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3 years ago

Why does salt dissolve in water act as an antifreeze?

Physics

2 answers:

aalyn [17]3 years ago

8 0

Answer:

Explanation:

An antifreeze is a substance which when added to a liquid reduces the freezing point of the liquid. Hence the solution can only freeze at a lower temperature compared to the solvent. Examples of antifreeze includes salt and propylene glycol. Antifreeze are used in low temperature environments to prevent water from freezing quicky. Hence the answer is A.

Have a great day Aylabailey4930

strojnjashka [21]3 years ago

4 0

Answer:

Because the freezing point of a solution increases above that of the solvent as solute is added.

Explanation:

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A 64 kg swimmer jumps, with a velocity of 4.2 m/s, off the front of a 25 kg kayak when the kayak is moving forward at a velocity

Crank

Answer:

3.88m/s

Explanation:

Using the law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and 2 are the initial velocities

v is the final velocity

Given

m1 = 64kg

u1 = 4.2m/s

m2 = 25kg

u2 = 3.2m/s

Required

Final velocity v

Substitute the given values into the formula

64(4.2)+25(3.2) = (65+25)v

268.8+80 = 90v

348.8 = 90v

v = 348.8/90

v = 3.88m/s

Hence the velocity of the kayak after the swimmer jumps off is 3.88m/s

8 0

3 years ago

Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are

Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

$d_{12} = \sqrt{3^{2}+3^{2} } = \sqrt{18}$ m : distance from q₁ to q₂

(d₁₂)² = 18 m²

$d_{13} =\sqrt{1^{2}+3^{2} } = \sqrt{10}$ m : distance from q₁ to q₃

(d₁₃)² = 10 m²

$d_{14} =\sqrt{3^{2}+2^{2} } = \sqrt{13}$ m : distance from q₁ to q₄

(d₁₄)² = 13 m²

K= 8.98755 × 10⁹ N *m²/C²

q₁= 7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45° , β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

$F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }$

$F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }$

Fn₁= 23.95*10⁻⁹ N

8 0

3 years ago

Does a compressed spring transfer elastic energy to its surroundings?

shusha [124]

Answer:

Yes

Explanation:

5 0

3 years ago

Over a 24-hour period, the tide in a harbor can be modeled by one period of a sinusoidal function. the tide measures 5.15 ft at

RSB [31]

<span>f(x) = 5.05*sin(x*pi/12) + 5.15

First, you need to determine the period of the function. The period will be the time interval between identical points on the sinusoidal function. For this problem, the tide is rising and at 5.15 at midnight for two consecutive days. So the period is 24 hours. Over that 24 hour period, we want the parameter passed to sine to range from 0 to 2*pi. So the scale factor for x will be 2*pi/24 = pi/12 which is approximately 0.261799388. The next thing to note is the magnitude of the wave. That will simply be the difference between the maximum and minimum values. So 10.2 ft - 0.1 ft = 10.1 ft. And since the value of sine ranges from -1 to 1, we need to divide that magnitude by 2, so 10.1 ft / 2 = 5.05 ft.

So our function at this point looks like f(x) = 5.05*sin(x*pi/12) But the above function ranges in value from -5.05 to 5.05. So we need to add a bias to it in order to make the low value equal to 0.1. So 0.1 = X - 5.05, 0.1 + 5.05 = X, 5.15 = X. So our function now looks like:
f(x) = 5.05*sin(x*pi/12) + 5.15

The final thing that might have been needed would have been a phase correction. With this problem, we don't need a phase correction since at X = 0 (midnight), the value of X*pi/12 = 0, and the sine of 0 is 0, so the value of the equation is 5.15 which matches the given value of 5.15. But if the problem had been slightly different and the height of the tide at midnight has been something like 7 feet, then we would have had to calculate a phase shift value for the function and add that constant to the parameter being passed into sine, making the function look like:
f(x) = 5.05*sin(x*pi/12 + C) + 5.15
where
C = Phase correction offset.

But we don't need it for this problem, so the answer is:
f(x) = 5.05*sin(x*pi/12) + 5.15

Note: The above solution assumes that angles are being measured in radians. If you're using degrees, then instead of multiplying x by 2*pi/24 = pi/12, you need to multiply by 360/24 = 15 instead, giving f(x) = 5.05*sin(x*15) + 5.15</span>

7 0

3 years ago

The freight cars a and b have a mass of 20 mg and 15 mg, respectively. if the cars collide and couple together, what is the velo

igomit [66]

Suppose car A is moving with a velocity Va, and car b with a velocity Vb,

According the principle of conservation of momentum:

Va x Ma + Vb x Mb = (Ma + Mb) V

V = (Va x Ma + Vb x Mb)/(Ma +Mb)

V = speed of cars after coupling

V = (Va x 20 mg + Vb x 15 mg)/(20 mg + 15 mg)

Put in the values of Va and Vb, and get the V

7 0

3 years ago

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