Ask question

Ask question

All categories

2 years ago

7

High‑speed ultracentrifuges are useful devices to sediment materials quickly or to separate materials. An ultracentrifuge spins

a small tube in a circle of radius 13.1 cm at 50000.0 rev/min. What is the centripetal force experienced by a sample that has a mass of 0.00310 kg?

1 answer:

pantera1 [17]2 years ago

3 0

Answer:

Fc = 7.14N

Explanation:

First of all, let's convert everything to the same unit system:

m = 0.0031kg R = 13.1cm * 1m / 100cm = 0.131m

ω = 50000 rev/min * 1rev /( 2π rad ) * 1min / 60s = 132.63 rad/s

Now we can calculate centripetal force as:

$Fc = m * \frac{V^2}{R} = m * \frac{(\omega*R)^2}{R}=m*R*\omega ^2$

Replacing the values we get the answer:

Fc = 7.14N

You might be interested in

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0

2 years ago

Lexy used the formula shown to calculate the force of gravity on a space shuttle. Fg = G What does 3 × 105 kg represent? the dif

STALIN [3.7K]

<h2>Answer:</h2>

<u>This term shows the </u><u>mass of the space shuttle</u>

<h2>Explanation:</h2>

We know that the mass of the Earth is 5.972 × 10^24 kg. Similarly the sum of mass of earth and the mass of shuttle must be a greater number as compared to the number given. It simply means that the mass of earth is itself 5.972 × 10^24 kg and the value given is 3 × 105 kg so it is obvious that if was the sum then it must be greater than the mass of earth. Therefore we can say that this not the mass of earth, neither the sum of mass of earth and shuttle, but this is only the mass of space shuttle which is the last multiple choice.

5 0

2 years ago

A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on th

son4ous [18]

Answer:

Part a)

$E = 8436.7 N/C$

Part b)

$E = 8436.7 N/C$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = \frac{Q}{A}$

$Q = -3.8 nC$

$A = \pi(0.09)^2$

$A = 0.025 m^2$

$\sigma = \frac{-3.8\times 10^{-9}}{0.025}$

$\sigma = -1.5 \times 10^{-7} C/m^2$

now the electric field is given as

$E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}$

$E = 8436.7 N/C$

Part b)

Now since the electric field is required at same distance on other side

so the field will remain same on other side of the plate

$E = 8436.7 N/C$

5 0

2 years ago

A cliff diver running 3.60 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2.00 s later

mart [117]

Explanation:

It is given that,

The horizontal speed of a cliff diver, $v_x=3.6\ m/s$

It reaches the water below 2.00 s later, t = 2 s

Let $d_x$ is the distance where the diver hit the water. It can be calculated as follows :

$d_x=v_x\times t\\\\=3.6\times 2\\\\=7.2\ m$

Let $d_y$ is the height of the cliff. It can be calculated using second equation of motion as follows :

$d_y=u_yt+\dfrac{1}{2}gt^2\\\\d_y=\dfrac{1}{2}\times 9.8\times 2^2\\\\=19.6\ m$

So, the cliff is 19.6 m high and it will hit the water at a distance of 19.6 m.

8 0

1 year ago

Gas a bG1 5.22 0.0289G2 1.05 0.0388G3 2.31 0.0467G4 4.05 0.0310Based on the given van der Waals constants for four hypothetical

inysia [295]

Answer:

Gas 2, Gas 3, Gas 4, Gas 5 is the order of decreasing strength of inter-molecular forces.

Explanation:

The strength increases as there is a decrease in the vanderwaals constant and vice versa.

3 0

2 years ago