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Stella [2.4K]
3 years ago
5

Thin Layer Chromatography consists of three parts: The analyte, the stationary phase, and mobile phase. Match each of these term

s to what it was in our experiment. Stationary Phase ____ a) The solvent
Mobile Phase ____ b) Silica

Analyte ____ c) One of the analgesiscs
Physics
1 answer:
Margaret [11]3 years ago
8 0

Answer:

Analyte⇒ one of analgesics

stationery phase⇒ silica

mobile phase⇒ solvent

Explanation:

during the thin layer chromatography non volatile mixtures are separated.The technique is performed on the plastic or aluminum foil that is coated with a thin layer.

You might be interested in
How will the force of friction affect a wood block being pushed over a table?
olasank [31]

Answer:

D

friction acts in the opposite direction of motion but does not affect the motion of the object

5 0
2 years ago
Our milky way galaxy is 100000 lyly in diameter. a spaceship crossing the galaxy measures the galaxy's diameter to be a mere 1.
Sidana [21]

The speed of the spaceship relative to the galaxy is 0.99999995c.

A light-year measures distance rather than time (as the name might imply). A light-year is a distance a light beam travels in one year on Earth, which is roughly 6 trillion miles (9.7 trillion kilometers). One light-year equals 5,878,625,370,000 miles. Light moves at a speed of 670,616,629 mph (1,079,252,849 km/h) in a vacuum.We multiply this speed by the number of hours in a year to calculate the distance of a light-year (8,766).

The Milky way galaxy is 100,000 light years in diameter.

The galaxy's diameter is a mere 1. 0 ly.

We know that ;

L = L_0 \sqrt{1-\frac{v^2}{c^2} }

L = 1 light year

L₀ = 100,000 light year

1 = 100,000 \sqrt{1-\frac{v^2}{c^2} }

1 = 100,000 \sqrt{1-\frac{v^2}{(3*10^8)^2} }

\frac{1}{100,000}  = \sqrt{1-\frac{v^2}{c^2} }

v = 0.999999995 c

Therefore, the speed of the spaceship relative to the galaxy is 0.99999995c.

Learn more about a light year here:

brainly.com/question/17423632

#SPJ4

5 0
2 years ago
A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles
Dimas [21]
A) The total energy of the system is sum of kinetic energy and elastic potential energy:
E=K+U= \frac{1}{2}mv^2 +  \frac{1}{2}kx^2
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
x=A=4.00 cm = 0.04 m
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
E=U= \frac{1}{2}kA^2 =  \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J

b) When the position of the object is 
x=1.00 cm = 0.01 m
the potential energy of the system is
U= \frac{1}{2}kx^2 =  \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J
and so the kinetic energy is
K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J
since the mass is m=50.0 g=0.05 kg, and the kinetic energy is given by
K= \frac{1}{2}mv^2
we can re-arrange the formula to find the speed of the object:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s

c) The potential energy when the object is at 
x=3.00 cm=0.03 m
is
U= \frac{1}{2}kx^2 =  \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J
Therefore the kinetic energy is
K=E-U=0.028 J-0.016 J = 0.012 J

d) We already found the potential energy at point c, and it is given by
U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J
5 0
4 years ago
A voltage of 12 cos(I000t+45) Vis applied to a circuit in which a resistor of 4 .n, aninductor of L H, and a capacitor of 100 μF
vlabodo [156]

Answer:

0.01 H

Explanation:

V = 12 cos (1000t + 45)

C = 100 micro farad

Let the inductance be L .

When the current and the voltage are in the same phase so it is the condition of resonance.

So capacitive reactance = inductive reactance

Xc = XL

1/ωC = ωL

L = 1 / ω²C

By comparisonV = Vo Cos (ωt + Ф)

ω = 1000 rad/s

L = 1 / (1000 x 1000 x 100 x 10^-6)

L = 1 / 100

L = 0.01H

thus, the inductance of the inductor is 0.01 H.

8 0
3 years ago
A car travelling at 55 mph for three and half hours would have travelled<br> how many miles
Mars2501 [29]

Answer:

192.5 miles

Explanation:

55mph = 55 miles in 1 hour

55 X 3.5 = 192.5

4 0
3 years ago
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