Answer:
-7.44°C
Explanation:
Calculate the molality of the solution. Use the density of the solvent(water) as a conversion factor in order to convert from millilitres of solvent to grams of solvent. Then convert grams into kilograms. Finally, use the molar mass of ethylene glycol as a conversion factor to convert from grams to moles of ethylene glycol.
m = 25.4 g C2H6O2/89.0 mL solv
= 4.6321 C10H8O
Compute the freezing-point depression.
ΔT_f=K_f*m ==> (1.86°C)*(3.9996 m)
=7.44°C
Compute the freezing point of the solution by subtracting the freezing-point depression to the freezing point of the pure solvent.
freezing point =0.0°C-ΔT_f
= -7.44°C
Answer:
1097.8 V/m
Explanation:
The equation that relates the intensity of an electromagnetic wave with the magnitude of the electric field is:
where
c is the speed of light
is the vacuum permittivity
E is the peak magnitude of the electric field
In this problem, we know the intensity:
I = 1600 W/m^2
So we can rearrange the formula to find E:
The correct answer is C.
We will use Boyle's law that states that for a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional.
P1 V1 = P2 V2
Where
P1 is initial pressure = 5 psi
V1 is initial volume = 20 cubic inch
P2 is final pressure = 10 psi
V2 is final volume = unknown
V2 = P1,V1 / P2
V2 = 20 × 5 / 10
V2 = 100/10
V2 = 10 cubic inches
Kinetic energy =(1/2) (mass) (speed²)
First object: KE = (1/2) (2 kg) (2m/s)² = 4 joules during the lift.
Second object: KE = (1/2) (4kg) (3 m/s)² = 18 joules during the lift.
The second object has more kinetic energy while it's being lifted
than the first object has while it's being lifted. Once they reach their
final heights and stop, neither object has any kinetic energy.
Answer:
the rate of acceleration of the train is 4 m/s²
Explanation:
Given;
initial velocity of the train, u = 10 m/s
change in time of motion, dt = 5 s
final velocity of the train, v = 30 m/s
The rate of acceleration of the train is calculated as;
Therefore, the rate of acceleration of the train is 4 m/s²
