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3 years ago

Freezing Point Depression: Can someone explain this formula to me? ΔTf = Kfcm

1 answer:

8 0

If the solution is treated as an ideal solution, the extent of freezing point depression depends only on the solute concentration that can be estimated by a simple linear relationship with the cryoscopic constant: ΔTF = KF · m · i ΔTF, the freezing point depression, is defined as TF (pure solvent) - TF (solution). KF, the cryoscopic constant, which is dependent on the properties of the solvent, not the solute. Note: When conducting experiments, a higher KF value makes it easier to observe larger drops in the freezing point. For water, KF = 1.853 K·kg/mol.[1] m is the molality (mol solute per kg of solvent) i is the van 't Hoff factor (number of solute particles per mol, e.g. i = 2 for NaCl).

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F all of the energy in a falling object's gravitational potential energy store is transferred to its kinetic energy store by the

stepladder [879]

Answer:

The options are not shown, so let's derive the relationship.

For an object that is at a height H above the ground, and is not moving, the potential energy will be:

U = m*g*H

where m is the mass of the object, and g is the gravitational acceleration.

Now, the kinetic energy of an object can be written as:

K = (1/2)*m*v^2

where v is the velocity.

Now, when we drop the object, the potential energy begins to transform into kinetic energy, and by the conservation of the energy, by the moment that H is equal to zero (So the potential energy is zero) all the initial potential energy must now be converted into kinetic energy.

Uinitial = Kfinal.

m*g*H = (1/2)*m*v^2

v^2 = 2*g*H

v = √(2*g*H)

So we expressed the final velocity (the velocity at which the object impacts the ground) in terms of the height, H.

5 0

3 years ago

Write the answer:<br>physics ... i need help

Mariana [72]

Answer:

6 gallons

Explanation:

At 30 mph, the fuel mileage is 25 mpg.

After 5 hours, the distance traveled is:

30 mi/hr × 5 hr = 150 mi

The amount of gas used is:

150 mi × (1 gal / 25 mi) = 6 gal

7 0

3 years ago

A 20-cm long spring is attached to the wall. When pulled horizontally with a force of 100N, the spring stretches to a length of

suter [353]

Answer:

Explanation:

Part 0

All the spring moves is 2 cm

x = 2 cm * [1 m / 100 cm ]

x = 0.020 meters

F = k*d

100N = k * 0.02 m

100 N / 0.02 = k

5000 N / m

Part A

The spring feels a force of 100 N - - 100N = 200 N because each person is pulling in the opposite direction.

F = k * x

200N = 5000 N/m * d

200 / 5000 = d

d = 0.04 meters.

Part B

10.2 kg must be converted to a force as experienced here on earth.

F = m * g

g = 9.81

m = 10.2

F = 10.2 * 9.81

F = 100.06 N

F = k * d

100.06 = 5000 * d

d = 100.06 / 5000

d = 0.02 meters.

3 0

3 years ago

72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. The doorknob is locat

expeople1 [14]

Answer: $I=2 kg-m^2$

Explanation:

Given

mass $m=72 kg$

Force $F=5 N$

door knob is located at a distance of r=0.8 m from axis

Angular acceleration of door $\alpha =2 rad/s^2$

Torque $T=I\alpha =F\times r$

where I=moment of inertia

$5\times 0.8=I\times 2$

$I=2 kg-m^2$

4 0

3 years ago

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by