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3 years ago

In gas chromatography, what are the advantages of (a) temperature programming? (check all that apply.)

1 answer:

3 0

The following statements apply:
1. Resolution of low boiling solutes is maintained.
2. Retention times of high boiling solutes are decreased.
Temperature programming refers to the process of increasing the temperature of gas chromatography column as a function of time. Temperature programming is usually applied to samples which contain a mixture of components whose boiling points are within narrow ranges

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A vertical spring with a spring constant of 420 N/m is mounted on the floor. From directly above the spring, which is unstrained

mixas84 [53]

Answer:

19.53 cm

Explanation:

The computation of the height is as follows:

Here we applied the conservation of the energy formula

As we know that

P.E of the block = P.E of the spring

m g h = ( 1 ÷ 2) k x^2

where

m = 0.15

g = 9.81

k = 420

x = 0.037

So now put the values to the above formula

(0.15) (9.81) (h) = 1 ÷2 × 420 × (0.037)^2

1.4715 (h) = 0.28749

h = 0.19537 m

= 19.53 cm

3 0

3 years ago

Two blocks of clay, one of mass 100 kg and one of mass 3.00 kg, undergo a completely inelastic collision. Before the collision o

adoni [48]

Answer:

a) v = 0.8 m / s , b) $v_{f}$ = 0.777 m / s , c) ΔK = 0.93 J

Explanation:

This exercise can be solved using the concepts of moment, first let's define the system as formed by the two blocks, so that the forces during the crash have been internal and the moment is conserved.

They give us the mass of block 1 (m1 = 100kg, its kinetic energy (K = 32 J), the mass of block 2 (m2 = 3.00 kg) and that it is at rest (v₀₂ = 0)

Before crash

po = m1 vo1 + m2 vo2

po = m1 vo1

After the crash

$p_{f}$ = (m1 + m2) $v_{f}$

a) The initial speed of the block of m1 = 100 kg, let's use the kinetic energy

K = ½ m v²

v = √2K / m

v = √ (2 32/100)

v = 0.8 m / s

b) The final speed,

p₀ = $p_{f}$

m1 v₀1 = (m1 + m2) $v_{f}$

$v_{f}$ = m1 / (m1 + m2) v₀₁

The initial velocity is calculated in the previous part v₀₁ = v = 0.8 m / s

$v_{f}$ = 100 / (3 + 100) 0.8

$v_{f}$ = 0.777 m / s

c) The change in kinetic energy

Initial K₀ = $K_{f}$

K₀ = 32 J

Final $K_{f}$ = ½ (m1 + m2) $v_{f}$ ²

$K_{f}$ = ½ (3 + 100) 0.777²

$K_{f}$ = 31.07 J

ΔK = $K_{f}$ - K₀

ΔK = 31.07 - 32

ΔK = -0.93 J

As it is a variation it could be given in absolute value

Part D

For this part s has the same initial kinetic energy K = 32 J, but it is block 2 (m2 = 3.00kg) in which it moves

d) we use kinetic energy

v = √ 2K / m2

v = √ (2 32/3)

v = 4.62 m / s

e) the final speed

v₀₂ = v = 4.62 m/s

p₀ = m2 v₀₂

m2 v₀₂ = (m1 + m2) $v_{f}$

$v_{f}$ = m2 / (m1 + m2) v₀₂

$v_{f}$ = 3 / (100 + 3) 4.62

$v_{f}$ = 0.135 m / s

f) variation of kinetic energy

$K_{f}$ = ½ (m1 + m2) $v_{f}$ ²

$K_{f}$ = ½ (3 + 100) 0.135²

$K_{f}$ = 0.9286 J

ΔK = 0.9286-32

ΔK = 31.06 J

4 0

3 years ago

Kepler's laws, satellites motion and weightlessness

Anon25 [30]

Answer:

First Kepler law states that Each planet describes an ellipsoidal motion about the sun as its single focus.

Second Kepler law states that An imaginary line joining a planet to the Sun sweeps out equal areas in equal time intervals.

Third Kepler law states that The squares of period of revolution of the planet around the sun are proportional to the cubes of mean distance between the planet and the sun.

Weightlessness is the condition where the body has zero gravity ( its acceleration is equal to the acceleration due to gravity )

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3 0

3 years ago

A man pushes a shopping cart at a constant speed. What happens if the man increases the force on the cart?

Bezzdna [24]

Answer:

The cart would speed up.

Explanation:

According to Newton's 1st law, object subjected to no force, or net force 0, would have a constant speed. In our case the cart is initially at constant speed, meaning the man exerts a force that is equal to friction force. If he increases the force on the cart, the net force would no longer be 0. The cart would gain an acceleration and increases its speed.

4 0

3 years ago

First consider a Carnot engine that operates between a hot reservoir at 58°C and a cold reservoir at -17°C. In running 24 minute

jok3333 [9.3K]

Answer:

Explanation:

Given that,

Hot reservoir temperature is

TH = 58°C

TH = 58 + 273 = 331 K

Cold reservoir temperature

TC = -17°C

TC = -17°C + 273 = 256K

In 24minutes 590 J of heat was removed from hot reservoir

t = 24mins

t = 24 × 60 = 1440 seconds

Then,

Power is given as

P = E / t

P = 590 / 1440

P = 0.41 W

Since this removed from the hot reservoir, then, QH = 0.41W

We want to find heat expelled to the cold reservoir QX

Efficiency is given as

η = 1 - TH/TC = 1 - QH/QC

1 - TH/TC = 1 - QH/QC

TH/TC = QH/QC

Make QC subject of formula

QC = QH × TC / TH

QC = 0.41 × 256 / 331

QC = 0.317 W

7 0

3 years ago