the two speeds:
v= \frac{v_1+v_2}{2}= \frac{104.5+56.2}{2}=80.35 km/h
7 0
3 years ago
Which of the following devices is used in this experiment to separate the different wavelengths emitted by the light source?
adelina 88 [10]

Answer:

A. Prism

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7 0
3 years ago
A van has a weight of 4000 lb and center of gravity at Gv. It carries a fixed 900 lb load which has a center of gravity at Gl. I
natulia [17]

Answer:

 x = 25 / μ     [ ft]

Explanation:

To solve this exercise we can use Newton's second law.

Let's set a reference system where the x axis is parallel to the road

Y axis  

       N_B + N_A - W_van - W_load = 0

       N_B + N_A = W_van + W_load

X axis

     fr = ma

     a = fr / m

the total mass is

        m = (W_van + W_load) / g

the friction force has the expression

      fr = μ N_{total}

      fr = μy (W_van + W_load)

we substitute

      a = μ (W_van + W_load)    \frac{g}{W_van + W_load}

      a = μ g

taking the acceleration let's use the kinematic relations where the final velocity is zero

       v² = v₀² - 2 a x

       0 = v₀² -2a x

        x = \frac{v_o^2}{2a}

        x = \frac{v_o^2}{2 \mu g}

        x = \frac{40^2}{2 \ 32 \  \mu}

        x = 25 / μ     [ ft]

5 0
3 years ago
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