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charle [14.2K]
3 years ago
9

Which mixtures could be separated using evaporation? Check all that apply. fat in milk salt in ocean water mud in pond water but

ter in cookie dough vinegar in salad dressing
Physics
2 answers:
Charra [1.4K]3 years ago
8 0

Answer:

B. salt in ocean water

C. mud in pond water

Your welcome :D

salantis [7]3 years ago
5 0
I would go with salt in ocean water as when you heat it and the water begins to evaporate it will leave the salt behind


Also water in mud as the water would evaporate and leave the mud residue behind


GOOD LUCK
BRAINLIEST IF HELPED
You might be interested in
A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,
Alla [95]

Answer:

Explanation:

Let the balls collide after time t .

distance covered by falling ball

s₁ = v₀ t + 1/2 g t²

distance covered by rising ball

s₂ = v₀ t - 1/2 g t²

Given ,

s₁ + s₂ = D

D = v₀ t + 1/2 g t² + v₀ t - 1/2 g t²

= 2v₀ t

t = D / 2v₀

s₂ = v₀ t - 1/2 g t²

= v₀ x D / 2v₀ - (1/2) x  g x D² / 4v₀²

= D / 2 - gD² / 8 v₀²

6 0
2 years ago
>
tester [92]

The pressure exerted on the block on the ground in N/m² is 200N/m².

<h3>What is pressure?</h3>

The pressure is the amount of force applied per unit area.

Given is a 5000 Newton block rests on the ground over 25 m² of area.

Pressure p = Force/Area

Put the values, we get

p = 5000 /25

p = 200 N/m²

Hence, pressure exerted on the block on the ground is 200 N/m².

Learn more about pressure.

brainly.com/question/12971272

#SPJ1

4 0
2 years ago
An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.275 T. If the kinetic energy of the electr
xxMikexx [17]

Answer:

Radius, r=2.14\times 10^{-5}\ m

Explanation:

It is given that,

Magnetic field, B = 0.275 T

Kinetic energy of the electron, E=4.9\times 10^{-19}\ J

Kinetic energy is given by :

E=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{2E}{m}}

v=\sqrt{\dfrac{2\times 4.9\times 10^{-19}}{9.1\times 10^{-31}}}            

v = 1037749.04 m/s

The centripetal force is balanced by the magnetic force as :

qvB\ sin90=\dfrac{mv^2}{r}

r=\dfrac{mv}{qB}

r=\dfrac{9.1\times 10^{-31}\times 1037749.04}{1.6\times 10^{-19}\times 0.275 }

r=2.14\times 10^{-5}\ m

So, the radius of the circular path is 2.14\times 10^{-5}\ m. Hence, this is the required solution.

3 0
3 years ago
Read 2 more answers
MATHPHYS CAN U HELP ME PLEASE
ludmilkaskok [199]

Explanation:

(1) The heat added to warm the ice to 0°C is:

q = mCΔT = (0.041 kg) (2090 J/kg/°C) (0°C − (-11°C)) = 942.59 J

The heat added to melt the ice is:

q = mL = (0.041 kg) (3.33×10⁵ J/kg) = 13,653 J

The heat added to warm the water to 100°C is:

q = mCΔT = (0.041 kg) (4186 J/kg/°C) (100°C − 0°C) = 17,162.6 J

The heat added to evaporate the water is:

q = mL = (0.041 kg) (2.26×10⁶ J/kg) = 92,660 J

The heat added to warm the steam to 115°C is:

q = mCΔT = (0.041 kg) (2010 J/kg/°C) (115°C − 100°C) = 1236.15 J

The total heat needed is:

q = 942.59 J + 13,653 J + 17,162.6 J + 92,660 J + 1236.15 J

q = 125,654.34 J

(2) When the first two are mixed:

m C₁ (T₁ − T) + m C₂ (T₂ − T) = 0

C₁ (T₁ − T) + C₂ (T₂ − T) = 0

C₁ (6 − 11) + C₂ (25 − 11) = 0

-5 C₁ + 14 C₂ = 0

C₁ = 2.8 C₂

When the second and third are mixed:

m C₂ (T₂ − T) + m C₃ (T₃ − T) = 0

C₂ (T₂ − T) + C₃ (T₃ − T) = 0

C₂ (25 − 33) + C₃ (37 − 33) = 0

-8 C₂ + 4 C₃ = 0

C₂ = 0.5 C₃

Substituting:

C₁ = 2.8 (0.5 C₃)

C₁ = 1.4 C₃

When the first and third are mixed:

m C₁ (T₁ − T) + m C₃ (T₃ − T) = 0

C₁ (T₁ − T) + C₃ (T₃ − T) = 0

(1.4 C₃) (6 − T) + C₃ (37 − T) = 0

(1.4) (6 − T) + 37 − T = 0

8.4 − 1.4T + 37 − T = 0

2.4T = 45.4

T = 18.9°C

(3) Heat gained by the ice = heat lost by the tea

mL + mCΔT = -mCΔT

m (3.33×10⁵ J/kg) + m (2090 J/kg/°C) (30.8°C − 0°C) = -(0.176 kg) (4186 J/kg/°C) (30.8°C − 32.8°C)

m (397372 J/kg) = 1473.472 J

m = 0.004 kg

m = 4 g

4 grams of ice is melted and warmed to the final temperature, which leaves 128 grams unmelted.

(4) The heat added to warm the ice to 0°C is:

q = mCΔT = (0.028 kg) (2090 J/kg/°C) (0°C − (-67°C)) = 3920.84 J

The heat added to melt the ice is:

q = mL = (0.028 kg) (3.33×10⁵ J/kg) = 9324 J

The heat added to warm the melted ice to T is:

q = mCΔT = (0.028 kg) (4186 J/kg/°C) (T − 0°C) = (117.208 J/°C) T

The heat removed to cool the water to T is:

q = -mCΔT = -(0.505 kg) (4186 J/kg/°C) (T − 27°C)

q = (2113.93 J/°C) (27°C − T) = 57076.11 J − (2113.93 J/°C) T

The heat removed to cool the copper to T is:

q = -mCΔT = -(0.092 kg) (387 J/kg/°C) (T − 27°C)

q = (35.604 J/°C) (27°C − T) = 961.308 J − (35.604 J/°C) T

Therefore:

3920.84 J + 9324 J + (117.208 J/°C) T = 57076.11 J − (2113.93 J/°C) T + 961.308 J − (35.604 J/°C) T

13244.84 J + (117.208 J/°C) T = 58037.418 J − (2149.534 J/°C) T

(2266.742 J/°C) T = 44792.58 J

T = 19.8°C

(5) Kinetic energy of the hammer = heat absorbed by ice

KE = q

½ mv² = mL

½ (0.8 kg) (0.9 m/s)² = m (80 cal/g × 4.186 J/cal × 1000 g/kg)

m = 9.68×10⁻⁷ kg

m = 9.68×10⁻⁴ g

(6) Heat rate = thermal conductivity × area × temperature difference / thickness

q' = kAΔT / t

q' = (1.09 W/m/°C) (4.5 m × 9 m) (10°C − 4°C) / (0.09 m)

q' = 2943 W

After 10.7 hours, the amount of heat transferred is:

q = (2943 J/s) (10.7 h × 3600 s/h)

q = 1.13×10⁸ J

q = 113 MJ

6 0
3 years ago
A scientist pours 0.120 L of solution A into an Erlenmeyer flask. She adds 2.345 L of solution B. How many significant figures a
nika2105 [10]
0.120L + 2.345L = 2.465L = 4 significant figures in the answer
6 0
3 years ago
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