All categories

3 years ago

Can you check this and tell me if I anything wrong

Physics

1 answer:

posledela3 years ago

6 0

All correct except Moves large molecules which should be Diffusion

You might be interested in

By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows. The pulley can be tre

tankabanditka [31]

Answer:

Mass of the pull is 77 kg

Explanation:

Here we have for

Since the rope moves along with pulley, we have

For the first block we have

T₁ - m₁g = -m₁a = -m₁g/4

T₁ = 3/4(m₁g) = 323.4 N

Similarly, as the acceleration of the second block is the same as the first block but in opposite direction, we have

T₂ - m₂g = m₂a = m₂g/4

T₂ = 5/4(m₂g) = 134.75 N

T₂r - T₁r = I·∝ = 0.5·M·r²(-α/r)

∴ $M = -\frac{2}{a} (T_2-T_1)$

$M = -\frac{2}{2.45} (134.75-323.4) = 77 \, kg$

Mass of the pull = 77 kg.

5 0

3 years ago

First extinguish a match or candle by blasting it violently. Why?

Andreas93 [3]

Answer:

The wax vapor on burning candles could reignite the flame.

Explanation:

7 0

3 years ago

A capacitor consists of two concentric cylinders. The inner cylinder has a radius of 0.001 m and the outer cylinder a radius of

AlexFokin [52]

Answer:

The capacitance is 1.75 nF

Explanation:

From the question we are given that

The inner radius is $r_{in} = 0.001$

The outer radius is $r_{out} = 0.0011 \ m$

Length of the capacitor is $L = 1m$

The dielectric constant is $Di = 2 \ for \ 0 < \phi < \pi$

The dielectric constant is $Di_2 = 4 \ for \ \pi < \phi < 2\pi$

Generally the capacitance of a capacitor can be mathematically represented as

$C = \frac{\pi \epsilon_0 Di_1 L}{ln\frac{r_{out}}{r_{in}} } + \frac{\pi \epsilon_0 Di_2L}{ln\frac{r_{out}}{r_{in}} }$

$= \frac{\pi \epsilon_0 L (Di_1 + Di_2)}{ln\frac{r_{out}}{r_{in}} }$

$= \frac{(3.142)(8.85*10^{-12})(1)(2+4)}{ln\frac{0.0011}{0.001} }$

$=1.75*10^{-9} F$

$1.75nF$

5 0

3 years ago

A golf ball with an initial angle of 34° lands exactly 240 m down the range on a level course.

g100num [7]

Answer:50.39 m/s,

40.46 m

Explanation:

Given

launch angle $=34^{\circ}$

Range=240 m

We know that Range $=\frac{u^2sin2\theta }{g}$

$240=\frac{u^2sin(68)}{9.81}$

u=50.39 m/s

(b)maximum height of projectile is given by

$H=\frac{u^2sin^2\theta }{2g}$

$H=\frac{50.39^2(sin34)^2}{2\times 9.81}$

H=40.46 m

3 0

3 years ago

Read 2 more answers

Hi please help with this question! Need the workings.

andreyandreev [35.5K]

<span>Mass of the copper penny m = 2.6 g Atomic mass of copper = 63.55, Atomic number = 29, So the number of neutrons = Atomic mass - Atomic number = 63 - 29 = 34 a. Neutron mass = 34 x (2.6 / 63.55) = 1.4 grams Copper atoms per mole = 6.040 x 10^23 atoms/mol moles of copper = 2.6 / 63.06 = 0.04123 mol Total atoms in the copper = 6.040 x 10^23 atoms/mol x 0.04123 mol = 0.25 x 10^23 atoms Number of electrons in the copper = 29 per atom Mass of the electron = 9.085 x 10^-28 g b. Electron mass = 0.25 x 10^23 x 29 x 9.085 x 10^-28 = 65.86 x 10^-5 g</span>

6 0

3 years ago